Q. Two parallel large thin metal sheets have equal surface charge densities $\left( \sigma = 26.4 \times 10^{- 12}Cm^{- 2} \right)$of opposite signs. The electric field between these sheets is
A) $1.5\ NC^{- 1}$
B) $1.5 \times 10^{- 10}NC^{- 1}$
C) $3\ NC^{- 1}$
D) $3 \times 10^{- 10}NC^{- 1}$
Explanation
(c) The situation is shown in the figure. Plate 1 has surface charge density $\sigma$ and plate 2 has surface charge density$–\sigma$. The electric fields at point $P$ due to two charged plates add up,
giving
[image:1_image254.emf]
$E = \frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}$
Given, $\sigma = 26.4 \times 10^{- 12}Cm^{- 2}\ $
$\varepsilon_{0} = 8.85 \times 10^{- 12}C^{2}N^{- 1}m^{- 2}$
Hence,
$E = \frac{26.4 \times 10^{- 12}}{8.85 \times 10^{- 12}} \approx 3NC^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.