PHYS 3.1: Introducing fields 
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PPLATO / FLAP (Flexible Learning Approach To Physics) 


We are all familiar with the effect of gravitational forces. They help to keep us ‘down to Earth’! On the surface of the Earth the strongest gravitational influence is that of the Earth itself, but gravitational forces are thought to be universal – they exist between all masses, wherever they may be. This is the basic message of the famous law of universal gravitation first propounded by Isaac Newton. When dealing with such a universal force it is useful to introduce the idea of a gravitational field. The gravitational force acts only where there is a body for it to act upon, but the gravitational field can exist at all points within a region of space. The gravitational field tells us the force that would act on a body if one were present and it characterizes the distribution of masses that cause such a force. If we know how the gravitational field varies from place to place, we can predict the behaviour of a body placed in the field.
Fields are of fundamental importance in physics. They arise naturally whenever we deal with quantities that are defined simultaneously at many different points or throughout an entire region. In this module we will begin by drawing the distinction between two quite separate kinds of field, scalar fields and vector fields. Gravitational fields and the closely similar electrostatic fields, used to describe electrical forces, are both examples of vector fields. It is these fields, and their associated forces, that occupy us throughout Subsections 2.2 to 2.5. In particular, we see how the fields are defined, how they are represented mathematically and pictorially using vectors and field lines, and we investigate the particular fields that correspond to some very simple distributions of mass and charge.
In the second part of the module (Section 3) we examine the work done on a particle when it is displaced from one point to another within a gravitational or electrostatic field. We learn how to calculate the potential energy that such a particle possesses by virtue of its position within a field, and we see how the gravitational (or electrostatic) force on a particle is related to the rate at which its gravitational (or electrostatic) potential energy changes with position. Finally we introduce gravitational and electric potential, which are scalar fields, and show how to represent fields by equipotential surfaces or contours.
Study comment Having read the introduction you may feel that you are already familiar with the material covered by this module and that you do not need to study it. If so, try the following Fast track questions. If not, proceed directly to the Subsection 1.3Ready to study? Subsection.
Study comment Can you answer the following Fast track questions? If you answer the questions successfully you need only glance through the module before looking at the Subsection 4.1Module summary and the Subsection 4.2Achievements. If you are sure that you can meet each of these achievements, try the Subsection 4.3Exit test. If you have difficulty with only one or two of the questions you should follow the guidance given in the answers and read the relevant parts of the module. However, if you have difficulty with more than two of the Exit questions you are strongly advised to study the whole module.
Question F1
The hydrogen atom may be crudely modelled as an electron orbiting a proton in a circular orbit of radius 5 × 10^{−11} m. Treating the proton and the electron as positive and negative point charges, respectively, estimate the ratio of the electrostatic force to the gravitational force between the two particles.
(Mass of electron m_{e} = 9.11 × 10^{−31} kg, mass of proton m_{p} = 1.67 × 10^{−27} kg, the magnitude of the charge on the electron and proton e = 1.60 × 10^{−19} C. Universal gravitational constant G = 6.67 × 10^{−11} N m^{2} kg^{−2}. Electrostatic constant 1/(4πε_{0}) = 8.98 × 10^{9} N m^{2} C^{−2}).
Answer F1
The attractive gravitational force between the electron and proton masses has magnitude F_{grav} = Gm_{p}m_{e}/r^{2} where m_{p} is the mass of the proton, m_{e} the mass of the electron, r the distance between them. The attractive electrostatic force between the charges has magnitude F_{el} = _{ }q_{1}q_{2}_{ }/(4πε_{0}r^{2}) where the charges q_{1} and q_{2} have the same magnitude as the charge on the proton. The ratio of these forces is given by
$\dfrac{F_{\rm el}}{F_{\rm grav}} = \dfrac{e^2}{4\pi\epsilon_0Gm_{\rm p}m_{\rm e}}$
Notice that the r^{2} terms have cancelled. The ratio of the forces is independent of r since both forces are inverse square law forces.
$\dfrac{F_{\rm el}}{F_{\rm grav}} = \dfrac{\left[(1.60\times 10^{19})^2\,{\rm C^2} \times 8.98\times 10^9\,{\rm N\,m^2\,C^{2}}\right]}{6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}}\times 1.67\times 10^{27}\,{\rm kg}\times 9.11\times 10^{31}\,{\rm kg}} = 2.27\times 10^{39}$
Figure 1 See Question F2.
Question F2
Figure 1 shows the field lines in a region containing two objects both of which are charged and have mass.
(a) With no more information than this, can you tell whether the field shown is the gravitational field between the two objects or the electric field?
(b) Given the additional information that the two objects carry charges of opposite sign, which type of field is shown?
(c) Which object has the greater mass?
Answer F2
(a) The field lines could represent either a gravitational field, or an electric field provided that both objects are negatively charged.
(b) Since the objects carry charges of the opposite sign, then the electric field lines should emerge from one of them (the positively charged object). Some of the field lines would go from the positive charge and end up on the negative charge. Since this isn’t the case, it must be the gravitational field which is shown.
(c) The more massive object is the one on the left as it has more field lines.
Question F3
Calculate the change in gravitational potential energy when a satellite of mass m_{s} = 4.00 × 10^{3} kg is put into an orbit of radius R_{s} = 4.20 × 10^{7} m about the Earth. Treat the Earth as a point mass situated at its centre, and take its radius to be R_{E} = 6.40 × 10^{6} m, its mass m_{E} = 6.00 × 10^{24} kg and G = 6.67 × 10^{−11} N m^{2} kg^{−2}.
Answer F3
The change in potential energy when the satellite is put into orbit is
E_{pot} = E_{grav}(R_{s}) − E_{grav}(R_{E})
where$E_{\rm grav} = \dfrac{Gm_{\rm E}m_{\rm s}}{r}$
so$\Delta E = Gm_{\rm E}m_{\rm s}\left[\dfrac{1}{R_{\rm E}}  \dfrac{1}{R_{\rm s}}\right] = Gm_{\rm E}m_{\rm s}\left[\dfrac{R_{\rm s}R_{\rm E}}{R_{\rm E}R_{\rm s}}\right]\\ \phantom{\Delta E} = \dfrac{(6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}} \times 6.00\times 10^{24}\,{\rm kg}\times 4.00\times 10^3\,{\rm kg}\times 35.6\times 10^6\,{\rm m})}{4.20\times 10^7\,{\rm m}\times 6.40\times 10^6\,{\rm m}}\\ \phantom{\Delta E} = \dfrac{6.67\times 6.00\times 4.00\times 35.6}{4.20\times 6.40} \times \dfrac{10^{11}\times 10^{24}\times 10^3\times 10^6}{10^{13}}\,{\rm N\,m} = 2.12\times 10^{11}\,{\rm J}$
Study comment: In order to study this module you will need to be familiar with the following terms: acceleration, Cartesian coordinate system, charge, displacement, energy, force, function, gradient (of a graph), mass, magnitude_of_a_vector_or_vector_quantitymagnitude (of a vector), Newton’s laws of motion, scalar, tangent_to_a_curvetangent (to a curve), vector, velocity and work. You should be familiar with vector notation, and be able to express a vector in terms of its (scalar) components_of_a_vectorcomponents, you should also be able to carry out vector addition. This module makes some use of calculus notation, but you are not required to have a detailed knowledge of the techniques of differentiation and integration. In particular it is assumed you know that a derivative such as dy/dx represents the rate of change of y with respect to x, and that it may be interpreted in terms of the gradient of a graph of y against x. It would also be an advantage if you were familiar with the idea that a definite integral such as $\displaystyle y=\int_a^b \dfrac{c}{x^2}\,dx$ represents the limit of a sum and may be interpreted as the area under a graph of c/x^{2} between x = a and x = b. However, this is not an essential prerequisite since it is explained at the appropriate point in the text. SI units are used throughout the module. If you need help with any of these concepts or procedures, refer to the Glossary which will indicate where in FLAP they are developed. The following questions will help you establish whether you need to review some of the above topics before embarking on this module.
Question R1
Write down in terms of kg, m, and s the SI units of (a) acceleration, (b) force, (c) energy. Where appropriate, give the name of the unit.
Answer R1
(a) acceleration = rate of change of velocity with respect to time, and velocity = rate of change of position with respect to time. Velocity has units m s^{−1} and acceleration has units m s^{−2}.
(b) Force = mass × acceleration so its units are kg m s^{−1} or newtons, N. (c) Energy = force × distance so its units are kg m^{2} s^{−2} or joules, J.
Consult SI units in the Glossary for further information.
Question R2
Draw a sketch showing the addition_of_vectorsaddition of two vectors of magnitude_of_a_vector_or_vector_quantitymagnitude 1 unit and 2 units, inclined at an angle of 60° to each other.
Figure 21 See Answer R2.
Answer R2
Figure 21 shows the parallelogram rule for the addition of vectors. You may have drawn the triangle rule instead. The important point is that directions must be taken into account when vectors are added.
Consult vector addition in the Glossary for further information.
Question R3
Plot a graph of y = 1/x over the range x = 0 to x = 3.
Figure 22 See Answer R3.
Answer R3
The graph of y = 1/x is shown in Figure 22 for positive x and y.
(This graph will be referred to in Subsection 2.2.)
Question R4
When a particle initially located at a point with position vector r_{1} is moved to a different point with position vector r_{2}, the particle is said to have undergone a displacement s_{12}. Express s_{12} in terms of r_{1} and r_{2}. What is the physical significance of the magnitude_of_a_vector_or_vector_quantitymagnitude of s_{12}?
Answer R4
The displacement may be written as
s_{12} = r_{2} − r_{1}
The magnitude_of_a_vector_or_vector_quantitymagnitude of s_{12}, which may be written _{ }s_{12}_{ } or simply s_{12}, represents the distance from the initial position to the final position.
Consult the relevant terms in the Glossary for further information.
What is a field? The term is used in many different ways in everyday English as shown by the following extract from Chambers Twentieth Century Dictionary:
... country or open country in general: a piece of ground enclosed for ... pasture or sport ... speciality: area of knowledge, interest, ...: the locality of a battle ...
Chambers 1981
The list continues with ‘a wide expanse’ (geologists refer to large areas covered with ice as ice fields), and ‘area visible to an observer at one time’ (we may speak of the field of view of an optical instrument such as a microscope or camera and also of its depth of field). This is followed by the dictionary’s closest approach to the meaning that will concern us in this module: ‘region of space in which forces are at work’.
In physics we are accustomed to dealing with systems that exert forces at all points throughout a region of space. The Earth is a good example of such a system. No humans, not even the astronauts who have walked on the Moon, have ever been entirely free of the gravitational force that the Earth exerts on objects in its vicinity. We describe this situation by saying that the Earth is the source of a gravitational field, and that any body located within this field will be acted upon by a gravitational force due to the Earth.
