First slide

Pontial due to

Question

Two thin spherical conducting shells are at a large distance apart. One of radius 10 cm carries a charge of +0.5 μC and the other of radius 20 cm carries a charge of+ 0. 7 μC. The charge on each, when they are connected by a suitable conducting wire, is respectively :

Moderate

A

+0.4 μC and +0.8 μC
B

+0.425 μC and +0.85 μC
C

+0.5 μC and 0.7 μC
D

0.6 μC and 0.6 μC

Solution

Common potential V = $\frac{total charge}{total capacity} = \frac{0.5 + 0.7}{C_{1} + C_{2}}$
For spherical capacitor, capacity C = $4 {πε}_{0} R$
$Q_{1} = C_{1} V = \frac{4 {πε}_{0} R_{1} (1.2)}{4 {πε}_{0} R_{1} (R_{1} + R_{2})} = 0.4 μC$
and $Q_{2} = C_{2} V = \frac{R_{2} (1.2)}{R_{1} + R_{2}} = 0.8 μC$

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