Question

A particle carrying charge +q is placed at the center of a thick-walled conducting shell that has inner radius R and outer radius 2R and carries charge −2q. A thin-walled conducting shell of radius 5R carries charge +2q and is concentric with the thick-walled shell. Define V = 0 at infinity. Calculate all distances from the particle at which the electrostatic potential is zero.

Answer 1

Answer:

10R/11 and 5R/2

Explanation:

The radius of the conducting shell = R,

Electrostatic potential inside the shell (r<R) = kq/R

Electrostatic potential outside the shell (r>R) = kq/r

If x is the point of zero potential

Electrostatic potential for inner shell, $V_{1} = \frac{kq}{X - R}$

Electrostatic potential for outer shell, $V_{2} = \frac{-2kq}{X - 2R}$

Electrostatic potential for the thin walled shell, $V_{3} = \frac{2kq}{X - 5R}$

$V_{1} + V_{2} + V_{3} = 0$

$\frac{kq}{X-R} - \frac{2kq}{X-2R} + \frac{2kq}{X-5R} = 0$

$\frac{1}{X-R} - \frac{2}{X-2R} + \frac{2}{X-5R} = 0\\(X-R) - 2(X-R)(X-5R)+2(X-R)(X-2R) = 0$

The values of X=r that satisfy the above equation are 10R/11 and 5R/2