In a similar way, a positive or negative electric charge will exert an electrical force on any other charge in its vicinity, so it may be said to produce an electric field, and a magnet may be said to produce a magnetic field. Gravitational and electrical forces only act on bodies with mass or charge, but the presence of a gravitational or an electric field in some region of space tells us that a force will act if an appropriate body is placed in that region. Later in this section we will replace these rather vague notions of gravitational fields and electric fields by precise definitions based on quantitative relationships between the fields and the forces they describe. However, before linking fields and forces in that way it is worth examining the general concept of a field in order to emphasize that fields are not only of use when describing force distributions.
If you watch television weather forecasts you will see several examples of fields that have nothing to do with forces. Figure 2 shows the distribution of ground–level temperatures across the British Isles at a certain time, this constitutes a temperature field. The word field is used in this context to emphasize that the temperature has some particular value at every point throughout the region of interest at the time in question.
Figure 2 The temperature field observed across the British Isles one day in January 1981.
In a similar way, a weather forecast might provide information about the distribution of air pressure, hours of sunshine in the past week or even inches of rainfall in the past month. In each of these cases it is possible to associate a value of the named physical quantity with each point in a given region at a particular time. It is this feature that is common to all fields and which we will adopt as the defining characteristic of a field.
A field is a physical quantity to which a definite value can be ascribed at each point in some region of space at a particular time.
The temperatures in Figure 2 are only shown at regularly–spaced intervals on a two–dimensional grid, but the implication that there is a unique temperature at every point is clear. We can express this mathematically by saying that the temperature is a function of position, and we can indicate that by writing the temperature T at a point with coordinates (x, y) as T_{ }(x, y). If we wanted to discuss temperatures at different heights z above the surface of the British Isles, we would write T_{ }(x, y, z), and if we wanted to discuss the way this temperature changed with time we might write T_{ }(x, y, z, t). From this point of view, a field is simply a function of position and (possibly) time.
Figure 3 Isotherms plotted using the same data as Figure 2.
Temperature is a scalar quantity – its value at any point in space and time is fully specified by a single number together with an appropriate unit of measurement, e.g. the kelvin (K), or the degree Celsius (°C). For this reason a temperature field is said to be a scalar field. Similarly, any other field whose value at any given point in space and time is specified by a single scalar quantity is said to be a scalar field.
Although a two–dimensional scalar field such as the temperature field T_{ }(x, y) can be represented by numbers as in Figure 2, it is generally much easier to interpret the alternative representation shown in Figure 3. In this case lines called isotherms i passing only through points of equal temperature have been used to represent the field. Other scalar fields may be depicted in similar ways (e.g. all points on an isobar have the same pressure, and all points on a contour line have the same altitude above sealevel). The ability to form a mental image of a field on the basis of such a pictorial representation is a useful skill to develop.
Figure 4 See Question T1.
Question T1
A flat sheet of metal is heated near its edge, and its temperature is measured at points in a grid–like array, as shown on Figure 4. Sketch on the figure the isotherms for 100°C, 200°C and 300°C. (Assume that the temperature varies smoothly across the sheet so that you can interpolate values, e.g. if two neighbouring grid points are at 170°C and 230°C then the temperature halfway between them is 200°C.)
Figure 23 See Answer T1.
Answer T1
Isotherms on a heated metal plate are shown in Figure 23.
Figure 5 The wind velocity field for the same day as illustrated in Figure 2. The symbol ∝ is read as ‘is proportional to’.
Fields that specify forces, such as gravitational and electric fields (defined more fully in Subsections 2.3 and 2.4, respectively) cannot be scalar fields since they must define a force which has direction as well as magnitude at every point. Since force is a vector quantity, any such field is said to be a vector field. Velocity is also a vector quantity, so the weather maps that use arrows to show wind velocities across the British Isles are pictorial representations of the wind velocity field (see Figure 5). Since vectors are conventionally denoted in print by a bold typeface, it is usual to also use bold type for vector fields. Thus, the ground–level wind velocity field of Figure 5 may be denoted υ_{ }(x, y), the gravitational field a height z above the surface of the Earth g_{ }(x, y, z) and the rapidly changing electric field inside a device by E_{ }(x, y, z, t).
✦ Classify the following as scalar fields or vector fields: the temperature throughout a room, the weight of a cubic centimetre of air at any point in the Earth’s atmosphere, wind speeds throughout central Scotland, the depth below sea–level of each point on the floor of the Atlantic Ocean.
✧ Weight is a force, so the weight field is a vector field, directed towards the centre of the Earth at every point. The others are all scalar fields; temperature, speed and depth below sea–level can all be specified by a single number and a unit at every point. (Note: Remember speed is a scalar but velocity is a vector quantity.)
Study comment One of our immediate aims is to write down the general definition of the gravitational field, however before doing so we need to introduce the vector description of the gravitational force. If you are already familiar with this way of describing gravitational forces you need only glance through the content of this section, pausing only to answer the text questions that it contains.
We are all familiar with the gravitational force of the Earth pulling downwards on objects situated on or above its surface. It is what keeps our feet on the ground. In a similar way objects on the Moon feel the gravitational force of the Moon.
Isaac Newton (1642–1727) i suggested that the force which one ‘heavenly’ body exerts on another is of the same nature as the gravitational force that the Earth exerts on a terrestrial object such as a person or an apple. In fact, according to Newton, gravitational forces act between all masses. This led him to formulate the following law:
law of universal gravitation
Every particle of matter in the Universe attracts every other particle with a force whose magnitude is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them.
Thus the magnitude of the gravitational force that any particle of mass m_{1} exerts on a particle of mass m_{2} a distance r away is given by
$F_{\rm grav} = \dfrac{Gm_1m_2}{r^2}$
where G = 6.673 × 10^{−11} N m^{2} kg^{−2} is the universal gravitational constant.
There are several points to note about the gravitational force between two particles:
It is a distant–action force, i.e. it always exists between two particles without their being in contact and
regardless of any intervening matter.
The force magnitude is proportional to the mass of each particle.
The force magnitude obeys an inverse square law, i.e. it is proportional to 1/r^{2} and thus decreases rapidly as the separation of the particles increases.
The force is always attractive, i.e. when two particles exert a gravitational force on each other the direction of the force on each particle is towards the other. (Since the magnitudes of these two oppositely directed forces are equal it is clear that they constitute a third law pair of the kind described by Newton’s third law of motion.) i
Question T2
Confirm that the units of G are N m^{2} kg^{−2}. i
Answer T2
Rearranging the equation for the gravitational force, we have
G = F_{grav}r^{2}/(m_{1}m_{1})
so the units of G are N m^{2}/kg^{2}, i.e. N m^{2} kg^{−2}.
Question T3
Sketch a graph of F_{grav} against r over the range r = 0 to r = 3 m.
$F_{\rm grav} = \dfrac{Gm_1m_2}{r^2}$
Figure 22 See Answer T3.
Figure 24 See Answer R3.
Answer T3
The graph of F_{grav} against r has the shape of 1/r^{2} against r, see the sketch in Figure 24. The graph falls off and rises much more steeply than the graph of 1/r against r (see Figure 22).
Figure 6 Alternative, but equivalent, ways of defining the position of point P, (a) by its coordinates (x, y, z), (b) by the position vector r.
By making use of vector notation we can express all the information about the magnitude and direction of the gravitational force in a single equation. This equation takes its simplest form if we imagine that the particle of mass m_{1} is located at the origin O of a Cartesian coordinate system, while the particle of mass m_{2} is at a point P with position coordinates (x, y, z). The point P can be specified by a position vector r (see Figure 6b) which may be written in terms of its components as (x, y, z), and which has magnitude r = _{ }r_{ } = $\sqrt{\smash[b]{x^2+y^2+z^2}}$. i
We can then introduce a unit vector r^{^}, defined by the relation
$\hat{\boldsymbol r} = \dfrac{\boldsymbol r}{\vert\,{\boldsymbol r}\,\vert}\quad\text{i.e.}\quad\hat{\boldsymbol r} = \dfrac{{\boldsymbol r}}{\sqrt{\smash[b]{x^2+y^2+z^2}}}$
The unit vector r^{^} points in the same direction as r but its magnitude is just 1; it is therefore a dimensionless quantity whose only purpose is to indicate the direction from the mass m_{1} at the origin to the mass m_{2} at the point P.
We can now write the gravitational force acting on m_{2} due to m_{1} as follows:
${\boldsymbol F}_{21} = \dfrac{Gm_1 m_2}{r^2}\hat{\boldsymbol r}$(1)
The negative sign in Equation 1 indicates that F_{21} points in the opposite direction to the unit vector r^{^}. It is worth noting that the force F_{12} on m_{1} due to m_{2} points in the same direction as r^{^}, and is given by F_{12} = −F_{21}.
Question T4
Suppose that the masses m_{1} and m_{2} are located at points specified by the position vectors r_{1} and r_{2}, respectively. Write down a formula similar to Equation 1,
${\boldsymbol F}_{21} = \dfrac{Gm_1 m_2}{r^2}\hat{\boldsymbol r}$(Eqn 1)
for the gravitational force on m_{2} due to m_{1}, taking care to define any new terms that you introduce.
[Hint: You may find it useful to recall that the displacement from m_{1} to m_{2} is s_{12} = r_{2} − r_{1}. See Question R4.]
Answer T4
Since the displacement from m_{1} to m_{2} is s_{12} = r_{2} − r_{1}, it follows that a unit vector from m_{1} to m_{2} is
$\hat{\boldsymbol s}_{12} = \dfrac{{\boldsymbol s}_{12}}{\left\vert\,{\boldsymbol s}_{12}\,\right\vert}=\dfrac{{\boldsymbol r}_2{\boldsymbol r}_1}{\left\vert\,{\boldsymbol r}_2{\boldsymbol r}_1\,\right\vert}$
Hence, from Newton’s law of universal gravitation, the gravitational force on m_{2} due to m_{1} may be written
${\boldsymbol F}_{21} = \dfrac{Gm_1 m_2}{\left\vert\,{\boldsymbol r}_2{\boldsymbol r}_1\,\right\vert^2}\hat{\boldsymbol s}_{12} = \dfrac{Gm_1 m_2}{\left\vert\,{\boldsymbol r}_2{\boldsymbol r}_1\,\right\vert^2}\left(\dfrac{{\boldsymbol r}_2{\boldsymbol r}_1}{\left\vert\,{\boldsymbol r}_2{\boldsymbol r}_1\,\right\vert}\right)$
Question T5
Consider two small lead spheres of masses m_{1} and m_{2}, where m_{1} is greater than m_{2}, isolated in space from all other bodies. Since the gravitational force between the spheres is attractive they will accelerate towards each other.
(a) Will their accelerations be equal? If not, which is larger?
(b) As the spheres move towards each other, will their accelerations remain constant? If not, will they increase or decrease?
Answer T5
(a) Acceleration = force/mass, so the less massive sphere (m_{2}) will have the larger acceleration.
(b) The force between the particles increases as they approach one another since F_{grav} is proportional to 1/r^{2}, so their accelerations will increase as r decreases.
Question T6
Two small objects of masses 1 g and 1 kg are placed 1 m apart. Given that G = 6.67 × 10^{−11} N m^{2} kg^{−2}, calculate the magnitude of the gravitational force that each of the masses exerts on the other.
