Q. A charge of $5\ C$ experiences a force of $5000\ N$ when it is kept in a uniform electric field. What is the potential difference between two points separated by a distance of $1cm$
A) $10\ V$
B) $250\ V$
C) $1000\ V$
D) $2500\ V$
Explanation
(a) $F = QE = \frac{QV}{d} \Rightarrow 5000\ \frac{5 \times V}{10^{- 2}} \Rightarrow V = 10\ volt$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.