Three identical charged balls each of charge 2 C are suspended from a common point P by silk threads of 2 m each (as shown in figure). They form an equilateral triangle of side 1m. The ratio of net force on a charged ball to the force between any two charged balls will be :

Electrostatics PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q56 with Solution

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Correct answer: D

Each ball has charge $q=2\,\text{C}$ and the distance between any two balls is the side of the equilateral triangle, $a=1\,\text{m}$ .

So the electrostatic force between any two balls is

F=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{a^2}

We do not need its numerical value, only the ratio.

Consider one ball, say ball $A$ .

It experiences repulsive forces due to the other two balls $B$ and $C$ . Since all charges are identical and all sides are equal:

The angle between these two forces is the angle of the equilateral triangle:

\angle BAC=60^\circ

Resultant of two equal forces $F$ making angle $60^\circ$ is

F_{\text{net}}=\sqrt{F^2+F^2+2F^2\cos 60^\circ}

F_{\text{net}}=\sqrt{2F^2+2F^2\cdot \frac12}

F_{\text{net}}=\sqrt{3F^2}=\sqrt3\,F

The force between any two charged balls is $F$ .

Hence,

F_{\text{net}}:F=\sqrt3 F:F=\sqrt3:1

So the correct option is D.