A ball of mass 1 g and charge $10^{- 8} C$ moves from a point A, where potential is 600 V to the point B
where potential is zero. Velocity of the ball at the point B is $20 {cms}^{- 1}$ . The velocity of the ball at the
point A will be

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$16.8 {cms}^{- 1}$

$22.8 {cms}^{- 1}$

$228 {cms}^{- 1}$

$168 {cms}^{- 1}$

answer is A.

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Detailed Solution

By using, $\frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) = QV$

$\frac{1}{2} \times 10^{- 3} [v_{1}^{2} - (0.2)^{2}] = 10^{- 8} (600 - 0)$

Velocity of the ball at the point A, $v_{1} = 22.8 {cms}^{- 1}$

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