MATH 2.6: Scalar product of vectors	PPLATO @
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1 Opening items

1.1 Module introduction

In this module we assume that you are already familiar with addition of vectors and that you know how to multiply a vector by a scalar. i Our main purpose is to introduce the concept and use of the scalar product of vectors, which is a way of multiplying two vectors together to produce a scalar.

Imagine an aircraft landing on a runway; as it touches down, the forward components_of_a_vectorcomponent of its velocity may be quite large (perhaps something well in excess of 100 mph), but the pilot’s main concern is to keep the vertical component of the velocity small. Similarly, there are many applications of physics where it is a particular component of a vector that is of greatest concern. As a trivial illustration, when a child slides down a chute in the playground he or she is moving under the influence of gravity; it is the component of the gravitational force (i.e. the child’s weight) along the line of the chute that determines how fast the child moves. To find the kinetic energy gained in the descent we need to multiply this component by the vertical component of the displacement. The natural way to deal with components of vectors is to use the mathematical device that we introduce here – the scalar product.

In Section 2 of this module we define the scalar (or dot) product and then deduce some of its elementary properties. The section continues by showing how the product may be expressed in terms of the Cartesian unit vector and ordered triple notations. The section ends with a discussion of the way in which scalar products may be used to find the projection of one vector on to another, the component of a vector in a given direction and the angle between two known vectors.

This module also introduces the idea of a vector field. You have probably seen a simple experiment in which iron filings are scattered on a piece of paper resting on a bar magnet; when you tap the paper the iron filings form into a pattern of curves. These curves show the magnetic field lineslines of ‘magnetic force’ that lead from one magnetic_polepole of the magnet to the other. Although this experiment only shows the lines of magnetic force that lie in the plane of the paper, the lines actually extend into three dimensions, and their shape is related to the shape of the magnet – they will be very different for a horse–shoe magnet, for example. These lines of force are usually taken as evidence that a magnetic field surrounds the magnet. This is the agency responsible for exerting magnetic forces on distant objects. Since this magnetic field gives rise to forces, it must itself be characterized by a magnitude and direction at each point, and is therefore an example of a vector field. In mathematical terms a vector field is perhaps most simply thought of as a function that associates a vector with each point in space. This idea of a ‘vector-valued function of position’ is very important in physics; vector fields play a fundamental role in the description of electrical and gravitational effects as well as in magnetism. In this module you will see that scalar products are as important to the description of vector fields as they are in the treatment of individual vectors.

In Section 3 we show how scalar products are related to various physical quantities such as work done, electric potential difference, power, and the flux of a vector field, using a surface integral, illustrated by Gauss’ law in electrostatics.

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.

1.2 Fast track questions

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.

1.3 Ready to study?

2 The scalar product

2.1 A review of vectors

In this module it is assumed that you are already familiar with unit vectors, modulus of a vector, vector addition and multiplication of a vector by a scalar (scaling), but to refresh your memory we list the main results for you here.

For any two vectors a and b, and any scalar α

If the vectors are expressed in Cartesian form a = a_xi + a_yj + a_zk and b = b_xi + b_yj + b_zk, then the Cartesian component vectors of a are a_xi, a_yj and a_zk while the Cartesian scalar components of a are a_x, a_y and a_z. i

Also

The above list describes two fundamental operations on vectors:

• vector addition, in which two vectors are combined in order to produce a third vector, as in Equation 4;
• scaling, in which a vector is combined with a scalar to produce another vector, as in Equation 5. We now consider an operation in which two vectors are multiplied to produce a scalar. This operation is known as the scalar (or dot) product, and we will see later that it allows us to manipulate a number of useful quantities that are easily expressed in vector form. i

2.2 Definition and properties of the scalar product

In physics we use vectors to represent vector quantities such as electric fields, gravitational fields or the velocity of a fluid, each of which has a definite magnitude and direction associated with it at every point. Very often we are interested in the component of a vector in a particular direction, as for example in the case of the velocity of an aircraft as its wheels touch the runway, when the vertical component is critical. Such components can be expressed using the scalar product.

More precisely, θ is defined to be the angle between the positive directions of the two vectors, such that 0 ≤ θ ≤ π.

From Equation 8 it can be seen that the scalar product is always a scalar.

