Question

A charge $Q$ is distributed over three concentric spherical shells of radii $a,b,c(a<b<c)$ such that their surface charge densities are equal to one another. The total potential at a point at distance $r$ from their common centre, where $r<a$, would be

• A. $\frac{Q}{12\pi {ϵ}_0}\frac{ab+bc+ca}{abc}$
• B. $\frac{Q(a^2+b^2+c^2)}{4\pi {ϵ}_0(a^3+b^3+c^3)}$
• C. $\frac{Q(a+b+c)}{4\pi {ϵ}_0(a^2+b^2+c^2)}$
• D. $\frac{Q}{4\pi {ϵ}_0(a+b+c)}$

Answer 1

The correct option is C $\frac{Q(a+b+c)}{4\pi {ϵ}_0(a^2+b^2+c^2)}$

Given:
Surface charge densities are equal to one another
i.e., ${\sigma }_a={\sigma }_b={\sigma }_c$
And, $a<b<c$


Potential at point $P,$

$V=\frac{k{Q}_a}{a}+\frac{k{Q}_b}{b}+\frac{k{Q}_c}{c}$

Since ${\sigma }_a={\sigma }_b={\sigma }_c$

$\therefore Q_a:Q_b:Q_c=a^2:b^2:c^2$

$\therefore Q_a=\left[\frac{a^2}{a^2+b^2+c^2}\right]Q$

$Q_b=\left[\frac{b^2}{a^2+b^2+c^2}\right]Q,$

$Q_c=\left[\frac{c^2}{a^2+b^2+c^2}\right]Q$ and

Now, putting all the values,
we get potential

$\therefore V=\frac{Q}{4\pi {ϵ}_0}\left[\frac{(a+b+c)}{a^2+b^2+c^2}\right]$

Hence, option $\left(D\right)$ is correct.