Two small spheres each of mass 10 mg are suspended from a point by threads 0.5m long. They are equally charged and repel each other to a distance of 0.20m. The charge on each of the sphere is $\frac{a}{21} \times 10^{- 8} C .$ the value of ‘a’ will be —— [given g = 10 $m s^{- 2}$ ]

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answer is 20.

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Detailed Solution

$At equilibrium,$

$\begin{array}{l} F = m g \tan θ \\ \Rightarrow 9 \times 10^{9} \times \frac{q^{2}}{0.04} = 10 \times 10^{- 3} \times 10^{- 3} \times 10 \times \frac{1}{\sqrt{24}} \\ \Rightarrow q = 0.952 \times 10^{- 8} C = \frac{20}{21} \times 10^{- 8} C \\ So, a = 20 \end{array}$