Various properties of the scalar product are obvious from the definition. First, it is clear that, for any vectors a and b, we have |a||b|cosθ = |b||a|cosθ, and therefore the order of the vectors does not affect the scalar product, and we say that the scalar product is commutative:

commutative property   a ⋅ b = b ⋅ a(9)

Consider next the outcome of taking the scalar product of a vector a with itself, i.e. a ⋅ a. From the definition (Equation 8), we have

since cos(0°) = 1. Thus the scalar product of a vector with itself yields the square of the magnitude of the vector. Alternatively, we can express Equation 10 in the form

magnitude   |a| = $\sqrt{{\boldsymbol a}\,{\boldsymbol \cdot}\,{\boldsymbol a}}$ = a(11)

where a is the magnitude or ‘length’ of the vector a. i

If the vectors a and b happen to be perpendicular (i.e. orthogonal) then the angle θ = 90° so that cosθ = 0, in which case

In general, the scalar product is positive if 0 ≤ θ < 90° (because cosθ is positive for this range of values of θ); on the other hand, the scalar product is negative if θ is in the range 90° < θ ≤ 180° (because cosθ is negative for these values of θ).

We require a further property of the scalar product, which can be easily determined from Figure 2. In this figure a, b and r are intended to represent arbitrary vectors, while c = a + b. The vectors a, b and c make angles θ_a, θ_b and θ_c respectively with r; it is clear from the figure that

|c|cos(θ_c) = |a|cos(θ_a) + |b|cos(θ_b)

Multiplying both sides of this equation by |r| we obtain

|r||c|cos(θ_c) = |r||a|cos(θ_a ) + |r||b|cos(θ_b)

which (using Equation 8)

a ⋅ b = |a||b|cosθ(Eqn 8)

may be rewritten in the form

r ⋅ c = r ⋅ (a + b) = r ⋅ a + r ⋅ b(14) i

We say that the scalar product is distributive over vector addition.

2.3 The scalar product in terms of unit vectors

In Equation 8 the scalar product of a and b was defined in terms of the magnitudes (i.e. the lengths) of the vectors and the angle between them – all of which are geometric concepts. The vectors can also be specified algebraically in Cartesian form

a = a_xi + a_yj + a_zk   and  b = b_xi + b_yj + b_zk

(with no reference to geometry), and this raises the possibility that the scalar product can also be determined in terms of the scalar components of a and b. This is indeed the case, as we shall see shortly. First we require some simple but very important results.

Let us now return to our problem of evaluating the scalar product of vectors a and b in terms of the scalar components. The scalar product is given by

a ⋅ b = (a_xi + a_yj + a_zk) ⋅ (b_xi + b_yj + b_zk)

and, from Equation 14 we may expand the brackets to obtain

a ⋅ b = (a_xi) ⋅ (b_xi + b_yj + b_zk) + (a_yj) ⋅ (b_xi + b_yj + b_zk) + (a_zk ) ⋅ (b_xi + b_yj + b_zk)

Expanding the right–hand side, all terms like i ⋅ i are equal to 1, while the terms like i ⋅ j are zero, thus

2.4 Using scalar products to find angles between vectors

We now have two rather different forms for the scalar product, Equations 8 and 15, and we can take advantage of these different expressions for a ⋅ b to find the angle between the vectors a and b. i

Suppose that we wish to find the angle between the two vectors a and b of Question T2, where a = 2i − j + 3k and b = i + 2j + 2k.

First we have

a ⋅ b = |a||b|cosθ(Eqn 8)

The left–hand side can be evaluated, using Equation 15, while on the right–hand side |a| and |b| can be evaluated using Equation 6. In Question T2 we found a ⋅ b = 6, |a| = 14 and |b| = 3, and it follows that

6 = 3 $\sqrt{14\os}$ cosθ, so that cosθ = 2/$\sqrt{14\os}$, and therefore θ ≈ 57.7°.

Notice that we have been able to determine this geometric property of the two vectors from a purely algebraic argument.

In general we have, from Equation 8,

$\cos\theta = \dfrac{{\boldsymbol a}\,{\boldsymbol \cdot}\,{\boldsymbol b}}{\lvert\,{\boldsymbol a}\,\lvert\,\lvert\,{\boldsymbol b}\,\rvert}$

so that:

We may extend this further and express θ in terms of the scalar components of a and b by use of Equations 6 and 15:

|a| = (a_x² + a_y² + a_z²)^1/2(Eqn 6)

a ⋅ b = a_xb_x + a_yb_y + a_zb_z(Eqn 15)

$\theta = \arccos\,\left[\dfrac{a_xb_x+a_yb_y+a_zb_z}{\sqrt{\left(a_x^2+a_y^2+a_z^2\right)}\sqrt{\left(b_x^2+b_y^2+b_z^2\right)}}\right]$(17)

2.5 Scalar products and projections

Let us now explore the relation between the scalar product of two vectors and the projection of one vector on to the other. Indeed, what is meant by the term ‘projection’ in this context? We can obtain some insight into the meaning by considering Figure 3.

