Question

If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then the value of $q$ is :

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

Answer 1

The correct option is D $-Q/4$

Let, LK=KM=$\frac{x}{2}$

LM=x

Force on the charge at "M" due to charge at "L" is

$F_1$=$\frac{K{Q}^2}{x^2}$

Force on the charge at "M" due to charge at "K" is

$E_2$=$\frac{KQq}{{x_2}^2}$

For the system to be in equilibrium

$F_1+F_2$=0

$\frac{K{Q}^2}{x^2}$+$\frac{4KQq}{x^2}$=0

so,

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