In the figure shown, conducting shells $A$ and $B$ have charges $Q$ and $2 Q$ distributed uniformly over $A$…

In the figure shown, conducting shells $A$ and $B$
have charges $Q$ and $2 Q$ distributed uniformly
over $A$ and $B$.Value of $V_{A}-V_{B}$ is
  1. $\frac{Q}{4 \pi \varepsilon_{0} R}$
  2. $\frac{Q}{8 \pi \varepsilon_{0} R}$
  3. $\frac{3 Q}{4 \pi \varepsilon_{0} R}$
  4. $\frac{3 Q}{8 \pi \varepsilon_{0} R}$

Solution

$V_{A}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{Q}{R}+\frac{2 Q}{2 R}\right]=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 Q}{R}$
$V_{B}=\frac{1}{4 \pi \varepsilon_{0}} \frac{3 Q}{2 R}, \quad V_{A}-V_{B}=\frac{Q}{8 \pi \varepsilon_{0} R}$
$-\quad$ (b)

Asked in: JEE Mains - Electrostatics - Test 3

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