First slide

Kirchhoff's rules for capacitors

Question

The left plate of the capacitor shown in the figure carries a charge +Q while the right plate is uncharged. The total final charge on the right plate after closing the switch S will be

Moderate

A

$\frac{Q}{2} + C ε$
B

$\frac{Q}{2} - C ε$
C

$- \frac{Q}{2}$
D

none of these

Solution

Electric field between the capacitor plates
$E = \frac{Q + x}{2 A ε_{0}} + \frac{x}{2 A ε_{0}} = \frac{1}{2 ε_{0} A} [Q + 2 x]$
Potential difference
$= E d = \frac{d}{2 ε_{0} A} [Q + 2 x] = ε$
$ε = \frac{Q + 2 x}{2 C} \Rightarrow x = \frac{Q}{2} - C ε$

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