A small positively charged ball B of mass m is suspended by an insulating thread of negligible mass. Another positively charged small ball C is moved slowly from a large distance until it is in the original position of the first ball. As a result the ball B rises by h. How much work (in joule) has been done by the agent moving the charge C? $(G i v e n m = 1 k g, g = 10 m / s^{2} a n d h = 10 c m)$

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answer is 3.

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Detailed Solution

For similar triangles ABD and ACE

$\frac{\frac{x}{2}}{l} = \frac{h}{x} \dots . (i)$

For equilibrium of ball B

$\Rightarrow \frac{m g}{\frac{k q^{2}}{x^{2}}} = \frac{\cos (\frac{θ}{2})}{\sin θ} = \frac{1}{2 \sin (\frac{θ}{2})} = \frac{1}{2 \frac{h}{x}} \Rightarrow \frac{k q^{2}}{x} = 2 m g h W = Δ U_{e l e c t r o s t a t i c} + Δ U_{g r a v i t a t i o n a l} = 2 m g h + m g h = 3 m g h = 3 \times 1 \times 10 \times 0.1 = 3 J$