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Physics Question on electrostatic potential and capacitance

A ball of mass 1 g and charge 10−810^{-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−1cms^{ -1}. The velocity of the ball at the point A will be

A

22.8 cms−1cms^{ -1}

B

228 cms−1cms^{ -1}

C

16.8 cms−1cms^{ -1}

D

168 cms−1cms^{ -1}

Answer

22.8 cms−1cms^{ -1}

Explanation

Solution

By using, 12 m(v12−v22)=QV\frac{1}{ 2} \, m ( v_1^2 - v_2^2 ) = QV 12×10−3 [v12−(0.2)2]=10−8 (600−0) \frac{ 1}{ 2} \times 10^{ - 3} \, [ v_1^2 - ( 0.2)^2 ] = 10^{ - 8 } \, ( 600 - 0) v1=22.8 cms−1 v_1 = 22.8 \, cms^{ - 1}.