A hollow metallic sphere of radius r is kept at potential of 1 Volt. The total electric flux coming out of the concentric spherical surface of radius R (> r) is

Understanding Electric Flux and Gauss's Law

This question asks us to determine the total electric flux exiting a spherical surface of radius R, which is concentric with a hollow metallic sphere of radius r. The metallic sphere is maintained at a constant electric potential of 1 Volt.

Calculating Charge on the Metallic Sphere

We know that a hollow metallic sphere acts as a conductor. The electric potential (V) on the surface and inside a charged conducting sphere of radius r holding a charge Q is given by the formula:

$$V=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}$$

Here:

- V is the potential (given as 1 Volt).
- r is the radius of the metallic sphere.
- Q is the charge on the metallic sphere.
- ${ϵ}_0$ is the permittivity of free space.

We can rearrange this formula to find the charge Q on the sphere:

$$Q=V\times (4\pi {ϵ}_{0}r)$$

Substituting the given potential $V=1$ Volt:

$$Q=1\times (4\pi {ϵ}_{0}r)$$

$$Q=4\pi {ϵ}_{0}r$$

So, the total charge on the hollow metallic sphere is $4\pi {ϵ}_{0}r$ Coulombs.

Applying Gauss's Law for Electric Flux

Gauss's Law is a fundamental principle in electrostatics that relates the electric flux (${\Phi }_E$) through a closed surface to the net charge enclosed ($Q_{enc}$) within that surface. The law states:

$${\Phi }_E=\frac{Q_{enc}}{{ϵ}_0}$$

We are interested in the electric flux through a concentric spherical surface of radius R, where R > r. This larger spherical surface acts as our Gaussian surface.

The charge enclosed ($Q_{enc}$) by this surface of radius R is simply the charge present on the inner metallic sphere of radius r, which we calculated as Q.

Therefore, $Q_{enc}=Q=4\pi {ϵ}_{0}r$.

Now, we can calculate the electric flux through the spherical surface of radius R using Gauss's Law:

$${\Phi }_E=\frac{4\pi {ϵ}_{0}r}{{ϵ}_0}$$

Simplifying the expression by canceling out ${ϵ}_0$:

$${\Phi }_E=4\pi {ϵ}_{0}r$$

This result shows that the electric flux depends on the charge enclosed and is independent of the radius of the outer Gaussian surface (as long as it encloses the charge).

Conclusion

The total electric flux coming out of the concentric spherical surface of radius R is $4\pi {ϵ}_{0}r$. This matches the first option provided.