JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Electrostatics

Two equal positive point charges are separated by a distance

2 a

. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge

q_{0}

becomes maximum is

\frac{a}{x}

. The value of

x

is ______.

Answer: 2

Solution

Each charge sits a distance

x^{2} + a^{2}

from the test charge on the perpendicular bisector, and by symmetry only the components along the bisector survive. The net force is

F = 2 \cdot \frac{k q q _{0}}{x ^{2} + a ^{2}} \cdot \frac{x}{x ^{2} + a ^{2}} = \frac{2 k q q _{0} x}{( x ^{2} + a ^{2} ) ^{3/2}} .

Setting

\frac{d F}{d x} = 0

gives

(x^{2} + a^{2}) - 3 x^{2} = 0

, that is

a^{2} - 2 x^{2} = 0

, so the force is largest at

x = \frac{a}{2} .

Comparing with

\frac{a}{x}

gives

x = 2

JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Electrostatics

Solution

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