Answer T6
Treating both the bodies as point masses
F_{grav} = Gm_{1}m_{2}/r^{2}
$= \dfrac{6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}}\times 1\,{\rm kg}\times 10^{3}\,{\rm kg}}{1\,{\rm m}^2}$
The forces exerted by each mass on the others are equal in magnitude but opposite in direction
i.e.F_{grav} = 6.67 × 10^{−14} N on each body.
Suppose that the 1 g mass in Question T6 is replaced by a 2 g mass. How will this affect the magnitude of the force acting on the mass? We could calculate the new magnitude using Equation 1,
${\boldsymbol F}_{21} = \dfrac{Gm_1 m_2}{r^2}\hat{\boldsymbol r}$(Eqn 1)
but there is really no need to do so. We know that the gravitational force on a particle is proportional to its mass, so doubling the mass will double the magnitude of the gravitational force that acts upon it. In fact, once we have calculated the force on one known mass at a particular point P due to any fixed arrangement of masses, we can immediately calculate the gravitational force on any other mass placed at P. The calculation is easy because once we know the gravitational force on one mass at P we can work out the gravitational force per unit mass placed at that point. If the coordinates of P are (x, y, z), we can denote the gravitational force per unit mass at P by g_{ }(x, y, z). The gravitational force on any mass m placed at a point P with coordinates (x, y, z) will then be
F_{grav} (on m at P) = mg(x, y, z)
Of course, P can be any point within a region, so we can regard this vector quantity g_{ }(x, y, z) that varies from place to place as a vector field, and it is then called the gravitational field. i
At any point, the value of the gravitational field g_{ }(x, y, z) is the gravitational force per unit mass that would act on a particle placed at that point.
The gravitational field at any point P can be determined experimentally by placing a test mass m at P, measuring the gravitational force on it, and then dividing that force by m. (A test mass m should be thought of as a particle of sufficiently small mass m that it will not significantly influence the bodies that give rise to the gravitational field and therefore will not disturb the field, though its mass should be sufficiently great that the force exerted on it by the field can be measured.) It should be noted that the above definition refers to the force that would act on a particle, so it makes sense to speak of the field even when there is no particle present to experience its effect.
When discussing the gravitational field, or any other field come to that, it is rather tedious to repeatedly write g_{ }(x, y, z), so we will follow the usual practice of representing the position coordinates x, y and z by the single vector quantity r, we may then indicate the gravitational field by the symbol g_{ }(r).
We then have the following general definition of the gravitational field:
${\boldsymbol g}({\boldsymbol r}) = \dfrac{{\boldsymbol F}_{\rm grav}({\rm on}~m~{\rm at}~{\boldsymbol r})}{m}$(2) i
where F_{grav} is the force that would act on a test mass m if it were located at the point with position vector r. Note that in writing the field as g_{ }(r) we are not implying that it points in the direction of r at every point, only that the magnitude and direction of g depend on position.
In the particular case that the source of the gravitational field is an isolated point mass m_{0} at the origin then, from Equations 1 and 2, we obtain:
${\boldsymbol g}({\boldsymbol r}) = \dfrac{Gm_0}{r^2}\hat{\boldsymbol r}$(3) i
Note the negative sign indicating that in this particular case g_{ }(r) is in the opposite direction to r^{^}. Also note that this is only an example of a gravitational field, albeit an important one. The general definition is still that of Equation 2.
✦ Write down an expression for the magnitude g_{ }(r) of the vector field g_{ }(r) given in Equation 3, at the point r.
${\boldsymbol g}({\boldsymbol r}) = \dfrac{Gm_0}{r^2}\hat{\boldsymbol r}$(Eqn 3)
✧ $g(\boldsymbol{r}) = \dfrac{Gm_0}{r^2}$^{}
Note that this is always a positive quantity as any magnitude must always be. i
Question T7
Calculate the magnitude of the gravitational field g_{ }(r) at a distance of 2 m from an isolated 5 kg mass. i Make sure you include appropriate SI units in your answer.
Answer T7
The magnitude of the gravitational field g_{ }(r) due to a mass m_{0} is given by Gm_{0}/r^{2}.
Therefore$g(r) = \dfrac{6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}}\times 5\,{\rm kg}}{(2\,{\rm m})^2}$
so$g(r) = \dfrac{6.67\times 5\times 10^{11}}{4}\,{\rm N\,kg^{1}} = 8.34\times 10^{11}\,{\rm N\,kg^{1}}$
Figure 7a Gravitational field lines for mass m.
Figure 7b Gravitational field lines for masses m and 2m close together, the neutral point will be closer to the mass m.
Figure 7c Gravitational field lines for the Earth’s gravitational field.
The gravitational field in a region, like the wind velocity field of Figure 5, can be represented pictorially by a set of arrows whose direction is that of the field and whose length is proportional to the field strength (i.e. magnitude) at the points where they are drawn. An alternative pictorial representation of a vector field is provided by field lines. These are directed lines, possibly curved, that have arrowheads on them drawn so that at any point the tangent to the field line represents the direction of the field at that point, and the spacing of the lines is related to the field strength (where the lines are close together the field is strong and where they are further apart the field is weaker).
Gravitational field lines surrounding an isolated point mass m are shown in Figure 7a. These extend radially in all directions. The direction of the field is shown by the arrowheads. A test mass placed at point A for example will experience a force towards the central mass. If the central mass had been 2m we would have drawn twice as many field lines since the field would have been twice as strong at every point.
What happens to the field lines if we bring two masses m and 2m closer together? As they approach, the field lines of each isolated mass will be modified by the presence of the other mass. This causes the field lines to curve (see Figure 7b). Where the lines are curved, as at point B, the direction of the field is along the tangent to the curve at that point. Between the masses it is as if the field lines ‘repel one another’, and there is a region where the field lines are far apart, indicating that the field is weak. There is even a point N, called a neutral point, at which the forces on a test mass due to m and 2m are equal and opposite so there is no resultant force at all and the gravitational field is zero at that point.
Figure 7c shows the gravitational field lines associated with the Earth. (You must of course imagine the field lines spreading out in three dimensions.) The magnitude of the field will be proportional to the density of field lines in any region. Notice that the lines are closer together near the surface of the Earth where the field is stronger. Further away from the Earth, the radial field lines will be modified by the field lines associated with the Moon.
Field lines are a useful aid to visualizing the distribution of a field in space. Be careful though not to fall into the trap of thinking that the field lines actually exist. They don’t!
At the neutral point N of Figure 7b we can determine the gravitational field due to two masses by working out the forces that each mass would separately exert on a test mass and adding those forces together vectorially.
However, there is really no need to worry about forces and test masses at all. We can find the combined effect of two gravitational fields by adding the fields themselves, provided we remember that the fields add vectorially at every point. The statement that gravitational fields may be added in this straightforward way is called the principle of superposition. i
Figure 8 Arrangement of two 4 kg masses about point O.
✦ Find the magnitude and direction of the gravitational field at point O due to two 4 kg masses placed at distances of 2 m from O as shown in Figure 8. (G = 6.67 × 10^{−11} N m^{2} kg^{−2}).
Figure 1 Vector sum of g_{1}(0, 0) and g_{2}(0, 0).
✧ Let the gravitational field due to the mass at x = 2 m, y = 0 be g_{1}(x, y), and let that due to the mass at x = 0, y = 2 m be g_{2}(x, y).
Then, at the origin, both fields have the same magnitude
g_{1}(0, 0) = g_{2}(0, 0) = $\dfrac{Gm}{r^2} = \dfrac{G\times 4\,{\rm kg}}{(2\,{\rm m})^2}$ = 6.67 × 10^{−11} N kg^{−1} (from Eqn 3)
$\boldsymbol{g}(\boldsymbol{r}) = \dfrac{Gm_0}{r^2}\hat{\boldsymbol{r}}$(Eqn 3)
but g_{1}(0, 0) is directed along the positive x–axis while g_{2}(0, 0) points along the positive y–axis, as shown in Figure 9.
Adding these two field values vectorially we see that their resultant g_{ }(0, 0) = g_{1}(0, 0) + g_{2}(0, 0) has magnitude:
$\sqrt{\smash[b]{[g_1(0,\,0)]^2+[g_2(0,\,0)]^2}} = \sqrt{2\os}g_1(0,\,0)$ = 9.43 × 10^{−11} N kg^{−1}
and points in a direction inclined at an (anticlockwise) angle ϕ to the x–axis, where:
tan_{ }ϕ = g_{2}(0, 0)/g_{2}(0, 0) = 1, so ϕ = arctan(1) = 45°.
From the resultant field at O we can calculate the magnitude of the gravitational force acting on, say, a 1 g test mass placed at O:
_{ }F (on 1 g at O)_{ } = test mass × field strength at O = 10^{−3} kg × g_{ }(0, 0)
_{ }F (on 1 g at O)_{ } = 10^{−3} kg × 9.43 × 10^{−11} N kg^{−1} = 9.43 × 10^{−14} N
The weight of a body near the surface of the Earth is usually taken to mean the gravitational force that the Earth exerts on the body. i You will almost certainly have seen calculations in which a body of mass m has been associated with a weight of magnitude W = mg, where g is called the magnitude of the acceleration due to gravity. The value of g varies a little from place to place, but is usually taken to be 9.81 m s^{−2} across the UK. In view of what has been said about gravitational fields in this subsection you should now be able to give a somewhat different interpretation to the equation W = mg. By rearranging the equation to obtain g = W/m, and then identifying W with the magnitude of the gravitational force on the body we find
$g = \dfrac{F_{\rm grav}}{m}$
However F_{grav}/m represents the magnitude of the Earth’s gravitational field; so, taking into account the downward direction of both the field and the acceleration due to gravity, we can say:
The gravitational field of the Earth is equal to the acceleration due to the Earth’s gravity at any point.
Question T8
Show that the SI units of gravitational field are exactly equivalent to those of acceleration.
Answer T8
1 N = 1 kg m s^{−1}, so
1 N kg^{−2} = 1 (kg m s^{−2}) kg^{−2} = 1 m s^{−2}.
The electrostatic force that exists between charged particles is similar in many ways to the gravitational force between masses, e.g. they are both actionatadistance forces. However, electric charge, which is measured in units of coulomb (C), may be positive or negative. This allows electrostatic forces to differ from gravitational forces in one major respect. Whereas the electrostatic force between oppositely charged particles is attractive, like the gravitational force, the force between similarly charged particles (both positive or both negative) is repulsive.
Coulomb’s law i is the electrostatic equivalent of Newton’s law of gravitation; it provides a quantitative description of the force between electrically charged particles. In its simplest vector form it says that the electrostatic force on a point charge q_{2} at a position r due to a point charge q_{1} at the origin is
${\boldsymbol F}_{21} = \dfrac{q_1 q_2}{4\pi\varepsilon_0r^2}\hat{\boldsymbol r}$(4) i
where r^{^} is a unit vector pointing from q_{1} to q_{2}, and ε_{0} = 8.854 × 10^{−12} C^{2} N^{−1} m^{−2} is a constant called the permittivity of free space.