Imagine parallel rayrays of light shining vertically downwards on to the x–axis. The quantity |b|cosθ which gives the size of the ‘shadow’ of vector b on the x–axis, is often termed the projection of b on to the x–axis. The projection in this case is the x–component b_x, of b, so that in this context the terms projection and component are synonymous.

We can develop this idea further by considering the situation shown in Figure 4.

Here the light is imagined to be shining in a direction at right angles to vector a, casting a shadow of b on to a. The quantity |b|cosθ is the projection of b on to a. We therefore have

projection of b on to a = |b|cosθ

and multiplying both sides of this equation by |a|

|a| × projection of b on to a = |a||b|cosθ = a ⋅ b

In words: the scalar product of vectors a and b is the magnitude of a times the projection of b on to a. Also, rearranging the equation gives

The projection of b on to a is sometimes known as the scalar component of b in the direction of a, or simply the components of a vectorcomponent of b in the direction of a.

In terms of the scalar components of a and b, Equation 18 becomes

Notice that in Equation 19 any vector in the direction of a will give precisely the same result. Moreover the vector a needs no units (since they cancel) and the dimensions of the projection of b on to a are the same as the dimensions of b.

3 Scalar products in physics

You may think of a vector as a mathematical object that is used to represent a specific physical quantity, such as the velocity of an object at a certain time, but in fact vectors are used in physics to model much more interesting structures. Suppose that we consider the gravitational force on a small object, of mass M say, at some point in the solar system and at some fixed instant of time (so that the planets are in fixed positions). The Sun and each of the planets will contribute to the net gravitational force on the object, but the magnitude and direction of this force will depend on the position of the object. Near to the Earth’s surface the vector that represents the force will point toward the centre of the Earth and will be of magnitude Mg, but close to the Sun it will point towards the centre of the Sun and be of much greater magnitude. i We might imagine such a vector associated with every point in the solar system, and a continuous distribution of vectors of this kind is usually known as a vector field.

Some typical examples of vector fields are indicated in Figure 5. In Figure 5a we see a uniform vector field.

When you first encounter the definition of the scalar product of two vectors a and b, you may well wonder why a ⋅ b is defined to be |a||b|cosθ rather than |a||b|, or |a||b|tanθ, or something else. The choice may appear to be totally arbitrary, but in fact the form |a||b|cosθ happens to be very useful, and it occurs frequently in mathematics and science, and particularly in the context of vector fields. Some of the more important applications of scalar products involve physical concepts beyond the scope of this module, but the following examples may convince you of the utility of the concept.

3.1 Work done by a force

In physics the scalar product may be used to model mathematically the energy transferred (i.e. the work done) by a force acting on an object when the object is displaced. The simplest case involves a constant force F and a linear displacement s, as shown in Figure 6. i

The work done by F may be defined as the component of F in the direction of the displacement multiplied by the magnitude of the displacement. The scalar component of F in the direction of s is |F |cosθ, where θ is the angle between F and s. Therefore the work done by F is given by |F||s|cosθ, which can be written as F ⋅ s. Therefore:

which represents the amount of energy transferred from one form to another. In the SI system of units, work done, or energy, has the units of force multiplied by the units of displacement i.e. newton metre (N m). This unit is also called the joule (J).

There are some important implications of Equation 20 to consider. If the vectors F and s are in the same direction then the angle θ is zero and cosθ = 1, so that the work done by F is given by:

work done = |F||s| = Fs

However, if the vectors F and s are orthogonal, then θ = 90° and cosθ = 0, so that the work done by F is zero. This is a very important result. It means that if an object is undergoing a displacement in a direction perpendicular to that of a force acting on the object, then that force does no work, and no energy is transferred as a result of the action of that force. For example, if a body is travelling in a circular path as a result of a force which always acts towards the centre of the circle (a centripetal force), then this centripetal force does no work on the body.

Imagine a satellite moving around the Earth in a circular path under the influence of gravity alone. In this case the centripetal force is supplied by gravity, and since the direction of motion is perpendicular to this force, the force of gravity does no work – there is no transfer of energy – and the speed of the satellite does not change.

