Q. Two plates are $2\ cm$ apart, a potential difference of $10\ volt$ is applied between them, the electric field between the plates is
A) $20\ N/C$
B) $500\ N/C$
C) $5\ N/C$
D) $250\ N/C$
Explanation
(b) $E = \frac{V}{d} = \frac{10}{2 \times 10^{- 2}} = 500\ N/C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.