Q. Two large metal plates are placed parallel to each other. The inner surfaces of plates are charged by $+ \sigma$ and $–\sigma\left( {cm}^{- 2} \right).$ The outer surfaces are neutral. The electric field in the region between the plates and outside the plates is
A) $\frac{2\sigma}{\varepsilon_{0}},\frac{\sigma}{\varepsilon_{0}}$
B) $\frac{\sigma}{\varepsilon_{0}},\ zero$
C) $\frac{2\sigma}{\varepsilon_{0}},\ zero$
D) $zero,\frac{2\sigma}{\varepsilon_{0}}$
Explanation
(b) The situation is as shown in the figure
[image:1_image238.emf]
Here, $E_{A} = \frac{\sigma}{2\varepsilon_{0}}$
and $E_{B} = \frac{\sigma}{2\varepsilon_{0}}$
Hence, the electric field between the plates
$E = E_{A} + E_{B} = \frac{\sigma}{2\varepsilon_{0}} + \frac{\sigma}{2\varepsilon_{0}} = \frac{\sigma}{\varepsilon_{0}}$
and electric field outside the plates will be zero.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.