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
- $\frac{Q}{4 \pi \varepsilon_{0} R}$
- $\frac{Q}{8 \pi \varepsilon_{0} R}$
- $\frac{3 Q}{4 \pi \varepsilon_{0} R}$
- $\frac{3 Q}{8 \pi \varepsilon_{0} R}$
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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