Chapter Two

Electrostatic Potential and Capacitance

2.1 Introduction

 In Chapters 6 and 8 (Class XI), the notion of potential energy was introduced. When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, that work gets stored as potential energy of the body. When the external force is removed, the body moves, gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic and potential energies is thus conserved. Forces of this kind are called conservative forces. Spring force and gravitational force are examples of conservative forces.

Coulomb force between two (stationary) charges is also a conservative force. This is not surprising, since both have inverse-square dependence on distance and differ mainly in the proportionality constants – the masses in the gravitational law are replaced by charges in Coulomb’s law. Thus, like the potential energy of a mass in a gravitational field, we can define electrostatic potential energy of a charge in an electrostatic field.

Consider an electrostatic field E due to some charge configuration. First, for simplicity, consider the field E due to a charge Q placed at the origin. Now, imagine that we bring a test charge q from a point R to a point P against the repulsive force on it due to the charge Q. With reference to Fig. 2.1, this will happen if Q and q are both positive or both negative. For definiteness, let us take Q, q > 0.

Two remarks may be made here. First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q from R to P, we apply an external force Fext just enough to counter the repulsive electric force FE (i.e, Fext= –FE). This means there is no net force on or acceleration of the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. In this situation, work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. If the external force is removed on reaching P, the electric force will take the charge away from Q – the stored energy (potential energy) at P is used to provide kinetic energy to the charge q in such a way that the sum of the kinetic and potential energies is conserved.

Thus, work done by external forces in moving a charge q from R to P is

${\mathrm{W}}_{\mathrm{RP}}={\int }_{\mathrm{R}}^{\mathrm{P}}{\mathrm{F}}_{\mathrm{ext}}.\mathrm{dr}$$=-{\int }_{\mathrm{R}}^{\mathrm{P}}{\mathrm{F}}_{\mathrm{E}}.\mathrm{dr}$

(2.1)

This work done is against electrostatic repulsive force and gets stored as potential energy.

At every point in electric field, a particle with charge q possesses a certain electrostatic potential energy, this work done increases its potential energy by an amount equal to potential energy difference between points R and P.

Thus, potential energy difference

(2.2)

(Note here that this displacement is in an opposite sense to the electric force and hence work done by electric field is negative, i.e., –W_RP.)

Therefore, we can define electric potential energy difference between two points as the work required to be done by an external force in moving (without accelerating) charge q from one point to another for electric field of any arbitrary charge configuration.

Two important comments may be made at this stage:

(i) The right side of Eq. (2.2) depends only on the initial and final positions of the charge. It means that the work done by an electrostatic field in moving a charge from one point to another depends only on the initial and the final points and is independent of the path taken to go from one point to the other. This is the fundamental characteristic of a conservative force. The concept of the potential energy would not be meaningful if the work depended on the path. The path-independence of work done by an electrostatic field can be proved using the Coulomb’s law. We omit this proof here.

(ii) Equation (2.2) defines potential energy difference in terms of the physically meaningful quantity work. Clearly, potential energy so defined is undetermined to within an additive constant.What this means is that the actual value of potential energy is not physically significant; it is only the difference of potential energy that is significant. We can always add an arbitrary constant α to potential energy at every point, since this will not change the potential energy difference:

Put it differently, there is a freedom in choosing the point where potential energy is zero. A convenient choice is to have electrostatic potential energy zero at infinity. With this choice, if we take the point R at infinity, we get from Eq. (2.2)

(2.3)

Since the point P is arbitrary, Eq. (2.3) provides us with a definition of potential energy of a charge q at any point. Potential energy of charge q at a point (in the presence of field due to any charge configuration) is the work done by the external force (equal and opposite to the electric force) in bringing the charge q from infinity to that point.

