JEE Main 2025 Jan 24 Shift 2, Physics Q18: Coulomb's Law And Electric Field

JEE Main2025Jan 24, Shift 2Physics

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 $\frac{1}{4 π ε _{0}}$ as $9 \times 1 0^{9}$ in SI units)

A
$2 cm$
B
$3 cm$
C
$4 cm$
D
$1 cm$

Solution

When two identical conducting spheres are brought into contact, the total charge distributes equally between them. So each sphere ends up with $\frac{Q}{2} = \frac{4 \times 1 0 ^{- 8}}{2} = 2 \times 1 0^{- 8} C$ .

Coulomb's law at separation $r$ :

$F = \frac{k Q _{1} Q _{2}}{r ^{2}} = \frac{k ( Q /2 ) ^{2}}{r ^{2}}$

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

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

$r = 2 \times 1 0^{- 2} m = 2 cm$

Correct option: (1)

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JEE Main 2025 Jan 24 Shift 2, Physics Q18: Coulomb's Law And Electric Field

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