2.7 Potential Energy of a System of Charges

Consider first the simple case of two charges q1and q2 with position r1 vector and r2 vector 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 vector. There is no external field against which work needs to be done, so work done in bringing q1 from infinity to r1 vector is zero. This charge produces a potential in space given by

V1 = (1/4πε0) (q1/r1p)


where r1P 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 vector is q2 times the potential at r2 vector due to q1:

work done on q2 = (1/4πε0) (q1q2/r12)


where r12 is the distance between points 1 and 2.

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

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 = (1/4πε0) (q1 q2/r12).. .. (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, Equation (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.

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 vector, r2 vector, r3 vector, respectively. To bring q1 first from infinity to r1 vector, no work is required. Next we bring q2 from infinity to r2 vector. As before, work done in this step is

q 2 V 1(r2 vector) = (1/4πε0) (q1 q2/r12) .. .. (2.23)

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

V1, 2 = (1/4πε0) (q1/r1p + q2/r2p) .. .. (2.24)

Work done next in bringing q3 from infinity to the point r3 vector is q3 times V1, 2 at r3 vector

q3V1, 2(r3 vector) = (1/4πε0) (q1q3/r13 + q2q3/r23) .. .. (2.25)

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

U = (1/4πε0) (q1q2/r12 + q1q3/r13 + q2q3/r23) .. .. (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, Equation (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 Equation (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?

Fig.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 × (q/4πε0 d) = −q^2/4πε0 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)

= +q × {+q/ (4πε0 d√2) + (−q) / (4πε0 d)}

= −q^2/4πε0 d (1 − 1/√2)

(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)

= −q {(+q)/4πε0 d + (−q)/4πε0 d √2 + q/4πε0 d}

= (−q^2/4πε0 d) × (2 − 1/√2)

Add the work done in steps (i), (ii), (iii) and (iv). The total work required is

= (−q^2/4πε0 d) × {(0) + (1) + (1 − 1/√2) + (2 − 1/√2)}

= (−q^2/4πε0 d) × (4 − √2)

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 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.