2.2 Electrostatic Potential

 Consider any general static charge configuration. We define potential energy of a test charge q in terms of the work done on the charge q. This work is obviously proportional to q, since the force at any point is qE, where E is the electric field at that point due to the given charge configuration. It is, therefore, convenient to divide the work by the amount of charge q, so that the resulting quantity is independent of q. In other words, work done per unit test charge is characteristic of the electric field associated with the charge configuration. This leads to the idea of electrostatic potential V due to a given charge configuration. From Eq. (2.1), we get:

Work done by external force in bringing a unit positive charge from point R to P

= $V_P-V_R\left(=\frac{U_P-U_R}{q}\right)$(2.4)

where V_P and V_R are the electrostatic potentials at P and R, respectively. Note, as before, that it is not the actual value of potential but the potential difference that is physically significant. If, as before, we choose the potential to be zero at infinity, Eq. (2.4) implies:

Work done by an external force in bringing a unit positive charge from infinity to a point = electrostatic potential (V) at that point.

In other words, the electrostatic potential (V) at any point in a region with electrostatic field is the work done in bringing a unit positive charge (without acceleration) from infinity to that point.

Figure 2.2 Work done on a test charge q by the electrostatic field due to any given charge configuration is independent of the path, and depends only on its initial and final positions.

The qualifying remarks made earlier regarding potential energy also apply to the definition of potential. To obtain the work done per unit test charge, we should take an infinitesimal test charge δq, obtain the work done δW in bringing it from infinity to the point and determine the ratio
δW/δq. Also, the external force at every point of the path is to be equal and opposite to the electrostatic force on the test charge at that point.

2.3 Potential due to a Point Charge

 Consider a point charge Q at the origin (Fig. 2.3). For definiteness, take Q to be positive. We wish to determine the potential at any point P with position vector r from the origin. For that, we must calculate the work done in bringing a unit positive test charge from infinity to the point P. For Q > 0, the work done against the repulsive force on the test charge is positive. Since work done is independent of the path, we choose a convenient path – along the radial direction from infinity to the point P.

Figure 2.3 Work done in bringing a unit positive test charge from infinity to the point P, against the repulsive force of charge Q (Q > 0), is the potential at P due to the charge Q.

At some intermediate point P′ on the path, the electrostatic force on a unit positive charge is (2.5)

where is the unit vector along OP′. Work done against this force from r′ to r′ + ∆r′ is

(2.6)

The negative sign appears because for ∆r′ < 0, ∆W is positive. Total work done (W) by the external force is obtained by integrating Eq. (2.6) from r′ = ∞ to r′ = r,

$W=-{\int }_{\infty }^r\frac{Q}{4\pi {\epsilon }_0{r}^{\text{'}2}}dr\text{'}={\left[\frac{Q}{4\pi {\epsilon }_{0}r\text{'}}\right]}_{\infty }^r=\frac{Q}{4\pi {\epsilon }_{0}r}$(2.7)

This, by definition is the potential at P due to the charge Q

(2.8)

Equation (2.8) is true for any sign of the charge Q, though we considered Q > 0 in its derivation. For Q < 0, V < 0, i.e., work done (by the external force) per unit positive test charge in bringing it from infinity to the point is negative. This is equivalent to saying that work done by the electrostatic force in bringing the unit positive charge form infinity to the point P is positive. [This is as it should be, since for Q < 0, the force on a unit positive test charge is attractive, so that the electrostatic force and the displacement (from infinity to P) are in the same direction.] Finally, we note that Eq. (2.8) is consistent with the choice that potential at infinity be zero.

Figure 2.4 Variation of potential V with r [in units of (Q/4πε₀) m^-1] (blue curve) and field with r [in units of (Q/4πε₀) m^-2] (black curve) for a point charge Q.

Figure (2.4) shows how the electrostatic potential (∝ 1/r) and the electrostatic field (∝ 1/r² ) varies with r.

2.4 Potential due to an Electric Dipole

 As we learnt in the last chapter, an electric dipole consists of two charges q and –q separated by a (small) distance 2a. Its total charge is zero. It is characterised by a dipole moment vector p whose magnitude is q × 2a and which points in the direction from –q to q (Fig. 2.5). We also saw that the electric field of a dipole at a point with position vector r depends not just on the magnitude r, but also on the angle between r and p. Further, the field falls off, at large distance, not as 1/r² (typical of field due to a single charge) but as 1/r³. We, now, determine the electric potential due to a dipole and contrast it with the potential due to a single charge.

