Two point charges (-q) and (+4q) are placed at separation 'r'. Where should a third charge be placed so that entire system of charges becomes in equilibrium?

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This question was previously asked in

CUET Physics 18th Aug 2022 Official Paper

Answer 1

Calculation:

Let a charge "Q" is placed at a distance of "x" from (-q).

Let the force exerted by "-q" be F₁ and "4q" be F₂on "Q".

To make the system in equilibrium, the net force acting on "Q" should be zero.

i.e., F₁ + F₂ = 0.-----(1)

F₁ = K(-q)Q/x²

F₂ = K(4q)Q/(r + x)²

Putting the value of F1 and F2 in equation (1), we get,

K(-q)Q/x2 + K(4q)Q/(r + x)2 = 0

(x + r)/x = 2

x = r

The charge "Q" is placed at a distance of "r" from charge "-q" on the extreme side of (-q).

The correct answer is option (1).

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