Q. A ball of mass 1 carrying a charge $10^{- 8}$ Cmoves from a point A at potential 600 V to a point B at zero potential. The change in its KE is
A) $- 6 \times 10^{- 6\ }$ erg
B) $- 6 \times 10^{- 6\ }$ J
C) $6 \times 10^{- 6\ }$ J
D) $6 \times 10^{- 6\ }$ erg
Explanation
(c) As work is done by the field, KE of the body increase by
KE =$\ W = E = q\left( V_{A} - V_{B} \right)$
${= 10}^{- 8}(600 - 0) = 6 \times 10^{- 6}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.