As before, we take the origin at the centre of the dipole. Now we know that the electric field obeys the superposition principle. Since potential is related to the work done by the field, electrostatic potential also follows the superposition principle. Thus, the potential due to the dipole is the sum of potentials due to the charges q and –q

$V=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q}{r_1}-\frac{1}{r_2}\right)$(2.9)

Figure 2.5 Quantities involved in the calculation of potential due to a dipole.

where r₁ and r₂ are the distances of the point P from q and –q, respectively.

Now, by geometry,

${\mathrm{r}}_1^2={\mathrm{r}}^2+{\mathrm{a}}^2-2\mathrm{ar} \mathrm{cos\theta }$

${\mathrm{r}}_2^2={\mathrm{r}}^2+{\mathrm{a}}^2+2\mathrm{ar} \mathrm{cos\theta }$(2.10)

We take r much greater than a $\left(\mathrm{r}>>\mathrm{a}\right)$ and retain terms only upto the first order in a/r

$r_1^2\cong r^2\left(1-\frac{2a \cos \theta }{r}\right)$(2.11)

Similarly,

$r_2^2\cong r^2\left(1+\frac{2a \cos \theta }{r}\right)$(2.12)

Using the Binomial theorem and retaining terms upto the first order in a/r ; we obtain,

$\frac{1}{r_1}\cong \frac{1}{r}{\left(1-\frac{2a \cos \theta }{r}\right)}^{-1/2}\cong \frac{1}{r}\left(1+\frac{a}{r}\cos \theta \right)$[2.13(a)]

$\frac{1}{r_2}\cong \frac{1}{r}{\left(1+\frac{2a \cos \theta }{r}\right)}^{-1/2}\cong \frac{1}{r}\left(1-\frac{a}{r}\cos \theta \right)$[2.13(b)]

Using Eqs. (2.9) and (2.13) and p = 2qa, we get

$V=\frac{q}{4\pi {\epsilon }_0}\frac{2a\cos \theta }{r^2}=\frac{p\cos \theta }{4\pi {\epsilon }_0{r}^2}$(2.14)

Now, p cos θ = $\mathrm{p}.\hat{\mathrm{r}}$

where is the unit vector along the position vector OP.

The electric potential of a dipole is then given by

$V=\frac{1}{4\pi {\epsilon }_0}\frac{p.\hat{r}}{r^2}; \left(r>>a\right)$; (r >> a) (2.15)

Equation (2.15) is, as indicated, approximately true only for distances large compared to the size of the dipole, so that higher order terms in a/r are negligible. For a point dipole p at the origin, Eq. (2.15) is, however, exact.

From Eq. (2.15), potential on the dipole axis (θ = 0, π ) is given by

(2.16)

(Positive sign for θ = 0, negative sign for θ = π.) The potential in the equatorial plane (θ = π/2) is zero.

 The important contrasting features of electric potential of a dipole from that due to a single charge are clear from Eqs. (2.8) and (2.15):

(i) The potential due to a dipole depends not just on r but also on the angle between the position vector r and the dipole moment vector p. (It is, however, axially symmetric about p. That is, if you rotate the position vector r about p, keeping θ fixed, the points corresponding to P on the cone so generated will have the same potential as at P.)

(ii) The electric dipole potential falls off, at large distance, as 1/r2, not as 1/r, characteristic of the potential due to a single charge. (You can refer to the Fig. 2.5 for graphs of 1/r² versus r and 1/r versus r, drawn there in another context.)

2.5 Potential due to a System of Charges

 Consider a system of charges q₁, q₂,…, q_n with position vectors r₁, r₂,…, r_n relative to some origin (Fig. 2.6). The potential V₁ at P due to the charge q₁ is

where r_1P is the distance between q₁  and P.

Similarly, the potential V₂ at P due to q₂ and V₃ due to q₃ are given by

,

where r_2P and r_3P are the distances of P from charges q₂ and q₃, respectively; and so on for the potential due to other charges. By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges

V = V₁ + V₂ + ... + V_n (2.17)

$=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1}{r_{1P}}+\frac{q_2}{r_{2P}}+......+\frac{q_n}{r_{nP}}\right)$(2.18)

 

Figure 2.6 Potential at a point due to a system of charges is the sum of potentials due to individual charges.

