2.5 POTENTIAL DUE TO A SYSTEM OF CHARGES

2.6

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

Consider a system of charges $q_{1}$ , $q_{2}$ ,…, $q_{n}$ with position vectors $r_{1}$ , $r_{2}$ ,…, $r_{n}$ relative to some origin (Fig. 2.6). The potential $V_{1}$ at $P$ due to the charge $q_{1}$ is $V_{1} = \frac{1}{4 π ε_{0}} \frac{q_{1}}{r_{1 P}}$ where $r_{1 P}$ is the distance between $q_{1}$ and $P$ .

Similarly, the potential $V_{2}$ at $P$ due to $q_{2}$ and $V_{3}$ due to $q_{3}$ are given by $V_{2} = \frac{1}{4 π ε_{0}} \frac{q_{2}}{r_{2 P}}, V_{3} = \frac{1}{4 π ε_{0}} \frac{q_{3}}{r_{3 P}}$

where $r_{2 P}$ and $r_{3 P}$ are the distances of $P$ from charges $q_{2}$ and $q_{3}$ , 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

$V = V_{1} + V_{2} + . . . + V_{n}$ (2.17)

$V = \frac{1}{4 π ε_{0}} (\frac{q_{1}}{r_{1 P}} + \frac{q_{2}}{r_{2 P}} + \dots \dots + \frac{q_{n}}{r_{n P}})$ (2.18)

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 $Δ v$ and carrying a charge $ρ Δ v$ . 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

$V = \frac{1}{4 π ε_{0}} \frac{q}{r} (r \geq R)$ (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

$V = \frac{1}{4 π ε_{0}} \frac{q}{R}$ (2.19(b))

Example 2.2

Two charges $3 * 10^{- 8} C$ and $- 2 * 10^{- 8} C$ are located $15 c m$ apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

2.7

VIEW SOLUTION

Let $P$ be the required point on the $x$ -axis where the potential is zero. If $x$ is the $x$ -coordinate of $P$ , obviously $x$ 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 $\frac{1}{4 π ε_{0}} [\frac{3 \times 10^{- 8}}{x \times 10^{- 2}} - \frac{2 \times 10^{- 8}}{(15 - x) \times 10^{- 2}}] = 0$ where $x$ is in cm. That is, $\frac{3}{x} - \frac{2}{15 - x} = 0$ which gives $x$ = 9 cm.

If $x$ lies on the extended line $O A$ , the required condition is $\frac{3}{x} - \frac{2}{x - 15} = 0$ which gives $x = 45 c m$ 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.

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 $A$ ?

VIEW SOLUTION

(a) As $V \propto \frac{1}{r}$ , $V_{P} > V_{Q}$ . 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 $A$ .