Question

Three concentric metallic spheres A, B and C have radii $a,b$ and $c(a<b<c)$ and surface charge densities on them are $\sigma ,-\sigma$, and $\sigma$ respectively. The values of $V_A$ and $V_B$ will be:

• A. $\frac{\sigma }{{ϵ}_0}(a-b+c),\frac{\sigma }{{ϵ}_0}(\frac{a^2}{b}-b+c)$
• B. $\frac{\sigma }{{ϵ}_0}(a-b+c),\frac{\sigma }{{ϵ}_0}\frac{a^2}{c}$
• C. $\frac{\sigma }{{ϵ}_0}(a+b-c),\frac{\sigma }{{ϵ}_0}(\frac{a^2}{c}-b+c)$
• D. $\frac{\sigma }{{ϵ}_0}(\frac{a^2}{c}-\frac{b^2}{c}+c),\frac{\sigma }{{ϵ}_0}(a-b+c)$

Answer 1

The correct option is A $\frac{\sigma }{{ϵ}_0}(a-b+c),\frac{\sigma }{{ϵ}_0}(\frac{a^2}{b}-b+c)$

The potential due to a metallic shell(of radius R,charge Q) is given by,
$V=\frac{kQ}{r}(r>R)$
$=\frac{1}{\left(4\pi {ϵ}_{0}r\right)}\times \sigma 4\pi R^2=\frac{\sigma R^2}{\left({ϵ}_{0}r\right)}$
$V=\frac{kQ}{R}(r\le R)$
$=\frac{1}{\left(4\pi {ϵ}_{0}R\right)}\times \sigma 4\pi R^2=\frac{\sigma R}{{ϵ}_0}$
Therefore,
potential on A $=\frac{\sigma c}{{ϵ}_0}-\frac{\sigma b}{{ϵ}_0}+\frac{\sigma a}{{ϵ}_0}=\frac{\sigma }{{ϵ}_0}(c-b+a)$
potential on B $=\frac{\sigma c}{{ϵ}_0}-\frac{\sigma b}{{ϵ}_0}+\frac{\sigma a^2}{\left({ϵ}_{0}b\right)}=\frac{\sigma }{{ϵ}_0}(c-b+\frac{a^2}{b})$.