If we have a continuous charge distribution characterised by a charge density ρ(r), we divide it, as before, into small volume elements each of size ∆v and carrying a charge ρ∆v. We then determine the potential due to each volume element and sum (strictly speaking , integrate) over all such contributions, and thus determine the potential due to the entire distribution.

We have seen in Chapter 1 that for a uniformly charged spherical shell, the electric field outside the shell is as if the entire charge is concentrated at the centre. Thus, the potential outside the shell is given by

[2.19(a)]

where q is the total charge on the shell and R its radius. The electric field inside the shell is zero. This implies (Section 2.6) that potential is constant inside the shell (as no work is done in moving a charge inside the shell), and, therefore, equals its value at the surface, which is

[2.19(b)]

Example 2.2 Two charges 3 × 10^–8 C and –2 × 10^–8 C are located 15 cm apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

Solution

Let us take the origin O at the location of the positive charge. The line joining the two charges is taken to be the x-axis; the negative charge is taken to be on the right side of the origin (Fig. 2.7).

Figure 2.7

Let P be the required point on the x-axis where the potential is zero. If x is the x-coordinate of P, obviouslyx must be positive. (There is no possibility of potentials due to the two charges adding up to zero for 
x < 0.) If x lies between O and A, we have

$\frac{1}{4\pi {\epsilon }_0}\left[\frac{3\times {10}^{-8}}{x\times {10}^{-2}}-\frac{2\times {10}^{-8}}{\left(15-x\right)\times {10}^{-2}}\right]=0$

where x is in cm. That is,

which gives x = 9 cm.

If x lies on the extended line OA, the required condition is

which gives

x = 45 cm

 

Thus, electric potential is zero at 9 cm and 45 cm away from the positive charge on the side of the negative charge. Note that the formula for potential used in the calculation required choosing potential to be zero at infinity.

Example 2.3 Figures 2.8 (a) and (b) show the field lines of a positive and negative point charge respectively.

Figure 2.8

(a) Give the signs of the potential difference V_P – V_Q; V_B – V_A.

(b) Give the sign of the potential energy difference of a small negative charge between the points Q and P; A and B.

(c) Give the sign of the work done by the field in moving a small positive charge from Q to P.

(d) Give the sign of the work done by the external agency in moving a small negative charge from B to A.

(e) Does the kinetic energy of a small negative charge increase or decrease in going from B to A?

Solution

(a) As, V_P > V_Q. Thus, (V_P – V_Q) is positive. Also V_B is less negative than V_A . Thus, V_B > V_A or (V_B – V_A) is positive.

(b) A small negative charge will be attracted towards positive charge. The negative charge moves from higher potential energy to lower potential energy. Therefore the sign of potential energy difference of a small negative charge between Q and P is positive.

Similarly, (P.E.)_A > (P.E.)_B and hence sign of potential energy differences is positive.

(c) In moving a small positive charge from Q to P, work has to be done by an external agency against the electric field. Therefore, work done by the field is negative.

(d) In moving a small negative charge from B to A work has to be done by the external agency. It is positive.

 

(e) Due to force of repulsion on the negative charge, velocity decreases and hence the kinetic energy decreases in going from B to A.

2.6.1 Relation between field and potential

 Consider two closely spaced equipotential surfaces A and B (Fig. 2.12) with potential values V and V + δ V, where δ V is the change in V in the direction of the electric field E. Let P be a point on the surface B. δ l is the perpendicular distance of the surface A from P. Imagine that a unit positive charge is moved along this perpendicular from the surface B to surface A against the electric field. The work done in this process is |E|δ l.

This work equals the potential difference VA–VB.

Thus,

|E|δ l = V – (V + δV)= – δV

i.e., $\left|\mathit{E}\right|=\frac{\delta V}{\delta l}$
(2.20)

Since δV is negative, δV = – |δV|. we can rewrite Eq (2.20) as

Figure 2.12 From the  potential to the field.

(2.21)

We thus arrive at two important conclusions concerning the relation between electric field and potential:

(i) Electric field is in the direction in which the potential decreases steepest.

(ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.

2.6 Equipotential Surfaces

 An equipotential surface is a surface with a constant value of potential at all points on the surface. For a single charge q, the potential is given by Eq. (2.8):

This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge.

Now the electric field lines for a single charge q are radial lines starting from or ending at the charge, depending on whether q is positive or negative. Clearly, the electric field at every point is normal to the equipotential surface passing through that point. This is true in general: for any charge configuration, equipotential surface through a point is normal to the electric field at that point. The proof of this statement is simple.

 

Figure 2.9 For a single charge q 

(a) equipotential surfaces are spherical surfaces centred at the charge, and 

(b) electric field lines are radial, starting from the charge if q > 0.

 

If the field were not normal to the equipotential surface, it would have non-zero component along the surface. To move a unit test charge against the direction of the component of the field, work would have to be done. But this is in contradiction to the definition of an equipotential surface: there is no potential difference between any two points on the surface and no work is required to move a test charge on the surface. The electric field must, therefore, be normal to the equipotential surface at every point. Equipotential surfaces offer an alternative visual picture in addition to the picture of electric field lines around a charge configuration.

For a uniform electric field E, say, along the x
-axis, the equipotential surfaces are planes normal to the x-axis, i.e., planes parallel to the y-z plane (Fig. 2.10). Equipotential surfaces for (a) a dipole and (b) two identical positive charges are shown in Fig. 2.11.

Figure 2.11 Some equipotential surfaces for (a) a dipole, (b) two identical positive charges.

2.7 Potential Energy of a System of Charges

  Consider first the simple case of two charges q₁and q₂ with position vector r₁ and r₂ relative to some origin. Let us calculate the work done (externally) in building up this configuration. This means that we consider the charges q₁ and q₂ initially at infinity and determine the work done by an external agency to bring the charges to the given locations. Suppose, first the charge q₁ is brought from infinity to the point  r₁. There is no external field against which work needs to be done, so work done in bringing q₁ from infinity to r₁ is zero. This charge produces a potential in space given by

$V_1=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1}{r_{1P}}$

where r_1P is the distance of a point P in space from the location of q₁. From the definition of potential, work done in bringing charge q₂ from infinity to the point r₂ is q₂ times the potential at r₂ due to q₁:

work done on q₂ =$\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_2}{r_{12}}$

where r₁₂ is the distance between points 1 and 2.

Since electrostatic force is conservative, this work gets stored in the form of potential energy of the system. Thus, the potential energy of a system of two charges q₁ and q₂ is

$U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q_1{q}_2}{r_{12}}$(2.22)

Obviously, if q₂ was brought first to its present location and q₁ brought later, the potential energy U would be the same. More generally, the potential energy expression, Eq. (2.22), is unaltered whatever way the charges are brought to the specified locations, because of path-independence of work for electrostatic force.

Equation (2.22) is true for any sign of q₁and q₂. If q₁q₂ > 0, potential energy is positive. This is as expected, since for like charges (q₁q₂ > 0), electrostatic force is repulsive and a positive amount of work is needed to be done against this force to bring the charges from infinity to a finite distance apart. For unlike charges (q₁ q₂ < 0), the electrostatic force is attractive. In that case, a positive amount of work is needed against this force to take the charges from the given location to infinity. In other words, a negative amount of work is needed for the reverse path (from infinity to the present locations), so the potential energy is negative.

Figure 2.13 Potential energy of a system of charges q₁ and q₂ is directly proportional to the product of charges and inversely to the distance between them.

 Equation (2.22) is easily generalised for a system of any number of point charges. Let us calculate the potential energy of a system of three charges q₁, q₂ and q3 located at r₁, r₂, r₃, respectively. To bring q₁ first from infinity to r₁, no work is required. Next we bring q₂ from infinity to r₂. As before, work done in this step is

(2.23)

The charges q₁ and q₂ produce a potential, which at any point P is given by

$V_{1,2}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q_1}{r_{1P}}+\frac{q_2}{r_{2P}}\right)$(2.24)

Work done next in bringing q₃ from infinity to the point r₃ is q₃ times V_1,2 at r₃

