First slide

Electrostatic potential

Question

Two spheres A and B of radii 17 cm each and having charges of 1 and 2 coulomb respectively are separated by a distance of 80 cm. The electric field at a point on the Iine joining the centres of two spheres is approximately zero at some distance from the sphere A. The electric potential at this point is

Moderate

A

$6 \cdot 56 \times 10^{10} V$
B

$8 \cdot 12 \times 10^{7} V$
C

$2 \cdot 03 \times 10^{9} V$
D

$1 .2 \times 10^{11} V$

Solution

Here,
$\begin{array}{l} |E_{1}| = |E_{2}| \\ \frac{1}{4 {πε}_{0}} \times \frac{1}{x^{2}} = \frac{1}{4 {πε}_{0}} \times \frac{2}{(0 \cdot 8 - x)^{2}} \end{array}$
or $\frac{1}{x^{2}} = \frac{2}{(0 \cdot 8 - x)^{2}}$
Solving, we get
$x \approx 0 .33 m = 33 cm$
Now the potential at this point is
$\begin{array}{l} V_{p} = \frac{1}{4 {πε}_{0}} \times \frac{q_{1}}{x} + \frac{1}{4 {πε}_{0}} \times \frac{q_{2}}{(0 \cdot 8 - x)} \\ = \frac{1}{4 {πε}_{0}} [\frac{q_{1}}{x} + \frac{q_{2}}{(0 \cdot 8 - x)}] \\ = 9 \times 10^{9} [\frac{1}{0 \cdot 33} + \frac{2}{0 \cdot 47}] = 6 \cdot 56 \times 10^{10} V . \end{array}$

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