First slide

Electric field

Question

Two charges Q₁ = 18 $μ$ C and Q₂= -2 $μ$ C are separated by a distance R, and Q₁ is on the left of Q₂. The distance of the point where the net electric field is zero is

Easy

A

between Q₁ and Q₂
B

left of Q₁ at R/2
C

right of Q₂ at R
D

right of Q₂ at R/2

Solution

Neutral point is obtained at point P located at distance x on right of Q₂
$∴ \frac{k |Q_{2}|}{x^{2}} = \frac{k |Q_{1}|}{(x + R)^{2}} p u t |Q_{2}| = 2 μ C a n d |Q_{1}| = 18 μ C \Rightarrow \frac{x + R}{x} = 3 \Rightarrow x = \frac{R}{2}$
$O R \Rightarrow x = \frac{d}{\sqrt{\frac{q_{2}}{q_{1}}} - 1} = \frac{R}{\sqrt{\frac{18}{2}} - 1} = \frac{R}{2}$

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