Q. A ball of mass $1\ g$ and charge $10^{- 8}C$ moves from a point $A.$ where potential is $600\ volt$ to the point $B$ where potential is zero. Velocity of the ball at the point $B$ is $20\ cm/s.$ The velocity of the ball at the point $A$ will be
A) $22.8\ cm/s$
B) $228\ cm/s$
C) $16.8\ m/s$
D) $168\ m/s$
Explanation
(a) By using $\frac{1}{2}m\left( v_{1}^{2} - v_{1}^{2} \right) = QV$
$\Rightarrow \frac{1}{2} \times 10^{- 3}\left\lbrack v_{1}^{2} - (0.2)^{2} \right\rbrack = 10^{- 8}(600 - 0)$
$\Rightarrow v_{1} = 22.8\ cm/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.