If the charges q_{1} and q_{2} have the same sign q_{1}q_{2}/(4πε_{0}r^{2}) will be positive and Equation 4 will imply that F_{21} points in the same direction as r^{^}, i.e. it is a repulsive force pushing q_{2} away from the origin. If q_{1} and q_{2} have opposite signs q_{1}q_{2}/(4πε_{0}r^{2}) will be negative and the force on q_{2} will point in the opposite direction to r^{^}, towards the origin. Thus, substituting values of q_{1} and q_{2} with their correct signs into Equation 4 gives the direction of F_{21} as well as its magnitude.
By analogy with the gravitational constant G in Newton’s law of gravitation, the constant 1/(4πε_{0}) in Coulomb’s law is sometimes called the electrostatic constant.
✦ The force F_{12} that acts on q_{1} due to q_{2} is equal in magnitude but opposite in direction to F_{21}. Write down an expression for the common magnitude F_{el} of these two forces and use it to determine appropriate SI units for 1/(4πε_{0}).
✧ It follows from Equation 4,
$\boldsymbol{F}_{21} = \dfrac{q_1q_2}{4\pi\epsilon_0r^2}\hat{\boldsymbol{r}}$(Eqn 4)
that
$F_{{\rm el}} = \dfrac{\vert q_1q_2\vert}{4\pi\epsilon_0r^2} = \dfrac{\vert q_1\vert\,\vert q_2\vert}{4\pi\epsilon_0r^2}$
(Note that we have to deal with the absolute values of the charges, as indicated by the vertical bars, since either of the charges may be negative but the magnitude F_{el} must be positive.)
Rearranging the equation for F_{el}, we see that the units of 1/(4πε_{0}) are those of F_{el}r^{2}/(q_{1}q_{2}), in other words, N m^{2} C^{−2}. (In SI units, 1/(4πε_{0}) = 8.988 × 10^{9} N m^{2} C^{−2}).
This answer shows that the force magnitude F_{el} varies as 1/r^{2}, i.e. like the magnitude of the gravitational force, it obeys an inverse square law.
A distribution of electric charges that might exert electrostatic forces on test charges will create an electric field in the same way that a distribution of masses creates a gravitational field.
At any point, the value of the electric field E_{ }(r) is the electrostatic force per unit positive charge that would act on a test charge placed at that point.
Note that the electric field may vary from point to point throughout a region, so we write it as E_{ }(r) or sometimes E_{ }(x, y, z) to emphasize its dependence on position. (Many authors will simply denote it by E and leave it to you to remember that it is a function of position.)
The general definition of an electric field given above may be also be expressed as an equation:
${\boldsymbol E}({\boldsymbol r}) = \dfrac{{\boldsymbol F}_{\rm el}({\rm on}~q~{\rm at}~{\boldsymbol r})}{q}$(5) i
where F_{el} (on q at r) represents the electrostatic force that would act on a test charge q if it were located at the point with position vector r. The similarity between electric field and gravitational field, as expressed by Equations 5 and 2, respectively, should be noted. However, the units of the two are quite different. The electric field is the force per unit positive charge whilst the gravitational field is the force per unit mass; it is only the gravitational field which has the same dimensions and units as acceleration.
It follows from Equations 4 and 5 that in the particular case that the sole source of the electric field in some region is an isolated point charge q_{0} at the origin:
${\boldsymbol E}({\boldsymbol r}) = \dfrac{q_0}{4\pi\varepsilon_0r^2}\hat{\boldsymbol r}$(6) i
It follows from Equation 5 that the SI units of electric field are N C^{−1}. However it is more conventional to express electric field strengths in terms of an equivalent unit, the volt per metre (V m^{−1}). The use of volts as a unit of measurement will be discussed more fully in Subsection 3.3.
Question T9
Consider two charges, −q situated at O and +q situated at P, where q is a positive quantity.
(a) In which direction does the electrostatic force on the charge at P act?
(b) Is it in the same direction as the electric field at P? Suppose we change the charge at O to +q and the charge at P to −q
(c) In which direction does the electrostatic force now act at point P?
(d) Is it in the same direction as the electric field at P?
Answer T9
(a) Using Equation 4, the electrostatic force is given by
${\boldsymbol F} = \dfrac{q^2}{4\pi\epsilon_0r^2}\hat{\boldsymbol r}$(Eqn 4)
Hence F is in the opposite direction to r^{^}, i.e. it points from P to O – it is an attractive force between the charges.
(b) From Equation 6,
${\boldsymbol E}({\boldsymbol r}) = \dfrac{q}{4\pi\epsilon_0r^2}\hat{\boldsymbol r}$(Eqn 6)
i.e.E_{ }(r) points in the direction from P to O, so it does point in the same direction.
(c) As in (a), we have an attractive force between the unlike charges. F is in the direction from P to O.
(d) The electric field due to a positive charge at the origin O is given by
${\boldsymbol E}({\boldsymbol r}) = +\dfrac{q}{4\pi\epsilon_0r^2}\hat{\boldsymbol r}$
and so the electric field is now in the opposite direction, along OP, i.e. in the same direction as r and in the opposite direction to F.
Figure 10 Arrangement of two charges +q and −q about O.
Question T10
Two charges +q and −q are situated as shown in Figure 10, where OA = OB. Find the direction of the electric field at point P.
Answer T10
The field E_{1} at P due to charge +q at A is along AP. The field E_{2} at P due to charge −q at B is of equal magnitude to that of E_{1} (since AP = BP) but points in the direction from P to B. The resultant at P is a vector pointing parallel to the x–axis (the components along y–axis cancel).
Figure 11a&b Field lines for (a) a negative point charge −q, (b) a positive point charge +q
Just as with gravitational fields, we can use field lines to represent an electric field. Figure 11a shows the electric field lines surrounding a negative point charge, and Figure 11b those surrounding a positive point charge. Note that electric fields, like gravitational fields, obey the principle of superposition.
✦ What is the difference between Figures 11a and 11b?
✧ In Figure 11a, the electrostatic force on a positive test charge would be attractive so the field lines converge on the negative charge (which acts as a sink of field lines). In Figure 11b, the electrostatic force on a positive test charge would be repulsive so the field lines diverge from the positive charge (which acts as a source of field lines). This highlights a difference between gravitational and electric field lines – since the gravitational force is always attractive, gravitational field lines always point into a mass.
Figure 11c&d Field lines for (c) point charges +q and −q close together, (d) point charges +q and +q close together. These are of course two–dimensional representations of the three–dimensional fields surrounding the charges.
If we bring two charges close together, the electric field lines of each isolated charge appear to be modified by the presence of the other, just as we found in the case of gravitational field lines. For example, in Figure 11c we illustrate a positive charge and a negative charge of equal magnitude. We see that some of the field lines emerge from +q and end on −q.
In Figure 11d two positive charges of equal magnitude are illustrated, and we see that the field lines appear to repel one another. It is also clear that there is a neutral point between the two charges where the resultant force is zero.
Note that electric fields, like gravitational fields, obey the principle of superposition. That is to say that if we know that a certain distribution of charge produces an electric field E_{1}(r), and that some other distribution of charge produces a field E_{2}(r), then when both charge distributions are present simultaneously the resultant electric field at any point r will be the vector sum E_{1}(r) + E_{2}(r).
Question T11
Consider three charges, −q, +2q, −q all in line, with equal spacing between them. Sketch the electric field lines surrounding the charges. (Remember that if the charge is doubled, then we must draw twice as many lines.)
Figure 25 See Answer T11.
Answer T11
Your sketch of the electric field lines surrounding charges −q, +2q and −q should resemble Figure 25, where ten lines have been drawn emerging from +2q, and five lines disappearing into each of the −q to indicate the relative strength of the fields. The pattern is symmetric since the negative charges are equidistant from the positive charge.
Question T12
Two positive and two negative charges of equal magnitude are arranged at the corners of a square with the two positive charges one above the other and the two negative charges one above the other.
(a) Draw arrows at several points in and around the square to represent the direction and relative strength of the electric field at those points due to the charges. Identify any positions where the field is zero. (Let the length of the arrow you draw at any point indicate the relative strength of the electric field at that point.)
(b) On a separate sketch draw the electric field lines associated with the arrangement of charges. Make sure your answers to (a) and (b) are consistent.
Figure 26b See Answer T12.
Figure 26a See Answer T12.
Answer T12
(a) The vector representation of the electric field produced by four charges is shown in Figure 26a. At each point you have to find the resultant of the four vectors corresponding to the fields due to the four charges, and the fields due to the charges which are furthest away will be smallest. At points X and Y (just outside the square) the resultant field is zero.
(b) The electric field lines surrounding the charges is shown in Figure 26b. Notice the direction of the arrows. This figure shows clearly the regions where the electric field is zero.
It is possible to indicate the presence of an electric field in some region by using small bits of thread suspended in oil or grass seeds scattered on a sheet of paper. Close to a charged body the threads (or seeds) will tend to form linear patterns similar to those of the field lines that might be used to represent the electric field in that region. i However, these lines should not be interpreted as being the field lines – field lines may be drawn at any point where the field is not zero and are simply a convenient way of depicting the field pictorially; they are not physically ‘real’ and are certainly not limited to those locations where the threads or seeds happen to be.
If a scalar quantity such as temperature has just a single value over an entire region then it may be represented by a scalar field of the form T_{ }(r) = constant. Such a field is said to be a uniform field. Similarly, a vector field is said to be uniform if it has the same magnitude and direction at all points. i
✦ What must the field lines look like for a uniform vector field?
✧ They must be equally spaced parallel lines – equally spaced to indicate that the field has the same magnitude everywhere, parallel to indicate that the field has the same direction everywhere.
Figure 12 A uniform gravitational field – one that doesn’t change with position.
Figure 7c Gravitational field lines for the Earth’s gravitational field.
Figure 13 The electric field due to an infinite sheet of uniform positive charge.
We have seen (in Figure 7c) that the gravitational field varies as we move away from the Earth. However, if we consider a small enough region the field lines within that region will be approximately parallel. The gravitational field within a room or building or even a town is, to all intents and purposes, the same everywhere, both in magnitude and direction. The gravitational field in such a region is therefore uniform, and a given mass will experience the same gravitational force at any point within the region. (See Figure 12). If $\hat{\boldsymbol z}$ is a unit vector in the vertically upwards direction, we may represent the uniform downward gravitational field in a room by ${\boldsymbol g}({\boldsymbol r}) = g\hat{\boldsymbol z}$, where the constant g is the local magnitude of the acceleration due to gravity.
A uniform electric field would exist at all points above an infinite plane sheet of uniformly distributed charge. At each point the field would be perpendicular to the sheet and would have magnitude σ/(2ε_{0}), where σ i is the charge per unit area of the sheet (measured in C m^{−2}). Another uniform electric field would exist at all points below the infinite sheet; it would have the same magnitude as the field above the sheet, but it would point in the opposite direction. Figure 13 shows the field lines around such a sheet of positive charge; if the charge had been negative all of the field lines would have to be reversed.
