Question

You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric field in the region other than this point charge. What is the electric flux through the surface of a sphere that has this charge at its center and that has radius 0.142m?

Answer 1

Answer:

The Electric flux will be $0.42\times10^6\ \rm N.m^2/C$

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

$E=1.36\times10^6\ \rm N/C$

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

$\int E.dA=\dfrac{q_{in}}{\epsilon_0}$

Let consider a  sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let $\phi$ be the flux of the Electric Field coming out\passing through it which is given  by

$\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C$

It can be observed that same amount of  flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is  $0.42\times10^6\ \rm N.m^2/C$