Question

Two concentric spheres kept in air have radii ‘$R$’ and ‘$r$’. They have similar charge and equal surface charge density ‘$\sigma$ ’. The electric potential at their common centre is
(${\epsilon }_0=$ permittivity of free space)

• A. $\frac{\sigma \left(R+r\right)}{{\epsilon }_0}$
• B. $\frac{\sigma \left(R-r\right)}{{\epsilon }_0}$
• C. $\frac{\sigma \left(R+r\right)}{2{\epsilon }_0}$
• D. $\frac{\sigma \left(R+r\right)}{4{\epsilon }_0}$

Answer 1

Answer: (A)

Potential at the centre due to sphere $1$
$V_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$
and due to sphere $2$ is
$V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
$\therefore$ Electric potential at the common centre
$V=V_1+V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}+\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$
$=\frac{q}{4\pi {\epsilon }_0}\left[\frac{1}{R}+\frac{1}{r}\right]=\frac{q}{4\pi {\epsilon }_0}\left[\frac{R+r}{Rr}\right]$
we have
$\frac{q}{4\pi Rr}=\sigma \therefore v=\frac{\sigma (R+r)}{{\epsilon }_0}$