First slide

Electrostatic potential

Question

Two electric charges $12 μC$ and $- 6 μC$ are placed 20 cm apart in air. There will be a point P on the line joining these charges and outside the region between them, at which the electric potential is zero. The distance of P from $- 6 μC$ charge is

Easy

A

0.10 m
B

0.15 m
C

0.20 m
D

0.25 m

Solution

Point P will lie near the charge which is smaller in magnitude i.e. $- 6 μC$ . Hence potential at P
$V = \frac{1}{4 {πε}_{0}} \frac{(- 6 \times 10^{- 6})}{x} + \frac{1}{4 {πε}_{0}} \frac{(12 \times 10^{- 6})}{(0 .2 + x)} = 0 \Rightarrow x = 0 .2 m$

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