Gujarati
1. Electric Charges and Fields
medium

Two charges $ + 4e$ and $ + e$ are at a distance $x$ apart. At what distance, a charge $q$ must be placed from charge $ + e$ so that it is in equilibrium

A

$x/2$

B

$2x/3$

C

$x/3$

D

$x/6$

Solution

(c) For equilibrium of $q$
|$F_1$| = |$F_2$|
Which gives ${x_2} = \frac{x}{{\sqrt {\frac{{{Q_1}}}{{{Q_2}}}} + 1}} = \frac{x}{{\sqrt {\frac{{4e}}{e}} + 1}} = \frac{x}{3}$

Standard 12
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.