Question

A ring of radius R is uniformly charged to linear charge density λ. An imaginary sphere of radius R is drawn with its centre on the circumference of the ring. Total electric flux passing through the sphere would be :-

Answer 1

As you know,

Linear charge density, λ = $\frac{Q_{in}}{L}$

So, $Q_{in}$ = λL  --(i)

L must be less than half of the circumference, i.e., $\pi R$.

Now, the Flux must be less than that of half of the original sphere.

From (i)

φ <  λ$\pi R$/ε0.

Answer 2

The total electric flux that passes through the sphere is $\frac{Q}{3\mathcal{E}_0}$.

Given:

A ring of radius R is uniformly charged to linear charge density $\lambda$. An imaginary sphere of radius R is drawn with its center on the circumference of the ring.

To Find:

Total electric flux passes through the sphere.

Solution:

Here, in the given figure, $OA=OO_1$ and $O_1A=O_1O$.

So, $OAO_1$ is an equilateral triangle.

here, in the figure,

$\angle AOO_1=60^{o}$

and,

$\angle AOB=120^{o}$.

In the figure, the arc $AO_1B$ in the ring shows an angle $120^{o}$ at the center O.

Now, one-third of the ring is inside the sphere.

So, the charge enclosed by the sphere is $\frac{Q}{3}$.

Here, by applying Gauss's law, we get the flux of the electric field passing through the surface of the sphere as $\frac{Q}{3\mathcal{E}_0}$.

Hence, the total electric flux that passes through the sphere is $\frac{Q}{3\mathcal{E}_0}$.

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