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 ' $-Q$ '.
$AD=l\cos 30^{\circ }=\frac{l\sqrt{3}}{2},AO=\frac{2}{3}AD=\frac{1}{\sqrt{3}}$
$2\left|{\vec{F}}_{CA}\right|\cos 30^{\circ }=\left|{\vec{F}}_{CO}\right|$
$2\times \frac{1}{4\pi {\epsilon }_0}\times \frac{q^2}{l^2}\times \frac{\sqrt{3}}{2}=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{(l/\sqrt{3}{)}^2}$
or $Q=-\frac{q}{\sqrt{3}}$