A solid sphere of radius 4a with centre at origin. Two charge, –2q at (–5a, 0) and 5q at (3a, 0) is placed. Flux through sphere is x_q/ε_o . Find x.

The correct answer is 5.

Let's solve the problem step-by-step using Gauss's Law.

Gauss's Law states that the net electric flux, $\Phi$, through any closed surface is equal to the net charge enclosed by the surface, $q_{\text{enclosed}}$, divided by the permittivity of free space, ${\epsilon }_o$:
$\Phi =\frac{q_{\text{enclosed}}}{{\epsilon }_o}$

In this problem, the closed surface is a solid sphere of radius R = 4a centered at the origin (0, 0).

Let's determine the position of the two charges relative to this sphere:

1. The first charge is $q_1=-2q$ located at (-5a, 0).
The distance of this charge from the origin is:
$d_1=|-5a|=5a$
Since $5a>4a$, this charge lies outside the sphere.

2. The second charge is $q_2=5q$ located at (3a, 0).
The distance of this charge from the origin is:
$d_2=\left|3a\right|=3a$
Since $3a<4a$, this charge lies inside the sphere.

According to Gauss's Law, only the charge inside the sphere contributes to the net electric flux through it. Therefore, the enclosed net charge is:
$q_{\text{enclosed}}=5q$

Thus, the flux through the sphere is:
$\Phi =\frac{5q}{{\epsilon }_o}$

We are given that the flux is equal to:
$\frac{xq}{{\epsilon }_o}$
By comparing the two expressions, we find:
$x=5$