Question

Three concentric spherical conductors are arranged as shown in the figure. The potential at point $\text{P}$ will be:

• A. $\frac{1}{4\pi {ϵ}_0}[\frac{Q_1}{r}+\frac{Q_2}{r}+\frac{Q_3}{r}]$
• B. $\frac{1}{4\pi {ϵ}_0}[\frac{Q_1+Q_2}{r}+\frac{Q_3}{c}]$
• C. $\frac{1}{4\pi {ϵ}_0}[\frac{Q_1}{a}+\frac{Q_2}{b}+\frac{Q_3}{c}]$
• D. $\frac{1}{4\pi {ϵ}_{0}c}\times [Q_1+Q_2+Q_3]$

Answer 1

The correct option is B $\frac{1}{4\pi {ϵ}_0}[\frac{Q_1+Q_2}{r}+\frac{Q_3}{c}]$


Electric potential at point $\text{P}$ can be calculated by using,

$V_P=V_A+V_B+V_C...\left(1\right)$

Where,
Potential at $\text{P}$ due to spherical conductor $\text{A}$,
$V_A=\frac{1}{4\pi {ϵ}_0}\frac{Q_1}{r}$

Potential at $\text{P}$ due to spherical conductor $\text{B}$,
$V_B=\frac{1}{4\pi {ϵ}_0}\frac{Q_2}{r}$

Potential at $\text{P}$ due to spherical conductor $\text{C}$,
$V_C=\frac{1}{4\pi {ϵ}_0}\frac{Q_3}{c}$

Substituting the values in $\left(1\right)$, we get

$V_P=\frac{1}{4\pi {ϵ}_0}[\frac{Q_1}{r}+\frac{Q_2}{r}+\frac{Q_3}{c}]$

$\therefore V_P=\frac{1}{4\pi {ϵ}_0}[\frac{Q_1+Q_2}{r}+\frac{Q_3}{c}]$

Hence, option (b) is the correct answer.