Question

A neutral spherical conductor having cavity A and B of radius '1cm' and '2cm' respectively. Radius of spherical conductor is 10 cm. If cavity 'A' contains charge 2µC and 'B' contains charge of -1µC. Electric potential at the centre of conductor is k x $10^4$ volt. Then what is the value of k.

Answer 1

9

Answer 2

The electric potential at any point inside a conductor in electrostatic equilibrium is constant and equal to the potential on its surface.

The conductor is neutral, and charges $q_A$ and $q_B$ are placed in the cavities. This induces charges $-q_A$ and $-q_B$ on the inner surfaces of the cavities. The remaining charge $q_{outer} = q_A + q_B$ resides on the outer surface of the conductor.

The potential at the center O is equal to the potential on the outer surface.

The outer surface is a sphere of radius R centered at O with charge $q_{outer}$ uniformly distributed on it.

The potential at the center of a uniformly charged spherical shell is $\frac{k q_{outer}}{R}$.

Thus, the potential at O is $V_O = \frac{k q_{outer}}{R}$.

Given $q_A = 2\mu C$, $q_B = -1\mu C$, $R = 10$ cm.

$q_{outer} = q_A + q_B = 2\mu C - 1\mu C = 1\mu C = 1 \times 10^{-6} C$.

$R = 10$ cm $= 0.1$ m.

$k = 9 \times 10^9 Nm^2/C^2$.

$V_O = \frac{(9 \times 10^9) \times (1 \times 10^{-6})}{0.1} = \frac{9 \times 10^3}{0.1} = 9 \times 10^4$ V.

The potential is given as $k \times 10^4$ volt.

Comparing, $k \times 10^4 = 9 \times 10^4$, so $k = 9$.