Question T13
Suppose we have an infinite sheet of positive charge of surface (charge) density σ and an infinite sheet of negative charge of surface (charge) density −σ, placed one above the other so that they are parallel. Add the fields of these two charge distributions together and hence determine the magnitude of the resultant field (a) between the parallel sheets, and (b) outside the sheets.
[Hint: Note that the units of surface charge density will be C m^{−2}, and remember that the fields add vectorially.]
Answer T13
(a) Between the parallel sheets the fields from the two sheets are in the same direction, so they reinforce to give a resultant uniform field of magnitude σ/ε_{0}.
(b) Outside the sheets, the fields are in opposite directions, so that they cancel to give zero resultant.
Note: You may be puzzled why the electric field above a charged plate does not depend on a distance from the plate. The explanation lies in the restriction to an ‘infinite plate’ – for which any actual distance above the plate is negligible in comparison to the size of the plate. In practice, these simple expressions are also approximately valid close to a real finite charged plate, providing the distance is small compared to the size of the plate and we are well away from the ends of the plates, where the field becomes nonuniform. A similar restriction of course applies also to the gravitational field of the Earth. It is only close to the Earth’s surface that the field is uniform.
Whether uniform or not, any field that does not vary with time is said to be a constant field. The electrostatic fields created by fixed distributions of charge are examples of constant fields.
If an appropriate test particle (a test mass or a test charge) is located at any point in a gravitational or an electric field, a force will act upon it due to the field. If the test particle is displaced from one position to another within the field then the force that the field exerts will do work on the test particle. i In the particular case that the field is uniform (and constant) the force that it exerts on a given test particle will be the same everywhere (and at all times), and may be represented by a single constant vector F. Under these circumstances, if the test particle undergoes a displacement s (from position r_{1} say to position r_{2}, so that s = r_{2} − r_{1}) then the work W done by the force may be expressed in any of the following equivalent ways:
W = magnitude of the force × component of the displacement in the direction of the force
W = magnitude of the displacement × component of the force in the direction of the displacement
W = F ⋅ s = Fs_{ }cos_{ }θ = F_{x}s_{x} + F_{y}s_{y} + F_{z}s_{z}
In the last of these expressions F ⋅ s denotes the so–called scalar product i of F and s, the angle θ is that between F and s, and the various subscripted quantities are the components of F and s in an arbitrarily chosen Cartesian coordinates system.
For the situations we consider in this module, force and displacement are always along the same line, so θ = 0° or 180°. For convenience we may choose this common line to be the x–axis of a Cartesian coordinate system, in which case the work done by the force is simply W = F_{x}s_{x}. In such (effectively) one–dimensional cases we can say that the work done is simply the product of the force and the displacement. If the displacement is small we may represent it by ∆x i and the correspondingly small amount of work done may then be represented by ∆W = F_{x }∆x.
The work done when a constant (onedimensional) force F_{x} acts over a displacement ∆x is ∆W = F_{x }∆x. i
When describing the (nearly) uniform gravitational field that acts over any small region near the surface of the Earth it is conventional to use a coordinate system in which the z–axis points vertically upwards. The force that the field exerts on a particle of mass m may then be represented by F_{z} = −mg, and the work done by that force when the mass undergoes a displacement ∆z is ∆W = −mg∆z. If the mass is raised, the displacement ∆z is positive and the work done by the gravitational force is negative; if the mass is lowered ∆z is negative, the gravitational force and the displacement are in the same direction, and the work done by the gravitational force is positive. Note that throughout this discussion the work we have been considering is that done by the gravitational force, not that done by any external agency working to overcome the gravitational force. If you had to raise an object you would have to do some positive work, but the work done by the gravitational force while you were raising the object would be negative.
The potential energy (E_{pot}) of a body describes its capacity to do work by virtue of its position in a field. By lifting a mass, we increase its potential energy since the gravitational force acting on the mass does positive work when the mass falls. If the mass had been lowered, i.e. displaced in the direction of the gravitational force, it would have lost part of its capacity to do work by falling and its potential energy would have been correspondingly reduced. It follows from these considerations that the change in potential energy when a test mass is displaced is generally equal to minus the work done by the gravitational force during the displacement, i so we may write
∆E_{pot} = mg∆z(7a)
Since in this case E_{pot} arises by virtue of position within a gravitational field, it is called gravitational potential energy, a name that is often abbreviated to gravitational energy, and may be denoted E_{grav}.
Note that Equation 7a relates changes in potential energy to changes in position; it does not completely determine the value of the potential energy at any point. This means we are free to choose a point at which E_{grav} = 0 for an object, and Equation 7a will then determine the potential energy of the object at all other points.
For example, you may choose to say that a body at a certain point on a horizontal table top has E_{grav} = 0, but it is then inevitable that the body will also have E_{grav} = 0 at every other point on the table top, that its gravitational potential energy will be increasingly positive as it is raised above the table top, and that its E_{grav} will become negative if the body falls from the table. You could decide, instead, to let E_{grav} = 0 at some point on the floor. E_{grav} would then be positive at all places above floor level, and it would increase with height.
The similarities between gravitational and electrostatic forces mean that all the preceding discussion may also be applied to uniform electric fields, provided we take due care over signs. Figure 14 shows a test charge q (which may be positive or negative) in a uniform constant electric field E_{x}. If we take the direction from left to right as the direction of increasing x, then the electrostatic force on the charge is specified by the single component F_{x} = qE_{x} and the work done by that force when the charge is displaced by an amount ∆x is ∆W = F_{x }∆x = qE_{x }∆x. It follows that the change in potential energy when the charge is displaced by ∆x is given by
∆E_{pot} = −qE_{x} ∆x(7b) i
In this case the potential energy is called the electrostatic potential energy, which is often abbreviated to electrical energy, and may be denoted E_{el}. As in the gravitational case, we can arbitrarily choose a point within the field at which we let E_{el} = 0, but having made that choice Equation 7b will automatically determine the value of E_{el} at every other point within the field.
Figure 14 See Answer T14.
Question T14
(a) Suppose we choose point C in Figure 14 as the place where E_{el} = 0. If q is positive, what is its electrical energy E_{el} at point A, and what can be said about E_{el} at point B?
(b) If we now define point A to be where E_{el} = 0, what can be said about E_{el} at points B and C?
Answer T14
(a) The displacement from C to A is ∆x = −d. Since E_{el} = 0 at C, it follows from Equation 7b that the electrostatic potential energy at A is E_{el} = −qE∆x = qEd. E_{el} will be positive at B, but its value will be less than qEd.
(b) The displacement from A to C is ∆x = +d. Since we now have E_{el} = 0 at A, it follows from Equation 7b,
∆E_{pot} = −qE_{x }∆x(Eqn 7b)
that the electrostatic potential energy at C is E_{el} = −qE∆x = −qEd. E_{el} will be
negative at B, but its value will be less negative than −qEd.
The general conclusion that may be drawn from this discussion of potential energy and uniform fields is as follows:
If a uniform field describes a constant (onedimensional) force F_{x} acting on a test particle, then the work done by that force when the test particle is displaced by an amount ∆x will be ∆W = F_{x }∆x, and the corresponding change in the potential energy of the test particle will be:
∆E_{pot} = −∆W = −F_{x }∆x(7c)
Equations 7a and 7b
∆E_{pot} = mg∆z(Eqn 7a)
∆E_{pot} = −qE_{x} ∆x(Eqn 7b)
are particular examples of this general relation.
Among the most common non–uniform fields are those due to an isolated mass or charge located at the origin (see Equations 3 and 6, respectively).
${\boldsymbol g}({\boldsymbol r}) = \dfrac{Gm_0}{r^2}\hat{\boldsymbol r}$(Eqn 3)
${\boldsymbol E}({\boldsymbol r}) = \dfrac{q_0}{4\pi\varepsilon_0r^2}\hat{\boldsymbol r}$(Eqn 6)
In such cases the force on a test particle may be described by a single non–zero component, F_{r}, that acts along a radial direction measured outward from the origin. (F_{r} will be positive if the force is directed away from the origin, and negative if it is directed towards the origin.) Because the field is nonuniform, the value of F_{r} may be expected to depend on the radial position coordinate of the test particle, r. In other words, the force on the test particle will depend on its distance from the origin, and may be represented by F_{r}(r).
✦ Write down an expression for the radial component of the gravitational force on a test particle of mass m at a radial distance r from an isolated point mass m_{0} located at the origin.
Figure 15 The graph of the gravitational force F_{r}(r) on a test particle at a radial distance r from a point mass m_{0} at the origin.
✧ Combining the general definition of the gravitational field, F_{ }(on m at r) = mg(r) (Equation 2), with the expression for the gravitational field of an isolated mass m_{0} at the origin (Equation 3)
$\boldsymbol{g}(\boldsymbol{r}) = \dfrac{Gm_0}{r^2}\hat{\boldsymbol{r}}$(Eqn 3)
we see that
$\boldsymbol{F}~({\rm on}~m~{\rm at}~\boldsymbol{r}) = \it m \left(\dfrac{Gm_0}{r^2}\hat{\boldsymbol{r}}\right)$
It follows from this that the radial component of the force depends on r, and can be written as
$F(r) = \dfrac{Gmm_0}{r^2}$
(This particular form of F_{r}(r) is shown graphically in Figure 15)
The integral in Equation 8,
$\displaystyle W = \int_{r_1}^{r_2} F_r(r)\,dr$(8)
can be interpreted as the area under this graph between r_{1} and r_{2}, which is negative for areas below the r–axis.
Although gravitational and electrostatic forces generally vary with position, if the (radial) displacement (∆r) of the test particle is small enough that F_{r}(r) does not change significantly, then we can still write the work done as ∆W = F_{r }∆r.
Suppose now the particle undergoes a large radial displacement from a point P_{1}, at a distance r_{1} from the origin, to a point P_{2}, at a distance r_{2} from the origin. We can estimate the work done by F_{r}(r) over such a displacement by dividing the radial path from P_{1} to P_{2} into a large number of very small steps and then adding together the small amount of work done in each of these steps. This estimate can be made into an accurate evaluation of the work done if we make even the largest of the steps ‘vanishingly’ small. This process of determining a quantity by considering a limit of a sum is known as definite integration and is fully explained elsewhere in FLAP. For our present purposes it is sufficient to note the following points:
Figure 15 The graph of the gravitational force F_{r}(r) on a test particle at a radial distance r from a point mass m_{0} at the origin. The integral in Equation 8 can be interpreted as the area under this graph between r_{1} and r_{2}, which is negative for areas below the r–axis.
The total work done by the force F_{r}(r) when the test particle is displaced from r_{1} to r_{2} is given by the definite integral:
$\displaystyle W = \int_{r_1}^{r_2} F_r(r)\,dr$(8) i
It is possible to interpret such an integral in terms of the ‘area under a graph’, measured in the scale units appropriate to the graph. In this case, given a graph of F_{r}(r) against r, the value of W is equal to the area between the graph and the r–axis between r_{1} and r_{2}, as indicated in Figure 15. In this case the curve representing F_{r}(r) happens to lie below the r–axis, so the area W is a negative quantity. (If the curve is plotted on graph paper this area can be found by counting squares, though doing so is tedious.)
