Question

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?

• 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 $l$ 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

Answer 1

Answer: (A)

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 ' $-Q$ '.
$A D=l \cos 30^{\circ}=\frac{l \sqrt{3}}{2}, A O=\frac{2}{3} A D=\frac{1}{\sqrt{3}}$
$2\left|\vec{F}_{C A}\right| \cos 30^{\circ}=\left|\vec{F}_{C O}\right|$
$ 2 \times \frac{1}{4 \pi \varepsilon_{0}} \times \frac{q^{2}}{l^{2}} \times \frac{\sqrt{3}}{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{(l / \sqrt{3})^{2}}$
or $Q=-\frac{q}{\sqrt{3}}$