Question

A charge ${}^{\prime }{q}^{\prime }$ is placed at the centre of the line joining two equal charges. The system will be in equilibrium if $q$ is equal to

• A. $-Q/2$
• B. $-Q/4$
• C. $-4Q$
• D. $+Q/2$

Answer 1

The correct option is B $-Q/4$

Force on one Q

$F=\frac{k{Q}^2}{4a^2}+\frac{kqQ}{a^2}$

For the system to be in equilibrium F=0

$\frac{k{Q}^2}{4a^2}+\frac{kqQ}{a^2}=0$

$\frac{k{Q}^2}{4a^2}=-\frac{kqQ}{a^2}$

$q=-\frac{Q}{4}$