In many cases the value of W may be found more simply by applying standard mathematical techniques that allow definite integrals to be evaluated once the function F_{r}(r) is specified. These are fully developed elsewhere in FLAP.
Let us apply these ideas to the case of a test particle of mass m in the gravitational field due to a mass m_{0} at the origin. (This situation is represented in Figure 15). In this case we know that
$F_r(r) = \dfrac{Gm m_0}{r^2}$
so$\displaystyle W = \int_{r_1}^{r_2} \dfrac{Gm m_0}{r^2}\,dr$
However G, m and m_{0} are all constants, so according to the rules of integration this may be written as
$\displaystyle W = Gm m_0\int_{r_1}^{r_2} \dfrac{1}{r^2}\,dr$
Now the result of evaluating this integral (as you can confirm using standard techniques if you know them, or by drawing the graph of 1/r^{2} and measuring the area between the curve and the r–axis from some value r_{1} to some other value r_{2} as just described) is
$W = Gmm_0\left(\dfrac{1}{r_2}\dfrac{1}{r_1}\right) = Gmm_0\dfrac{r_1r_2}{r_1 r_2}$
In this case, F_{r}(r) was negative because the gravitational force on the test mass pointed towards the origin, so over a positive displacement, in which r_{2} is greater than r_{1}, we would expect the work done by the gravitational force to be negative. This is exactly what the above expression indicates. (Remember, W is the work done by the gravitational force when the particle is displaced, not the work you would do if you had to cause the displacement.)
Using Equation 8,
$\displaystyle W = \int_{r_1}^{r_2} F_r(r)\,dr$(Eqn 8)
we can now find a general expression for the change in the potential energy of a test particle when it is displaced from r_{1} to r_{2}. We know from the definition of potential energy, that the change in the potential energy will be equal to minus the work done by the field force over the displacement.
Thus in general terms:
$\displaystyle \Delta E_{\rm pot}= W = \int_{r_1}^{r_2} F_r(r)\,dr$(9) i
Note that this is an expression for the change in the potential energy of the test particle. As usual, we are free to choose an arbitrary point in the field at which E_{pot} = 0, but having made that choice the expression for ∆E_{pot} will determine the value of E_{pot} at every other point.
Applying Equation 9 to the particular case of the gravitational field due to a point mass m_{0} at the origin we see that:
$\Delta E_{\rm grav} = Gmm_0\left(\dfrac{1}{r_2}\dfrac{1}{r_1}\right)$i
Figure 16 Gravitational potential energy, E_{grav}, of a mass m at a distance r from a mass m_{0}.
In this particular case it is conventional to choose the point at which the mass m has E_{grav} = 0 to be a point at infinity (∞), i.e. one for which r → ∞. i Taking this as the starting point for a displacement we can determine the full value of the gravitational potential energy of a mass m at any distance r from the mass of m_{0} at the origin by letting r_{1} → ∞ and r_{2} = r. It follows that:
$E_{\rm grav} ({\rm of}~m~{\rm at}~r) =  \dfrac{Gmm_0}{r}$(10) i
Note that in this situation the gravitational potential energy of the mass m is negative. E_{grav} is a small negative quantity for large r and becomes more negative as r decreases (see Figure 16).
We have dealt here with gravitational forces, but the above ideas may also be applied to electrostatic forces.
Question T15
Apply Equation 9,
$\displaystyle \Delta E_{\rm pot}= W = \int_{r_1}^{r_2} F_r(r)\,dr$(Eqn 9)
to two charges q and q_{0} separated by a distance r to find the equivalent electrical expression to Equation 10,
$E_{\rm grav} ({\rm of}~m~{\rm at}~r) =  \dfrac{Gmm_0}{r}$(Eqn 10)
Sketch E_{pot} (of q at r) as a function of r if q and q_{0} are both positive.
Figure 27 See Answer T15.
Answer T15
Equation 9 is
$\displaystyle \Delta E_{pot} = W = \int_{r_1}^{r_2} F_r(r)\,dr$(Eqn 9)
in this case $F_r(r) = \dfrac{qq_0}{4\pi\epsilon_0r^2}$ (from Equation 4)
hence we can write the electrostatic potential energy ∆E_{el} as
$\displaystyle \Delta E_{el} = \dfrac{qq_0}{4\pi\epsilon_0r^2}\int_{r_1}^{r_2} \dfrac{1}{r^2}\,dr = \dfrac{qq_0}{4\pi\epsilon_0r^2}\left(\dfrac{1}{r_2}\dfrac{1}{r_1}\right)$
If we take E_{el} = 0 at infinity, and set r_{1} → ∞ and r_{2} = r, we find
E_{el} (of q at r) = $\dfrac{qq_0}{4\pi\epsilon_0r}$
Hence the potential energy E_{pot} = E_{el} is positive for two positive charges, and varies as 1/r, as shown in Figure 27.
We have seen that a definite integral of the force F_{r}(r) described by a field allows us to deduce the associated potential energy E_{pot}(r), but we can reverse that process to reveal another important link between (conservative) forces and their associated potential energy functions. We saw earlier in this subsection that even in a non–uniform field the work done over a small displacement is given by an expression of the form ∆W = F_{r }∆r. It follows that if F_{r}(r) is the force at the starting point of the displacement then, at least approximately
∆E_{pot} ≈ −F_{r}(r)∆r i
Rearranging this and considering the limit in which ∆r becomes vanishingly small, so that the approximation becomes increasingly accurate, we have
$F_r(r) = \lim_{\Delta r \rightarrow 0}\left(\dfrac{\Delta E_{\rm pot}}{\Delta r}\right)$ i
i.e.$F_r(r) = \dfrac{dE_{\rm pot}}{dr}$(11) i
In words, the radial component of the force on a test particle at a distance r from the mass of m_{0} (or charge q_{0}) at the origin is equal to minus the derivative of the particle’s potential energy at that point. Since dE_{pot}/dr may be interpreted as the gradient of the graph of E_{pot} against r, we may describe Equation 11 as:
The component of the force in any direction is equal to minus the gradient of the potential energy in that direction.
The results obtained in this subsection have been derived in the context of the purely radial field of a point mass or charge at the origin, but this last result is actually much more general as we have stated. In any (conservative) field the component of the force that the field exerts in any direction on a test particle at any point is given by minus the gradient of the corresponding potential energy in that direction. i In these general cases the potential energy of the test particle is not restricted to depending on r alone (that was a particular property of the field of a point source) so we should represent it as E_{pot}(r) rather than E_{pot}(r). Nonetheless at any point we can still identify a direction in which the potential energy is declining most rapidly, and that is the direction in which the force on the test particle will act. The existence of such forces explains the tendency of a system to minimize its potential energy if it is free to do so.
Figure 16 Gravitational potential energy, E_{grav}, of a mass m at a distance r from a mass m_{0}.
✦ By considering the gradient of Figure 16, describe the way that F_{r}(r) depends on r, paying due attention to its direction.
Figure 15 The graph of the gravitational force F_{r}(r) on a test particle at a radial distance r from a point mass m_{0} at the origin.
✧ The gradient of Figure 16 is positive and very large at small r, giving a large negative value for Fr in that region, the gradient then decreases as r increases, tending to zero as r tends to ∞.
Correspondingly, the curve of F_{r} remains negative but tends to zero as r tends to ∞. These negative values indicate that F_{r} is always directed towards the origin, implying that the gravitational force is always attractive.
(The curve you have been asked to describe is the graph of F_{r} against r shown in Figure 15).
The integral in Equation 8 can be interpreted as the area under this graph between r_{1} and r_{2}, which is negative for areas below the r–axis.
Figure 14 A test charge q charge in a uniform electric field E_{x}. The points A, B and C represent three of the charge’s possible positions. A and C are separated by a distance d.
Question T16
Sketch graphs to show how force and potential energy vary with position in a uniform electrostatic field (e.g. show how F_{x} and E_{el} for a positive charge q depend on x in Figure 14).
Figure 28 See Answer T16.
Answer T16
The graphs of E_{el} and F_{x} for the uniform electric field illustrated Figure 14 are shown Figure 28. In this case, if we let E_{el} = 0 at point A, where x = 0, then E_{el} (of q at x) = −qEx. The gradient dE_{el}/dx = −qE, so the electrostatic force is F_{x} = −dE/dx = qE and points in the positive x direction if q is positive. (This applies equally to a uniform gravitational field.)
One of the main reasons for introducing the gravitational and electrostatic fields in the way that we did, as the force per unit mass and the force per unit positive charge on a test particle, was that the resulting quantities g_{ }(r) and E_{ }(r) are independent of the test charge. The force on the test particle is proportional to the mass or charge of the test particle, but the field that it helps to define does not depend on those properties. This subsection introduces a widely used quantity called the potential, that is related to the potential energy in the same way that the field is related to the force.
If the gravitational potential energy of a test mass m at some point r in a gravitational field is E_{grav} (of m at r), then the gravitational potential V_{grav}(r) at that point is the gravitational potential energy per unit mass at that point:
$V_{\rm grav}({\boldsymbol r}) = \dfrac{E_{\rm grav} ({\rm on}~m~{\rm at}~r)}{m}$(12a) i
Similarly, we can define the electric potential V_{el}(r) at any point in an electrostatic field as the electrostatic potential energy per unit charge at that point, i.e:
$V_{\rm el}({\boldsymbol r}) = \dfrac{E_{\rm el} ({\rm on}~q~{\rm at}~r)}{q}$(12b) i
✦ What are appropriate SI units of (a) gravitational potential and (b) electrical potential?
✧ The units are those of (a) energy/mass, i.e. J kg^{−1} and (b) energy/charge, i.e. J C^{−1}.
Electric potential is such a widely used quantity that its SI unit has a special name, the volt (V). i Thus, 1 V = 1 J C^{−1}. It is a common practice to refer to the difference in electric potential between two points as the ‘voltage’ or the ‘voltage difference’ between those points.
✦ A particle of charge e = 1.60 × 10^{−19} C is released from a point in an electric field at which the electric potential is V_{1} = 100 V. If the only forces on the particle are those due to the electric field, how much work will have been done on the particle when it arrives at a point where the potential is V_{2} = −100 V?
✧ It follows from Equation 12b,
$V_{{\rm el}}(\boldsymbol{r}) = \dfrac{E_{{\rm el}}~({\rm on}~q~{\rm at}~\boldsymbol{r})}{q}$(Eqn 12b)
that the particle’s potential energy will have changed by an amount equal to the change in potential multiplied by the charge of the particle, therefore
∆E_{el} = (V_{2} − V_{1}) × e = −200 V × 1.60 × 10^{−19} C = −3.20 × 10^{−17} J
However the work done by the field force, W, is minus the change in potential energy, i.e.
W = +3.20 × 10^{−17} J
There are two main reasons why the concept of potential is useful and important. First, the potential energies E_{grav} and E_{el} in Equations 12a and 12b are proportional to m and q, respectively, so the potentials V_{grav} and V_{el} defined by those equations do not depend on m and q. Hence the potentials (unlike the potential energies) are independent of the test charges used to determine them and characterize the fields themselves. Secondly, the potential is a scalar field, so it is easier to deal with mathematically than a vector field. (We will exploit this simplicity later.)
