Question

A hollow conducting sphere has inner radius $5\text{cm}$ and outer radius $8\text{cm}$, then find maximum possible potential of conductor, if breakdown electric field of air is $3\times {10}^6\text{V/m}$

Answer 1

Let the charge given to the conductor is $Q.$
All the charge will accumulate at the outer surface of the thick sphere.
Electric field near the surface is $E=\frac{KQ}{R^2}$
For maximum charge, it should be equal to breakdown electric field.
$\Rightarrow \frac{KQ}{(8\times {10}^{-2}{)}^2}=3\times {10}^2$
Now , maximum possible potential,
$\Rightarrow V=ER=3\times {10}^6\times 8\times {10}^{-2}$
$\Rightarrow V=3\times {10}^6\times 8\times {10}^{-2}$
$\therefore V=24\times {10}^4=240\text{kV}$