JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Electrostatics

A small uncharged conducting sphere is placed in contact with an identical sphere but having

4 \times 1 0^{- 8}

C charge and then removed to a distance such that the force of repulsion between them is

9 \times 1 0^{- 3} N

. The distance between them is (Take

\frac{1}{4 π ε _{0}}

9 \times 1 0^{9}

in SI units)

Answer: (A)

Solution

Two identical conducting spheres share charge equally on contact. One sphere initially carries

q = 4 \times 1 0^{- 8} C

and the other is uncharged; after contact, each carries

\frac{q}{2} = 2 \times 1 0^{- 8} C .

Once separated to distance

r

, Coulomb's law gives the repulsive force:

F = \frac{k ( q /2 ) ^{2}}{r ^{2}} .

Substituting

F = 9 \times 1 0^{- 3} N

and

k = 9 \times 1 0^{9} N m^{2} / C^{2}

9 \times 1 0^{- 3} = \frac{9 \times 1 0 ^{9} \times ( 2 \times 1 0 ^{- 8} ) ^{2}}{r ^{2}} = \frac{9 \times 1 0 ^{9} \times 4 \times 1 0 ^{- 16}}{r ^{2}} = \frac{3.6 \times 1 0 ^{- 6}}{r ^{2}} .

r^{2} = \frac{3.6 \times 1 0 ^{- 6}}{9 \times 1 0 ^{- 3}} = 4 \times 1 0^{- 4} m^{2} ⟹ r = 2 \times 1 0^{- 2} m = 2 cm .

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option