Question T17
Write down expressions for the gravitational potential due to a point mass at the origin, and for a point charge at the origin. Use the first expression to calculate the gravitational potential at a distance of 10 m from a 5 kg mass.
Answer T17
$E_{\rm grav} = \dfrac{Gmm_0}{r}\quad\text{and}\quad E_{\rm el} = \dfrac{qq_0}{4\pi\epsilon_0r}$
but V_{grav}(r) = E_{grav}/m and V_{el}(r) = E_{el}/q
so$V_{\rm grav}(r) = \dfrac{Gm_0}{r}\quad\text{and}\quad V_{\rm el} = \dfrac{q_0}{4\pi\epsilon_0r}$
Note that in these cases the potential depends on $r=sqrt{x^2+y^2+z^2}$ and not on any other combination of x, y and z. So we are justified in representing these potentials as V_{ }(r) rather than using the more general expression V_{ }(r).
If m_{0} = 5 kg,
$V{\rm grav}(10\,{\rm m}) = \rm \dfrac{6.67\times 10^{11}\,N\,m^2\,kg^{2}\times 5\,kg}{10\,m} = \dfrac{6.67\times 5\times 10^{11}\,N\,m^2\,kg^{1}}{10}$
i.e.V_{grav}(10 m) = −3.34 × 10^{−11} J kg^{−1}
In the context of gravitation, the description of a force as minus the gradient of potential energy
$F_r(r) = \dfrac{dE_{\rm pot}}{dr}$(Eqn 11)
implies that the ‘force per unit mass’ (i.e. the gravitational field) must be minus the gradient of the ‘potential energy per unit mass’ (i.e. the gravitational potential). A similar relationship links the electrostatic field and its potential. Hence, for the simple cases we have been considering in this module, where the position may be specified by a single variable such as r or x, we can write for a gravitational field:
$g_x(x) = \dfrac{dV_{\rm grav}(x)}{dx}\quad\text{or}\quad g_r(r) = \dfrac{dV_{\rm grav}(r)}{dr}$(13)
and for an electrostatic field:
$E_x(x) = \dfrac{dV_{\rm el}(x)}{dx}\quad\text{or}\quad E_r(r) = \dfrac{dV_{\rm el}(r)}{dr}$(14)
Like the relationship between force and potential energy, these relationships may be generalized to the case where the potential is a function of r (i.e. a function of x, y and z) rather than a function of a single variable. We will not explore this case mathematically, but as before we will state that generally:
The component of the field in any direction is minus the gradient of the potential in that direction.
The direction of the field is the direction in which the potential decreases most rapidly. i
Table 1 summarizes the relationship between the main characteristics of the fields that we have explored in this module.
Table 13 The relationship between forces, fields, potential energies and potentials in (effectively) one–dimensional cases.
Question T18
Make a table listing expressions for the field and potential of a point mass m_{0}, and the associated force and potential energy for a test mass m at a distance r from m_{0}. Check the entries in your table by confirming that they are related in the fashion indicated by Table 1.
Answer T18
Table 2 summarizes the equations for the gravitational effect of a point mass.
Scalars  Vectors 

$E_{\rm grav} = \dfrac{Gmm_0}{r}$  ${\boldsymbol F}({\boldsymbol r}) = \dfrac{Gmm_0}{r^2}\hat{\boldsymbol r}$ 
$V_{\rm grav}(r) = \dfrac{Gm_0}{r}$  ${\boldsymbol g}({\boldsymbol r}) = \dfrac{Gm_0}{r^2}\hat{\boldsymbol r}$ 
Note that as required,
$F_r = \dfrac{dE_{\rm grav}}{dr} = Gmm_0\dfrac{d}{dr}\left(\dfrac1r\right) = \dfrac{Gmm_0}{r^2}$
and similarly
$g_r = \dfrac{dV_{\rm grav}}{dr} = Gm_0\dfrac{d}{dr}\left(\dfrac1r\right) = \dfrac{Gm_0}{r^2}$
We can now describe the effect of an isolated mass m_{0} (or charge q_{0}) in terms of either its field or its potential. In general, which should we prefer? Both have their uses. If we want to know the force on an object, then it is more direct to think about the field. If we want to discuss work or energy it is better to deal with the potential. However, if we want to find the combined effect of several masses or charges, it is often easier to work with the potential, even if we are ultimately interested in the forces that such a combination exerts. The reason for this is that the potentials may be combined using scalar addition, but combining the fields involves the more complicated process of vector addition.i So, to find the combined field it is often quicker to combine the potentials and then take the negative gradient of the combined potential.
Figure 8 Arrangement of two 4 kg masses about point O.
✦ Calculate the potential at point O due to the two 4 kg masses shown in Figure 8.
✧ The gravitational potential due to each mass m at a distance r from that mass is V_{ }(r) = −Gm/r. The potentials of the two masses simply add as scalars at every point, so the combined potential at O due to both masses together is
V_{ }(0) = 2 × −6.67 × 10^{−11} N m^{2} kg^{−2} × 4 kg/2 m = −4 × 6.67 × 10^{−11} N m kg^{−1} = −2.67 × 10^{−10} J kg^{−1} i
Figure 17 Equipotential contours (full curves) surrounding two masses m close together. Gravitational field lines (dashed lines).
Figure 18 Gravitational equipotentials (full curves) for a point mass.
We could extend the above example by finding a general expression for V_{ }(r) as a function of position, then differentiating to find the field. Alternatively, the potential can be calculated at positions on a grid surrounding the masses. We can then produce a ‘map’ of the potential by drawing equipotential surfaces (these ‘surfaces’ are just lines or equipotential_contourcontours in two dimensions) over which the potential has a fixed constant value. Equipotentials are usually drawn so that for consecutive surfaces (or contours) the potential changes by a fixed amount. i
Figure 17 shows some equipotential contours (solid curves) surrounding two masses.
Equipotentials have two important features, which we will illustrate with the example of an isolated point mass. Since the potential depends only on r, the equipotentials are a set of concentric spheres in this case (Figure 18). There are a couple of points worth noting about Figure 18:
Along any equipotential surface V is, by definition, constant. Therefore, along an equipotential, F_{grav} must do no work and the field component must be zero. The direction of the field at any point will be that in which the potential is decreasing most rapidly, and that will always be perpendicular to the equipotential surfaces. Equipotentials can therefore be used to deduce the pattern of field lines, as shown in Figure 17, or vice versa.
✦ Describe the electrostatic equipotentials near a large flat sheet of uniformly distributed charge.
✧ They are a set of equally–spaced planes, parallel to the sheet.
Figure 25 See Answer T11.
Question T19
Sketch the equipotentials for the charges and field lines shown in Figure 25 (see Answer T11).
Figure 29 See Answer T19.
Answer T19
The equipotentials for the three charged particles are shown in Figure 29. The solid lines are the equipotentials. The field lines (from Figure 25) are shown dashed.
A field is a physical quantity to which a definite value can be ascribed at each point in some region of space at a particular time. For a scalar field (e.g. T_{ }(r)) we need to specify only the numerical value (and the appropriate unit of measurement) of the quantity at all points in the region. For a vector field (e.g. E_{ }(r)) we generally need to specify the three components of a vector at all points in the region.
According to Newton’s law of gravitation the gravitational force on a point mass m_{2} due to a point mass m_{1} is given by
${\boldsymbol F}_{21} = \dfrac{Gm_1 m_2}{r^2}\hat{\boldsymbol r}$(Eqn 1)
where G is the universal gravitational constant, r is the distance from m_{1} to m_{2} and r^{^} is a unit vector that points from m_{1} to m_{2}. (A unit vector parallel to a given vector r is defined by $\hat{\boldsymbol r} = {\boldsymbol r}/\vert\,{\boldsymbol r}\,\vert$, and has magnitude 1).
According to Coulomb’s law the electrostatic force on a point charge q_{2} due to a point charge q_{1} is given by
${\boldsymbol F}_{21} = \dfrac{q_1 q_2}{4\pi\varepsilon_0r^2}\hat{\boldsymbol r}$(Eqn 4)
where ε_{0} is the permittivity of free space, r is the distance from q_{1} to q_{2}, and r^{^} is a unit vector that points
from q_{1} to q_{2}.
The gravitational field g_{ }(r) at a point is the gravitational force per unit mass on a test mass placed at that point:
${\boldsymbol g}({\boldsymbol r}) = \dfrac{{\boldsymbol F}_{\rm grav}({\rm on}~m~{\rm at}~{\boldsymbol r})}{m}$(Eqn 2)
The electric field E_{ }(r) at a point is the electrostatic force per unit positive charge on a test charge placed at that point:
${\boldsymbol E}({\boldsymbol r}) = \dfrac{{\boldsymbol F}_{\rm el}({\rm on}~q~{\rm at}~{\boldsymbol r})}{q}$(Eqn 5)
The gravitational and electric fields due to a point particle of mass m_{0} and/or charge q_{0} at the origin are, respectively,
${\boldsymbol g}({\boldsymbol r}) = \dfrac{Gm_0}{r^2}\hat{\boldsymbol r}$(Eqn 3)
and${\boldsymbol E}({\boldsymbol r}) = \dfrac{q_0}{4\pi\varepsilon_0r^2}\hat{\boldsymbol r}$(Eqn 6)
When vector fields combine according to the principle of superposition, the resultant field is the vector sum of the individual fields at each point.
A vector field can be represented by directed field lines. The (directed) tangent to a field line represents the direction of the field and the density of the lines in any region represents the strength of the field in that region.
In a uniform fielduniform (gravitational or electric) field, the field is everywhere the same in magnitude and direction. The uniform gravitational field in any small region close to the Earth’s surface may be written g_{ }(r) = −gz^{^}, where g is the magnitude of the acceleration due to gravity, and a uniform electric field in the x–direction may be written E_{ }(r) = E_{x}x^{^}.
If a force described by a (conservative) field does work W on a test particle when that particle is displaced, then the Section 3potential energy that the particle possesses by virtue of its position in the field will change by an amount
∆E_{pot} = −W
The work done by a uniform force F_{x} over a displacement ∆x is W = F_{x }∆x, i and that done by a varying
force F_{r}(r) (that depends on a single variable r) over a displacement from r = r_{1} to r = r_{2} is given by
$\displaystyle W = \int_{r_1}^{r_2} F_r(r)\,dr$(Eqn 8)
A point at which E_{pot} = 0 may be chosen arbitrarily. Once that choice has been made, E_{pot} at all other places can be deduced from the appropriate expression for ∆E_{pot}.
