Question

Above an infinitely large plane carrying charge density $\sigma$, the electric field points up and is equal to $\frac{\sigma }{2{ϵ}_0}$. What is the magnitude and direction of the electric field below the plane?

• A. $\sigma /2{ϵ}_0$, down
• B. $\sigma /2{ϵ}_0$, up
• C. $\sigma /{ϵ}_0$, down
• D. $\sigma /{ϵ}_0$, up

Answer 1

The correct option is A $\sigma /2{ϵ}_0$, down

Electric field due to an infinite plane sheet of charge can be found using the gauss law. Consider a cylindrical gaussian surface as shown in the figure.

Using Gauss Law,

$2E\Delta A=\frac{\sigma \Delta A}{{\epsilon }_o}$

$E=\frac{\sigma }{2{\epsilon }_o}$

Hence, electric field is $\frac{\sigma }{2{\epsilon }_o}$ downwards.