Q. A thin spherical conducting shell of radius R has a charge q. Another charge $\mathcal{Q}$is placed at the centre of the shell. The electrostatic potential at a point P at a distance R/2 from the centre of the shell is
A) $\frac{2\mathcal{Q}}{4\pi\varepsilon_{0}R}$
B) $\frac{2\mathcal{Q}}{4\pi\varepsilon_{0}R} - \frac{2q}{4\pi\varepsilon_{0}R}$
C) $\frac{2\mathcal{Q}}{4\pi\varepsilon_{0}R} + \frac{q}{4\pi\varepsilon_{0}R}$
D) $\frac{(q + \mathcal{Q)}}{4\pi\varepsilon_{0}}\frac{2}{R}$
Explanation
(c) At $P$ due to shell, potential
$V_{1} = \frac{q}{4\pi\varepsilon_{0}R}$
at $P$ due to $\mathcal{Q}$, potential
$V_{2} = \frac{\mathcal{Q}}{4\pi\varepsilon_{0}\frac{R}{2}} = \frac{2\mathcal{Q}}{4\pi\varepsilon_{0}R}$
$\therefore$Net potential at P
$V = V_{1} + V_{2} = \frac{q}{4\pi\varepsilon_{0}R} + \frac{2\mathcal{Q}}{4\pi\varepsilon_{0}R}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.