First slide

Capacitors and capacitance

Question

A parallel plate capacitor is connected to a battery. The plates are pulled apart with uniform speed. If x is the separation between the plates, then the rate of change of electrostatic energy of the capacitor is proportional to

Moderate

A

$x^{2}$
B

x
C

$\frac{1}{x}$
D

$\frac{1}{x^{2}}$

Solution

The capacitance of capacitor C is $ε_{0} A / x .$ Charge on capacitor
$q = C V = \frac{ε_{0} A V}{x}; U = \frac{1}{2} q V = \frac{1}{2} \frac{ε_{0} A V^{2}}{x}$
So $\frac{d U}{d t} = \frac{1}{2} ε_{0} A V^{2} (- \frac{1}{x^{2}}) \frac{d x}{d t}$
$= \frac{1}{2} ε_{0} A V^{2} v (- \frac{1}{x^{2}})$
From the earlier expressions,
$\frac{d U}{d t} \propto \frac{1}{x^{2}}$

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