In the field of a mass m_{0} or charge q_{0} at r = 0, with E_{pot} = 0 as r → ∞
$E_{\rm grav} ({\rm of}~m~{\rm at}~r) =  \dfrac{Gmm_0}{r}$(Eqn 10)
and$E_{\rm el} ({\rm of}~q~{\rm at}~r) =  \dfrac{qq_0}{4\pi\varepsilon_0r}$
In a uniform gravitational field g_{ }(r) = −g$\hat{\boldsymbol z}$, with E_{grav} = 0 at z = 0
E_{grav} (of m at z) = mgz
In a uniform electrostatic field E_{ }(r) = E_{x}$\hat{\boldsymbol x}$, with E_{el} = 0 at x = 0
E_{el} (of q at x) = −qE_{x}x
The gravitational or electrostatic force on a test particle is equal to minus the gradient of the particle’s Section 3potential energy. So, in cases where the force may be specified by a single components_of_a_vectorcomponent that depends on a single variable such as r:
$F_r(r) = \dfrac{dE_{\rm pot}}{dr}$(Eqn 11)
The gravitational potentialgravitational (electric potentialor electric) electric potentialpotential at any point is equal to the potential energy per unit mass (or charge) at that point.
$V_{\rm grav}({\boldsymbol r}) = \dfrac{E_{\rm grav} ({\rm on}~m~{\rm at}~r)}{m}$(Eqn 12a)
and$V_{\rm el}({\boldsymbol r}) = \dfrac{E_{\rm el} ({\rm on}~q~{\rm at}~r)}{m}$(Eqn 12b)
In the case of a point mass or charge at r = 0
$V_{\rm grav}(r) = \dfrac{Gm_0}{r}\quad\text{and}\quad V_{\rm el}(r) = \dfrac{q_0}{4\pi\varepsilon_0r}$
The gravitational fieldgravitational or electrostatic field is equal to minus the gradient of its potential. So, in cases where the force may be specified by a single component that depends on a single variable such as r:
$g_x(x) = \dfrac{dV_{\rm grav}(x)}{dx}\quad\text{or}\quad g_r(r) = \dfrac{dV_{\rm grav}(r)}{dr}$(Eqn 13)
and$E_x(x) = \dfrac{dV_{\rm el}(x)}{dx}\quad\text{or}\quad E_r(r) = \dfrac{dV_{\rm el}(r)}{dr}$(Eqn 14)
An equipotential surface (or contour) is one on which the potential has a constant value. Equipotential surfaces (or contours) are always at right angles to field lines, and are closer together where the field is stronger.
Study comment You may now wish to take the following Exit test for this module which tests these Achievements. If you prefer to study the module further before taking this test then return to the topModule contents to review some of the topics.
Having completed this module, you should be able to:
Define the terms that are emboldened and flagged in the margins of the module.
Distinguish between scalar and vector fields, and give examples of each.
List the properties of gravitational forces.
Determine the gravitational force between two point masses.
Determine the gravitational field due to a simple arrangement of masses.
Apply the principle of superposition to gravitational and electric fields.
Determine the force on an appropriate test particle in a given gravitational or electric field.
Determine the electrostatic force between two point charges.
Determine the electrostatic field due to a simple arrangement of charges.
Draw field lines for simple arrangements of point masses or charges.
In simple cases, calculate the work done and the corresponding change in potential energy when a mass or charge is displaced in a gravitational or electrostatic field.
Determine the potential due to a simple arrangement of masses or charges.
Sketch equipotential surfaces (or contours) for a uniform gravitational or electrostatic field and for simple arrangements of point masses or charges.
Relate the force that a field exerts on a particle to the gradient of the corresponding potential energy, and relate a field to the gradient of its potential.
Study comment Having completed this module, you should be able to answer the following questions, each of which tests one or more of the Achievements.
Question E1 (A2)
(a) Is the electric field surrounding a point charge a scalar or a vector field?
(b) Do the contours on an Ordnance Survey map represent a field? If so, what kind of field?
Answer E1
(a) It is a vector field. If we change the sign of one of the charges we change the direction of the electrostatic force and hence of the electric field.
(b) Ordnance Survey maps show contours where the land is all at a certain height above sea level, i.e. they represent a (scalar) height field.
(Reread Subsections 2.1 and 2.4 if you had difficulty with this question.)
Question E2 (A4)
What is the gravitational force between two protons separated by a distance of 2.0 × 10^{−15} m? Take the mass of a proton as 1.67 × 10^{−27} kg.
Answer E2
Using Equation 1:
${\boldsymbol F}_{21} = \dfrac{Gm_1 m_2}{r^2}\hat{\boldsymbol r}$(Eqn 1)
${\boldsymbol F}_{\rm grav} = \dfrac{Gm_1 m_2}{r^2} = \dfrac{Gm_p^2}{r^2}$
i.e.${\boldsymbol F}_{\rm grav} = \dfrac{6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}}\times (1.67\times 10^{27}\,{\rm kg})^2}{(2.0\times 10^{15}\,{\rm m})^2}$
so${\boldsymbol F}_{\rm grav} = \dfrac{6.67\times(1.67)^2}{4}\times \dfrac{10^{11}\times10^{54}}{10^{30}}\,{\rm N} = 4.65\times 10^{35}\,{\rm N}$
(Reread Subsection 2.2 if you had difficulty with this question.)
Figure 19 See Question E3.
Question E3 (A5, A6 and A7)
Calculate the resultant gravitational field at point O due to two isolated masses one of mass 3 kg at a distance of 1 m and one of mass 4 kg at a distance of 2 m arranged as shown in Figure 19. Hence find the force on a 1 g mass placed at O.
Answer E3
The field at O due to the 3 kg mass is along x–axis and has magnitude
$g_1 = \dfrac{Gm_1}{r_1^2}$
The field at O due to the 4 kg mass is along y–axis and has magnitude
$g_2 = \dfrac{Gm_2}{r_2^2}$
The resultant field at O has magnitude
$g_{\rm res} = \sqrt{\smash[b]{g_1^2+g_2^2}} = \sqrt{\left(\dfrac{Gm_1}{r_1^2}\right)^2+\left(\dfrac{Gm_2}{r_2^2}\right)^2}\\ \phantom{g_{\rm res}} = 6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}} \sqrt{\left(\dfrac{3\,{\rm kg}}{1\,{\rm m^2}}\right)^2+\left(\dfrac{4\,{\rm kg}}{4\,{\rm m^2}}\right)^2}\\ \phantom{g_{\rm res}} = 6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}} \sqrt{\smash[b]{10\,({\rm kg\,m^{2}})^2}} = 2.11\times 10^{10}\,{\rm N\,kg^{1}}$
The resultant field is directed at an (anticlockwise) angle θ to the x–axis, where
θ = arctan(1/3) = 18.4°
So, the force acts at 18.4° to the x–axis and has magnitude
mg_{res} = 10^{−3} kg × 2.11 × 10^{−10} N kg^{−1} = 2.11 × 10^{−13} N
(Reread Subsections 2.2 and 2.3 if you had difficulty with this question.)
Question E4 (A6 and A10)
Sketch in some field lines surrounding the Earth–Moon pair taking into account the fact that the Earth is approximately 80 times more massive than the Moon, and hence its gravitational field (at any given radius) correspondingly stronger. Mark on your sketch the point N where the resultant gravitational field is zero.
Figure 30 See Answer E4.
Answer E4
Your sketch of the gravitational field lines around the Earth–Moon pair should resemble the schematic representation given in Figure 30. In three dimensions, there should be 80 times as many field lines from the Earth as from the Moon. The two fields cancel at point N.
(Reread Subsection 2.3 if you had difficulty with this question.)
Question E5 (A4, A5 and A6)
Use the data given in the margin i to calculate the distance of point N (in Question E4) from the Earth. Treat the Earth and Moon as point masses. [Hint: If you call the distance of point N from the Earth r_{1}, and the distance of point N from the Moon r_{2}, you can calculate the ratio r_{1}/r_{2} directly using m_{E} and m_{M}. You can then calculate the actual value of r_{1} using the Earth–Moon distance R.]
Answer E5
The magnitude of the gravitational field due to the Earth at N is given by
$g_{\rm E} = \dfrac{Gm_{\rm E}}{r_1^2}$
The magnitude of the gravitational field due to the Moon at N is given by
$g_{\rm M} = \dfrac{Gm_{\rm M}}{r_2^2}$
At N these are equal, so
$\dfrac{m_{\rm E}}{r_1^2} = \dfrac{m_{\rm M}}{r_2^2}$
$\left[\dfrac{r_1}{r_2}\right]^2 = \dfrac{m_{\rm E}}{m_{\rm M}}$
$\dfrac{r_1}{r_2} = \sqrt{\dfrac{m_{\rm M}}{r_2^2}} = \sqrt{\dfrac{6.0\times 10^{24}\,{\rm kg}}{7.4\times 10^{22}\,{\rm kg}}}= \sqrt{600/7.4\os} = 9.0$
so thatr_{1}/(r_{1} + r_{2}) = 9/10, and hence r_{1} = 9/10 × R.
r_{1} = 0.9 × 3.8 × 10^{8} m = 3.42 × 10^{8} m
(Reread Subsections 2.2 and 2.3 if you had difficulty with this question.)
Figure 20 See Question E6.
Question E6 (A6, A9 and A10)
Two positive and two negative charges of equal magnitude are arranged at the corners of a square as shown in Figure 20. Draw the field lines associated with the arrangement. What is the magnitude of the electric field at the centre of the square? Indicate the direction of the resultant field at the point A halfway down one side of the square.
Figure 31 See Answer E6.
Answer E6
Your drawing should resemble Figure 31b. The field is zero at the centre of the square, where the fields due to the like charges from opposite corners cancel. The four arrows drawn at point A (in Figure 31a) represent the four fields contributing to the resultant field at point A, which points ‘downwards’ along the side of the square.
(Reread Subsections 2.3 and 2.4 if you had difficulty with this question.)
Question E7 (A12)
Calculate the gravitational potential at a distance of 10 m from a 2 kg mass.
Answer E7
The gravitational potential V_{ }(r) due to a point mass m is given by
$V(r) = \dfrac{Gm}{r} = \dfrac{6.67\times 10^{11}\,{\rm N\,m^2\,kg^{2}}\times 2\,{\rm kg}}{10\,{\rm m}}$ = −1.3 × 10^{−11} N m kg^{−1}
i.e.V_{ }(r) = −1.3 × 10^{−11} J kg^{−1}
(Reread Subsection 3.3 if you had difficulty with this question.)
Figure 20 See Question E6.
Question E8 (A13)
Draw the equipotentials for the charges illustrated in Figure 20.
Figure 31 See Question E6.
Figure 32 See Question E6.
Answer E8
The equipotentials (solid lines) for the charges illustrated in Figure 20 are given in Figure 32. (The dashed lines represent the field lines as shown in Figure 31).
(Reread Subsection 3.4 if you had difficulty with this question.)
Study comment This is the final Exit test question. When you have completed the Exit test go back and try the Subsection 1.2Fast track questions if you have not already done so.
If you have completed both the Fast track questions and the Exit test, then you have finished the module and may leave it here.
Study comment Having seen the Fast track questions you may feel that it would be wiser to follow the normal route through the module and to proceed directly to the following Ready to study? Subsection.
Alternatively, you may still be sufficiently comfortable with the material covered by the module to proceed directly to the Section 4Closing items.