Three positive charges of equal magnitude q are placed at the vertices of an equilateral triangle of side l. How can the system of charges be placed in equilibrium?

Moderate

A

by placing a charge $Q = - q / \sqrt{3}$ at the centroid of the triangle
B

by placing a charge $Q = q / \sqrt{3}$ at the centroid of the triangle
C

by placing a charge Q = q at a distance I from all the three charges
D

by placing a charge Q = -q above the plane of the triangle at a distance l from all the three charges

Solution

To keep the system in equilibrium, net force experienced by charges at 'A' ,'B' , and 'C' should be zero. For this, another charge of opposite sign
should be placed at the centroid of the triangle. Let this charge be '-Q'.
$\begin{array}{l} A D = l \cos 30^{\circ} = \frac{l \sqrt{3}}{2}, A O = \frac{2}{3} A D = \frac{l}{\sqrt{3}} \\ F o r e q u i l i b r i u m o f c h a r g e a t C \\ 2 |{\vec{F}}_{C A}| \cos 30^{\circ} = |{\vec{F}}_{C O}| \\ \Rightarrow 2 \times \frac{1}{4 π ε_{0}} \times \frac{q^{2}}{l^{2}} \times \frac{\sqrt{3}}{2} = \frac{1}{4 π ε_{0}} \frac{|Q| q}{(l / \sqrt{3})^{2}} \\ \Rightarrow |Q| = \frac{q}{\sqrt{3}} \\ T h u s, Q = - \frac{q}{\sqrt{3}} \end{array}$

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