- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
normal
There are four concentric shells $A, B, C $ and $D $ of radii $ a, 2a, 3a$ and $4a$ respectively. Shells $B$ and $D$ are given charges $+q$ and $-q$ respectively. Shell $C$ is now earthed. The potential difference $V_A - V_C $ is :
A
$\frac{{Kq}}{{2a}}$
B
$\frac{{Kq}}{{3a}}$
C
$\frac{{Kq}}{{4a}}$
D
$\frac{{Kq}}{{6a}}$
Solution
Let charge on $C$ be $q:$
$V_{C}=\frac{k q}{3 a}+\frac{k q^{\prime}}{3 a}-\frac{k q}{4 a}$
$V_{C}=0(\text { earthed })$
$\frac{k q}{3 a}+\frac{k q^{\prime}}{3 a}-\frac{k q}{4 a}=0$
$q^{\prime}=\frac{-q}{4}$
$V_{A}=\frac{k q}{2 a}-\frac{k q / 4}{3 a}-\frac{k q}{4 a}=\frac{k q}{6 a}$
Therefore, $V_{A}-V_{C}=\frac{k q}{6 a}$
Standard 12
Physics
Similar Questions
hard
