Three charges, + Q, q and + Q are placed with q at mid point of the line joining the other two charges. For the system of charges to be in equilibrium, the value of q is

- Q/4

The correct answer is Option (2) → - Q/4

Three charges: +Q —— q —— +Q ; q at the midpoint.

Geometry:

$AO = OB = a$

Force on q due to left +Q:

$F_1 = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{a^2}$

Force on q due to right +Q:

$F_2 = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{a^2}$

Net force on q:

$F_1 - F_2 = 0$

Now, for equilibrium of left +Q (at A):

Force due to q:

$F_q = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{a^2}$

Force due to right +Q (at B):

$F_Q = \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{(2a)^2} = \frac{1}{4\pi\varepsilon_0}\frac{Q^2}{4a^2}$

Condition for equilibrium (equal and opposite):

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

Therefore,

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

Final Answer:

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