On the other hand, suppose that the path of the satellite is elliptical, so that its velocity may have a component directed towards (or away from the Earth), then the scalar product of the force vector with a small displacement vector is no longer zero – the force of gravity does some work i.e. there is a transfer of energy – and the speed of the satellite changes. i

Finally, let us examine the consequence on the scalar product of taking a value of θ in the range 90° < θ ≤ 180°.

In this range cosθ is negative and so the scalar product, and hence the work done by the force F, is negative. i

3.2 Work done by a constant gravitational force

In the neighbourhood of a particular point on the Earth’s surface we may assume that the force on a mass M due to gravity is of constant magnitude Mg (where g is the acceleration due to gravity, and is approximately (9.8 m s⁻²) acting vertically downwards.

In Figure 7 a heavy block of wood slides down a frictionless plane inclined at an angle α to the horizontal. i

The normal force R, exerted on the block by the plane, does no work during the displacement since R and s are orthogonal. Consider the work done by the weight W as the block undergoes a displacement s.

The angle between W and s is (90° − α), and therefore the work done by W is given by

W ⋅ s = |W||s|cos(90° − α) = |W||s|sinα i

but|s|sinα = (h₁ − h₂) and therefore  W ⋅ s = |W|(h₁ − h₂).

This represents the work done by W as the body is displaced such that it ‘falls’ through a vertical distance (h₁ − h₂). This may be thought of as the loss in gravitational potential energy of the block being compensated by an increase in kinetic energy. If n^{^} is a unit vector in the upward vertical direction then

W = −Mgn^{^}

where M is the mass of the block, so that |W| = Mg and therefore the work done by the force of gravity on the block is Mg(h₁ − h₂).

3.3 Work done by an electric force

We now turn our attention to the case of an electrostatic_forceelectric force, and incorporate the concept of electric field into our discussion. An electric field is said to exist at a given point if an electric charge placed at that point experiences an electric force. We may define the electric field E at any point in terms of the force F experienced by a charge q located at that point:

${\boldsymbol E} = \dfrac1q{\boldsymbol F}$

so that F = qE(21)

If E is the same at all points then the field is described as a uniform field. If the charge undergoes a linear displacement s in a uniform electric field E, then the work done by the electric force is given by:

3.4 Power

If you lift a heavy object then you do work, and in fact you do the same amount of work whether you lift it quickly or slowly. The difference is that you need more power to lift it rapidly than to lift it slowly. The concept of power is closely related to the work done by a force, and power is defined as the rate of doing work (i.e. the rate of transferring energy).

The SI unit of power is the joule per second (J s⁻¹), also called the watt (W). i

In assessing the performance of a machine in doing work, the rate at which work is done is a more important parameter than the total work done by the force provided by the machine.

We consider a situation where an object is being moved by a machine, so that the velocity v of the object is some function of time. Suppose that at some particular time the machine exerts a force F on the object and causes a displacement ∆s in a time interval ∆t, then the work done by F in this time interval is given by

∆W = F ⋅ ∆s

The average power delivered by the force F over this time interval is given by

$\langle P\rangle = \dfrac{\text{work done}}{\text{time taken}} = \dfrac{\Delta W}{\Delta t} = \dfrac{{\boldsymbol F}\,{\boldsymbol \cdot}\,\Delta{\boldsymbol s}}{\Delta t} = {\boldsymbol F}\,{\boldsymbol \cdot}\,\dfrac{\Delta{\boldsymbol s}}{\Delta t}$

As the time interval ∆t tends to zero, the magnitude of the displacement ∆s also tends to zero, and the expression $\dfrac{\Delta{\boldsymbol s}}{\Delta t}$ tends to the velocity v so that

3.5 Flux of a vector field across a surface

Most of the vector quantities that we encounter in physics have magnitudes and directions that have intuitive links to their physical nature. For example, it is natural to choose an arrow to describe a velocity, since we get a sense of an object moving faster in a particular direction as the length of the arrow increases. However, there are some physical quantities for which vectors do not immediately come to mind, but these quantities can be associated with a magnitude and a direction. The area of a surface is such a quantity.

Figure 8 shows the outward pointing unit normal vectors on three faces of a rectangular box, so that, for example, the unit vector n^{^}₁ is associated with face 1 of area bc. The vector S₁ = bcn^{^}₁ encapsulates two important pieces of information about face 1 of the box: its area and the orientation of the surface. Similarly we can construct vectors S₂ = can^{^}₂ and S₃ = abn^{^}₃ to represent the areas of faces 2 and 3 of the box.

