JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Electrostatics

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Two point charges Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is :

(A) $x = d$
(B) $x = \frac{d}{2}$
(C) $x = \frac{d}{2}$
(D) $x = \frac{d}{2 2}$ ✓ Correct

Answer: (D)

Solution

The net force on charge q at distance x from the midpoint on the perpendicular bisector is:

F_{n e t} = 2 F cos θ = \frac{2 K Qq x}{( x ^{2} + \frac{d ^{2}}{4} ) ^{3/2}}

For maximum force, we differentiate with respect to x and set equal to zero:

\frac{d F _{n e t}}{d x} = 0

Solving this condition yields:

x^{2} = \frac{d ^{2}}{8}

Therefore:

x = \frac{d}{2 2}

Answer: (D)

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Single correct option

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