Three equal point charges, each \( +q \), are fixed at the corners of an equilateral triangle. A point charge \( Q \) is placed at the centroid of the triangle. For each of the four charges to be in equilibrium, \( Q \) must be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -q/\sqrt{3} \)
In equilibrium the net force on every charge is zero.
The charge \( Q \) at the centroid is at the same distance from the three corner charges. By symmetry the three forces on it add up to zero, whatever its value.
The deciding condition is the force on each corner charge. There the attraction of \( Q \) has to balance the repulsion of the other two charges \( +q \).
Let the side of the triangle be \( a \). Each of the other two corner charges repels a corner charge with a force of magnitude \( kq^{2}/a^{2} \). These two forces are inclined at \( 60^\circ \) to each other. By the superposition principle, their resultant, directed away from the centroid, is:
\( \begin{aligned} F_1 &= 2\,\frac{kq^{2}}{a^{2}}\cos30^\circ \\ &= \frac{\sqrt{3}\,kq^{2}}{a^{2}} \end{aligned} \)
The centroid is at a distance \( r=a/\sqrt{3} \) from each corner. So the force of \( Q \) on a corner charge has the magnitude:
\( \begin{aligned} F_2 &= \frac{kq\,|Q|}{r^{2}} \\ &= \frac{3kq\,|Q|}{a^{2}} \end{aligned} \)
This force must point towards the centroid. So \( Q \) must attract \( +q \), and \( Q \) must be negative. For balance, \( F_2=F_1 \):
\( \begin{aligned} \frac{3kq\,|Q|}{a^{2}} &= \frac{\sqrt{3}\,kq^{2}}{a^{2}} \\ |Q| &= \frac{q}{\sqrt{3}} \end{aligned} \)
Hence, \( Q=-q/\sqrt{3} \).