Question

Four charges, each equal to -Q, are placed at the corners of a square and a charge +q placed at its centre. If the system is in equilibrium, the value of q is

• A. $\frac{Q}{4}(1+2\sqrt{2})$
• B. $-\frac{Q}{4}(1+2\sqrt{2})$
• C. $-\frac{Q}{2}(1+2\sqrt{2})$
• D. $\frac{Q}{2}(1+2\sqrt{2})$

Answer 1

The correct option is A

$\frac{Q}{4}(1+2\sqrt{2})$

Let the side of the square be a OA=OC =r =$\frac{a}{\sqrt{2}}$

1 Stability of charge +q at the centre
Charges - Q at corners A and C will attract charge + q with equal and opposite forces. Similarly charges -Q at corners B and D will attract charge + q with equal and opposite force. Hence no net force acts on charge -q.
2. Stability of charge -Q at any corner
Let us find the forces on charge - Q at corner A. This charge will experience four forces: (i) Force of repulsion Fi due to charge - Q at B
(ii) Force of repulsion F2 due to charge - Q at D
(iii) Force of repulsion F3 due to charge - Q at C
(iv) Force of attraction F' due to charge + q at 0.
Now $F_1=F_2=\frac{Q^2}{4\pi {ϵ}_0{a}^2}$
and $F_3=\frac{Q^2}{4\pi {ϵ}_0(2r{)}^2}=\frac{Q^2}{4\pi {ϵ}_0\left(2a^2\right)}$
and $F^{\prime }=\frac{1}{4\pi {ϵ}_0}.\frac{q\left(Q\right)}{r^2}=\frac{2qQ}{4\pi {ϵ}_0{a}^2}$
The resultant of $F_1$ and $F_2$ is given by
$F=\sqrt{F_1^2+F_2^2}=\sqrt{2}{F}_1=\frac{\sqrt{2}{Q}^2}{4\pi {ϵ}_0{a}^2}$
The force F and $F_3$ act along AP. Hence the net force acting on charge -Q at A due to charges -Q at B, C and D is
$F^{\prime \prime }=F+F_3$
$=\frac{\sqrt{2}{Q}^2}{4\pi {ϵ}_0{a}^2}+\frac{Q^2}{4\pi {ϵ}_0\left(2a^2\right)}=\frac{Q^2(1+2\sqrt{2})}{4\pi {ϵ}_0\left(2a^2\right)}$
For equilibrium, F' - F" = 0 or F'' = F', i.e.,
$=\frac{Q^2(1+2\sqrt{2})}{4\pi {ϵ}_0\left(2a^2\right)}=\frac{2qQ}{4\pi {ϵ}_0{a}^2}$
or $q=\frac{Q}{4}(1+2\sqrt{2})$
Hence the correct choice is (a).