For convenience we will now suppose that the unit normal vectors are in the directions of the Cartesian coordinate axes, so that n^{^}₁ = i, n^{^}₂ = j, n^{^}₃ = k. Now imagine that the box is just a cage, and that the edges are made of wire. The cage is now immersed in a stream of water that flows uniformly along the y–axis (i.e. in the direction of n^{^}₂ in Figure 8). How much water flows across the three faces in one second? Clearly the water is travelling parallel to faces 1 and 3, so no water crosses them; the amount of water crossing face 2 will depend on its area and the velocity of the stream. If the velocity of the stream is v(m s⁻¹) and the dimensions of the cage are measured in metres, then the volume of water crossing face 2 in one second will be ac|v|.

Now for a more difficult question. Suppose that the stream is no longer directed along the y–axis, but v is some arbitrary vector, what volume of water crosses face 2 in one second now?

Suppose that v = v_xi + v_yj + v_zk then it is only the component vector v_yj of the velocity that is relevant, since the remaining components are parallel to face 2. The volume of water crossing face 2 in one second is the area of the face (i.e. ac) times the magnitude of this component, which gives ac|v_y|.

In general we might be given a flat surface of any shape, with area A say, perpendicular to some unit vector n^{^}, and we wish to know the volume of liquid that crosses the area in one second. The vector S = An^{^} then contains all the information we need about the area, and it is the component of v in the direction of n^{^} that is relevant.n^{^}

In fact:

$\displaystyle \left\{\text{the volume of fluid that crosses} \atop \text{the surface S in one second}\right\} = \left\{\text{area of the} \atop \text{surface}\right\}\left\{\text{the component of }{\boldsymbol v}\text{ in} \atop \text{the direction of S}\right\} = \lvert\,{\boldsymbol S}\,\rvert \dfrac{{\boldsymbol S}\,{\boldsymbol \cdot}\,{\boldsymbol v}}{\lvert\,{\boldsymbol S}\,\rvert} = {\boldsymbol v}\,{\boldsymbol \cdot}\,{\boldsymbol S}$ i

In the case that we have been considering the vector field v is uniform and the surface S is flat, but more often the vector field varies from point to point and the direction of the normal changes as we move over a curved surface. Since the normal to the surface varies, it is not sensible to refer to the entire surface as a vector, so let us call it S. Although such a general case appears to be very much more difficult, we can in fact approach it in a very similar fashion. We simply divide the surface up into many small regions ∆S that are almost flat, then add up all the contributions v ⋅ ∆S from these small regions. As the size of each region shrinks our approximation improves, until in the limit (of zero area) we obtain an exact value for the volume of liquid that flows across the surface S in one second. Such a limit of a sum is known as a surface integral and is often written in the form $\int_S {\boldsymbol v}\,{\boldsymbol \cdot}\,d{\boldsymbol S}$ although you may also see the notation $\iint_S {\boldsymbol v}\,{\boldsymbol \cdot}\,d{\boldsymbol S}$ i

The surface integral $\int_S {\boldsymbol v}\,{\boldsymbol \cdot}\,d{\boldsymbol S}$ is often known as the flux_of_a_vector_fieldflux of the vector field v over the surface S, and such expressions occur in many branches of physics, but particularly in the study of electricity and magnetism, in addition to the study of fluids. Fortunately we are very rarely required to calculate such expressions; they occur most frequently in the development of the theory.

The flux of an electric field across a spherical surface

We can specify the position of a point P relative to the origin by a vector BI, and then the electric field E at r due to a point charge q placed at the origin is given by

${\boldsymbol E} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q\hat{\boldsymbol r}}{r^2}$(24) i

This means, from Equation 21,

F = qE(Eqn 21)

that a point charge q₁ placed at P will experience a force

${\boldsymbol F} = q_1{\boldsymbol E} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q q_1 \hat{\boldsymbol r}}{{\boldsymbol r}\,{\boldsymbol \cdot}\,{\boldsymbol r}}$(25)

4 Closing items

4.1 Module summary

1

The Section 2scalar product of vectors a and b is defined as:

a ⋅ b = |a||b|cosθ(Eqn 8)

where θ is the angle between the positive directions of a and b.

2

The Section 2scalar product is commutative. That is:

a ⋅ b = b ⋅ a(Eqn 9)

3

The Equation 10scalar product of a vector with itself gives the square of the Equation 11magnitude of the vector, i.e. a ⋅ a = |a|².

