Question

Two concentric spheres kept in air have radii ‘$R$’ and ‘$r$’. They have similar charge and equal surface charge density ‘$σ$ ’. The electric potential at their common centre is
($\varepsilon_0 =$ permittivity of free space)

• A. $\frac{\sigma\left(R+r\right)}{\varepsilon_{0}}$
• B. $\frac{\sigma\left(R-r\right)}{\varepsilon_{0}}$
• C. $\frac{\sigma\left(R+r\right)}{2\varepsilon_{0}}$
• D. $\frac{\sigma\left(R+r\right)}{4\varepsilon_{0}}$

Answer 1

Answer: (A)

Potential at the centre due to sphere $ 1$
$V_{1}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}$
and due to sphere $2$ is
$V_{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$
$\therefore $ Electric potential at the common centre
$V=V_{1}+V_{2} =\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{R}+\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r} $
$=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{R}+\frac{1}{r}\right]=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{R+r}{R r}\right]$
we have
$\frac{q}{4 \pi R r}=\sigma \therefore v=\frac{\sigma(R+r)}{\varepsilon_{0}}$