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
$A C=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$
and $O A=\frac{1}{2} A C=\frac{a}{\sqrt{2}}$
Now, $F_{B}=F_{D}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{a^{2}}$
$F_{C}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{(\sqrt{2} a)^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q^{2}}{2 a^{2}}$
and $ F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q q^{2}}{(a / \sqrt{2})^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2 Q q}{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
$Q=\left(\frac{1+2 \sqrt{2}}{4}\right) q \text { (negative) }$