The magnitude of the vector a is given by

|a| = (a ⋅ a)^1/2 = a(Eqn 11)

4

The scalar product of two orthogonal vectors is zero i.e. a ⋅ b = 0 when θ = 90°.

5

For values of θ in the range 0 ≤ θ < 90° the scalar product is positive, while for 90° < θ ≤ 180° the scalar

product is negative.

6

The Subsection 2.2scalar product in Cartesian scalar component notation is given by:

a ⋅ b = a_xb_x + a_yb_y + a_zb_z(Eqn 15)

7

The Equation 6magnitude of a vector a is given in scalar component notation by:

|a| = (a_x² + a_y² + a_z²)^1/2(Eqn 6)

8

The Subsection 2.5projection of a vector b on to a vector a is the scalar quantity

$\dfrac{{\boldsymbol a\,{\boldsymbol \cdot}\,}{\boldsymbol b}}{\lvert\,{\boldsymbol a}\,\rvert}$(Eqn 18)

9

The angle θ between two vectors a and b is:

$\theta = \arccos\,\left(\dfrac{{\boldsymbol a}\,{\boldsymbol \cdot}\,{\boldsymbol b}}{\lvert\,{\boldsymbol a}\,\lvert\,\lvert\,{\boldsymbol b}\,\rvert}\right)$

i.e.$\theta = \arccos\,\left[\dfrac{a_xb_x+a_yb_y+a_zb_z}{\sqrt{\left(a_x^2+a_y^2+a_z^2\right)}\sqrt{\left(b_x^2+b_y^2+b_z^2\right)}}\right]$(Eqn 17)

10

The Subsection 3.1work done by a constant force F acting on a body which undergoes a linear displacement s is F ⋅ s. In the general case where F is variable and the path is non-linear, then the work done by F as the body goes from A to B is given by the line integral

$\displaystyle \int_{\rm A}^{\rm B} {\boldsymbol F}\,{\boldsymbol \cdot}\,d{\boldsymbol S}$

11

If a body of weight W moves under gravity so that its vertical height above a given level changes from h₁ to h₂, then the work done by W is |W|(h₁ − h₂ ).

12

In a uniform electric field E, the work done by the electric force on a charge q which undergoes a linear displacement s is

F ⋅ s = qE ⋅ s(Eqn 22)

In the general case, where E is non–uniform and the path is non-linear, the work done by the electric force as the charge moves from A to B is given by the line integral

work done = $\displaystyle q\int_{\rm A}^{\rm B} {\boldsymbol E}\,{\boldsymbol \cdot}\,d{\boldsymbol S}$

13

The electrical potential difference between points A and B in an electric field E is given by

V_B − V_A = $-\displaystyle \int_{\rm A}^{\rm B} {\boldsymbol E}\,{\boldsymbol \cdot}\,d{\boldsymbol S}$

14

The surface integral $\int_S {\boldsymbol E}\,{\boldsymbol \cdot}\,d{\boldsymbol S}$ is known as the Subsection 3.5flux of the vector field E across the surface S.

15

line integralLine integrals and surface integralsurface integrals are merely limits of sums.

4.2 Achievements

Having completed this module, you should be able to:

A1

Define the terms that are emboldened and flagged in the margins of this module.

A2

Define the scalar product of two vectors.

A3

List and explain the properties of the scalar product.

A4

Evaluate the scalar product in terms of the Cartesian scalar components of the vectors.

A5

Express the magnitude of a vector in terms of the Cartesian scalar components of the vector.

A6

Find the projection of one vector on to another.

A7

Use the scalar product to find the angle between two vectors.

A8

Find the work done by a constant force which acts on a body undergoing a linear displacement.

A9

Give a general expression for the work done by a variable force acting on a body which moves along a non–linear path. (You are not expected to be able to calculate line integrals.)

A10

Evaluate the work done by the weight of a body when it moves under gravity from one vertical level to another.

A11

Find the work done by the electric force on a charge which undergoes a linear displacement in a uniform electric field.

A12

Give a general expression for the work done by an electric force acting on a charge which moves in an electric field, and explain how the electric potential difference is related to this expression.

A13

Use the scalar product of force and velocity to find the power associated with a force which acts on a moving body.

A14

Appreciate that the flux of a vector field through a surface is the limit of a sum which is usually written in the form of a surface integral. (You are not expected to be able to calculate surface integrals.)

Study comment You may now wish to take the 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.

4.3 Exit test

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.

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