First slide

Potential energy

Question

Two equal point charges are fixed at x = - $α$ and x = + $α$ on the X-axis . Another point charge Q is placed at the origin. The change in electrical potential energy of Q when it is displaced by a small distance x along the X-axis, is approximately proportional to

Moderate

A

X
B

X²
C

X³
D

1/X

Solution

When Q is at position x, then
$U = \frac{qQ}{4 {πε}_{0}} [\frac{1}{(a + x)} + \frac{1}{(a - x)}]$
or $U = \frac{2 qQa}{4 {πε}_{0} (a^{2} - x^{2})}$
$\begin{array}{l} ∴ \frac{dU}{dx} = \frac{2 aqQ}{4 {πε}_{0}} [\frac{2 x}{{(a^{2} - x^{2})}^{2}}] \\ = \frac{2 aqQ}{4 {πε}_{0}} \times \frac{2 x}{a^{4}} (∵ (a^{2} - x^{2}) \approx a^{2}) \end{array}$
Now $dU \propto dx or U \propto x^{2} .$

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