Question

It is required to hold four equal point charges $+q$ in equilibrium at the corners of a square. Find the point charge that will do this, if placed at the centre of the square.

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

Answer 1

Answer: (B)

Suppose that the charge $Q$ placed at the centre of the square keeps all the four charges in equilibrium. It follows that the charge $Q$ has to be negative in nature.
If the net force on a charge at any corner (say $A$ ) is zero, then by symmetry it follows that the net force experienced by the charges at other corners will also be zero. The charge at corner $A$ will experience force $F_B,F_C$ and $F_D$ due to the charges at corners $B,C$ and $D$. and a force $F$ due to the charge at the centre of the square.

If each side of the square is of length $a$, then
$AC=\sqrt{a^2+a^2}=\sqrt{2}a$
and $OA=\frac{1}{2}AC=\frac{a}{\sqrt{2}}$
Now, $F_B=F_D=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{a^2}$
$F_C=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{(\sqrt{2}a{)}^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q^2}{2a^2}$
and $F=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{Q{q}^2}{(a/\sqrt{2}{)}^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{2Qq}{a^2}$
For net force on the charge at $A$ to be zero,
$F_B+F_C\cos 45^{\circ }=F\cos 45^{\circ }$ (i)
and $F_D+F_C\sin 45^{\circ }=F\sin 45^{\circ }$ (ii)
Substituting for $F_B,F_C$ and $F$ in the equation (i), we get
$Q=\left(\frac{1+2\sqrt{2}}{4}\right)q\text{ (negative) }$