A small uncharged conducting sphere is placed in contact with an identical sphere but having charge and then removed a distance such that the force of repulsion between them is . The distance between them is (Take as in SI units) [2025]

Question

A small uncharged conducting sphere is placed in contact with an identical sphere but having $4 \times 10^{- 8} C$ charge and then removed to a distance such that the force of repulsion between them is $9 \times 10^{- 3} N$ . The distance between them is (Take $\frac{1}{4 π ε_{0}}$ as $9 \times 10^{9}$ in SI units) [2025]

Smart Achievers · Accepted Answer

(1)

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

$9 \times 10^{- 3} = \frac{9 \times 10^{9} \times (4 \times 10^{- 8}) \times 4 \times 10^{- 8}}{4 \times r^{2}}$

$r^{2} = \frac{9 \times 10^{9} \times 16 \times 10^{- 16}}{4 \times 9 \times 10^{- 3}} = 4 \times 10^{- 4}$

$r = 2 \times 10^{- 2} m = 2 cm$

Solution

1 2 cm

2 3 cm

3 4 cm

4 1 cm

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