Question

The electric field in a region of space is given by $ E = {E_0}\hat i + 2{E_0}\hat j $ ​ where $ {E_0} = 100\,N/C $. The flux of this field through a circular surface of radius $ 0.02\,m $ parallel to the Y-Z plane is nearly :
A) $ 0.125\,N{m^2}/C $
B) $ 0.002\,N{m^2}/C $
C) $ 0.005\,N{m^2}/C $
D) $ 3.14\,N{m^2}/C $

Answer 1

Hint
Gauss’s law tells us how much flux passes through a given surface. Only the component of the electric field having a direction parallel to the area vector of the surface will contribute to the flux.
Formula used: $ \phi = E.A $ where $ \phi $ is the flux through a surface, $ E $ is the electric field in the region, $ A $ is the area of the surface.

Complete step by step answer
We’ve been given that the electric field in a region is $ E = {E_0}\hat i + 2{E_0}\hat j $ and we want to find the flux passing through a circular surface that is parallel to the Y-Z plane. The area vector of this surface will be normal to the circular surface and hence will lie in the direction of the x-axis.
Since the flux through the surface is calculated as $ \phi = E.A $ only the component of the electric field parallel to the area vector of the surface i.e. the x component will contribute to the flux from the surface.
Hence, we can calculate the flux passing from a circular area of radius $ 0.02\,m $ as:
$ \phi = {E_0} \times (\pi R^2) \\ $
$ = \,100 \times  \times \pi \times R^2 \\ $
Substituting the value of $ R = 0.02\,m $, we get
 $ \phi = 0.125\,N{m^2}/C $
So the flux through the circular surface of the radius $ 0.02\,m $ will be $ \phi = 0.125\,N{m^2}/C $ which corresponds to option (A).

Note
While calculating the flux using Gauss’s law, we must remember that the operation between the electric field and the area vector of the given surface is not direct multiplication but a dot product. So we must only take the direction of the electric field that will have a direction parallel to the area vector of the given surface.