Question

Above an infinitely large plane carrying charge density $\sigma$ , the electric field points up and is equal to $\dfrac{\sigma }{{2{\varepsilon _0}}}$ . What is the magnitude and direction of the electric field below the plane?
A. $\dfrac{\sigma }{{2{\varepsilon _0}}}$ down
B. $\dfrac{\sigma }{{2{\varepsilon _0}}}$ up
C. $\dfrac{\sigma }{{{\varepsilon _0}}}$ down
D. $\dfrac{\sigma }{{2{\varepsilon _0}}}$ up

Answer 1

To find electric fields we can use Gauss law. Consider a Gaussian surface, find the electric field
above and below that Gaussian surface. A Gaussian surface is a closed surface used to calculate flux.
The given surface is a plane hence, the field above and below will be equal.

Complete step by step answer:
We are given the electric field above the plane and we are required to find the electric field below the surface. We know that the flux of the electric field out of a closed surface is proportional to the electric charge enclosed by the surface. For an infinitely large plane, the electric field will be perpendicular to the surface.
For the infinitely large plane let’s consider a cylindrical Gaussian surface.
Now the flux of the electric field out of this cylindrical surface will be proportional to the electric charge enclosed by the surface. Flux is the number of electrical lines of force that intersect a given area.
So, we can have:
$E_2 \Delta A = \dfrac{{\sigma \Delta A}}{{{\varepsilon _0}}}$
Where $E$ is the electric field.
$\Delta A$ denotes the enclosed area
$\sigma$ is the given charge density
${\varepsilon _0}$ is a constant
Please note that the area is taken twice. As the area above the plane and below the plane will be of the same magnitude.
Solving the above equation, we have
$E = \dfrac{\sigma }{{2{\varepsilon _0}}}$
This will give the magnitude of the electric field, down the plane.
This field has direction downwards.

So, the correct answer is “Option A”.

Note:
Please note that only the ends of the cylindrical Gaussian surface will contribute to the electric flux. While calculating the electric field only the charge enclosed in the gaussian surface is considered. Gauss law can also be used to find the electric field between infinitely long parallel plates.