Question

A charge q is placed at the center of the line joining two equal charges Q . The system of the three charges will be equilibrium, If q is equal to :-

• A. $-\frac{Q}{2}$
• B. $-\frac{Q}{4}$
• C. $+\frac{Q}{4}$
• D. $+\frac{Q}{2}$

Answer 1

The correct option is B −Q4-\dfrac { Q }{ 4 } −4Q​

The system is in equilibrium which means that the charge q is in equilibrium.

Charge q is in equilibrium since charges A and B exert equal and opposite forces on it.

So for the equilibrium of charge q at B

$F_{BC}+F_{AB}=0$

We can also write

$K\frac{qQ}{{\left(\frac{l}{2}\right)}^2}+K\frac{QQ}{l^2}=0$

After solving this equation we get

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

Hence the value of q should be $\frac{-Q}{4}$