$q_3{V}_{1,2}\left(r_3\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q_1{q}_3}{r_{13}}+\frac{q_2{q}_3}{r_{23}}\right)$(2.25)

The total work done in assembling the charges at the given locations is obtained by adding the work done in different steps [Eq. (2.23) and Eq. (2.25)],

$U=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left(\frac{q_1{q}_2}{r_{12}}+\frac{q_1{q}_2}{r_{13}}+\frac{q_2{q}_3}{r_{23}}\right)$(2.26)

Again, because of the conservative nature of the electrostatic force (or equivalently, the path independence of work done), the final expression for U, Eq. (2.26), is independent of the manner in which the configuration is assembled. The potential energy is characteristic of the present state of configuration, and not the way the state is achieved.

 

Figure 2.14 Potential energy of a system of three charges is given by Eq. (2.26), with the notation given 

in the figure.

 

Example 2.4 Four charges are arranged at the corners of a square ABCD of side d, as shown in Fig. 2.15.(a) Find the work required to put together this arrangement. (b) A charge q₀ is brought to the centre E of the square, the four charges being held fixed at its corners. How much extra work is needed to do this?

Figure 2.15

Solution

(a) Since the work done depends on the final arrangement of the charges, and not on how they are put together, we calculate work needed for one way of putting the charges at A, B, C and D. Suppose, first the charge +q is brought to A, and then the charges –q, +q, and –q are brought to B, C and D, respectively. The total work needed can be calculated in steps:

(i) Work needed to bring charge +q to A when no charge is present elsewhere: this is zero.

(ii) Work needed to bring –q to B when +q is at A. This is given by (charge at B) × (electrostatic potential at B due to charge +q at A)

$=+q\times \left(\frac{q}{4{\mathrm{\pi \epsilon }}_{0}d}\right)=-\frac{q^2}{4{\mathrm{\pi \epsilon }}_{0}d}$

(iii) Work needed to bring charge +q to C when +q is at A and –q is at B. This is given by (charge at C) × (potential at C due to charges at A and B)

$$\begin{gathered} =+q\left(\frac{+q}{4{\mathrm{\pi \epsilon }}_{0}d\sqrt{2}}+\frac{-q}{4{\mathrm{\pi \epsilon }}_{0}d}\right) \\ =\frac{q^2}{4{\mathrm{\pi \epsilon }}_{0}d}\left(1-\frac{1}{\sqrt{2}}\right) \end{gathered}$$

(iv) Work needed to bring –q to D when +q at A,–q at B, and +q at C. This is given by (charge at D) × (potential at D due to charges at A, B and C)

$$\begin{gathered} =-q\left(\frac{+q}{4{\mathrm{\pi \epsilon }}_{0}d}+\frac{-q}{4{\mathrm{\pi \epsilon }}_{0}d\sqrt{2}}+\frac{q}{4{\mathrm{\pi \epsilon }}_{0}d}\right) \\ =\frac{-q^2}{4{\mathrm{\pi \epsilon }}_{0}d}\left(2-\frac{1}{\sqrt{2}}\right) \end{gathered}$$Add the work done in steps (i), (ii), (iii) and (iv). The total work 
required is

$$\begin{gathered} =\frac{-q^2}{4{\mathrm{\pi \epsilon }}_{0}d}\left\{\left(0\right)+\left(q\right)+\left(1-\frac{1}{\sqrt{2}}\right)+\left(2-\frac{1}{\sqrt{2}}\right)\right\} \\ =\frac{-q^2}{4{\mathrm{\pi \epsilon }}_{0}d}\left(4-\sqrt{2}\right) \end{gathered}$$

The work done depends only on the arrangement of the charges, and not how they are assembled. By definition, this is the total electrostatic energy of the charges.

(Students may try calculating same work/energy by taking charges in any other order they desire and convince themselves that the energy will remain the same.)

 

(b) The extra work necessary to bring a charge q0 to the point E when the four charges are at A, B, C and D is q₀ × (electrostatic potential at E due to the charges at A, B, C and D). The electrostatic potential at E is clearly zero since potential due to A and C is cancelled by that due to B and D. Hence, no work is required to bring any charge to point E.