Question

Q.4. Two charges of magnitudes -2Q and + Q are located at points (a,0) and (4a,0) respectively. What is the electric flux due to these charges through a sphere of radius ’3a’ with its center at the origin?

Answer 1

Answer: $\phi = \dfrac{-2Q}{\epsilon_{0}}$

Explanation:

Given that,

The magnitude of two charges

$Q_{1} = -2Q$

$Q_{2} = +Q$

Located points = (a,0), (4a,0)

Radius of the sphere r = 3a

Using Gauss law

The net electric flux of the closed surface is equal to the$\dfrac{1}{\epsilon_{0}}$ times of the total charge in the closed surface.

$\phi_{E} = \dfrac{Q}{\epsilon_{0}}$

Now, the enclosed charge is

$Q = -2Q$

Because -2Q charge is located in the sphere and the other charge lies outside the surface and thus, there would be no electric flux due to it in the sphere.

So, the electric flux due to these charges through a sphere

$\phi = \dfrac{-2Q}{\epsilon_{0}}$

Hence, the electric flux is $\phi = \dfrac{-2Q}{\epsilon_{0}}$.