First slide

Electrostatic potential

Question

A charge Q is distributed over two concentric conducting thin spherical shells radii r and R (R=n r). If the surface charge densities on the two shells are equal, the electric potential at the common centre is:

Moderate

A

$\frac{1}{4 π ε_{0} r} \frac{(1 - n)}{(1 - n^{2})} Q$
B

$\frac{1}{4 π ε_{0} r} \frac{(1 + n)}{(1 + n^{2})} Q$
C

$\frac{1}{4 π ε_{0} r} \frac{(1 + 2 n) Q}{(1 + {(2 n)}^{2})}$
D

$\frac{1}{4 π ε_{0} r} \frac{(1 - 2 n)}{(1 - {(2 n)}^{2})} Q$

Solution

$\begin{array}{l} L e t s a y i n s i d e s h e l l h a s c h a r g e q_{1} a n d o u t s i d e s h e l l h a s c h a r g e q_{2} . \\ S u r f a c e c h a r g e d e n s i t i e s a r e e q u a l f o r b o t h s h e l l s . \\ \Rightarrow \frac{q_{1}}{4 π r^{2}} = \frac{q_{2}}{4 π R^{2}} \end{array}$
$\begin{array}{l} \Rightarrow \frac{q_{1}}{q_{2}} = \frac{r^{2}}{R^{2}} \to (1) \\ & \\ q_{1} + q_{2} = Q \to (2) \\ ∴ q_{1} = \frac{r^{2}}{r^{2} + R^{2}} Q & q_{2} = \frac{R^{2}}{R^{2} + r^{2}} Q \\ P o t e n t i a l a t c o m m o n c e n t r e = V = \frac{K q_{1}}{r} + \frac{K q_{2}}{R} = K \frac{Q r}{r^{2} + R^{2}} + K \frac{Q R}{R^{2} + r^{2}} = \frac{K Q (r + R)}{r^{2} + R^{2}} = \frac{K Q (r + n r)}{r^{2} + {(n r)}^{2}} = \frac{K Q (1 + n)}{r (1 + {(n)}^{2})} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App