Animated Solution for Physics - Electrostatics: For spherical symmetrical charge distribution, variation of electric potential with distance from centre is given in diagram. Given that $V=\frac{q}{4\pi {\epsilon }_0{R}_0}$ for $r\le R_0$ and $V=\frac{q}{4\pi {\epsilon }_{0}r}$ for $r\ge R_0$ Then, which option(s) are correct

Imagine you are an explorer mapping out the electrical landscape of a mysterious spherical object. You are given a map—a graph showing how the electric potential $V$ changes as you move away from the center.

The graph reveals a fascinating secret: from the center all the way out to a radius $R_0$, the potential is perfectly flat and constant. It doesn't change at all! But the moment you step outside this boundary, the potential starts dropping off, following a $1/r$ curve, exactly like the potential of a simple point charge.

What does a flat, constant potential mean for the electric field? We know that the electric field is the negative gradient of the potential, mathematically written as $E=-\frac{dV}{dr}$.

If the potential $V$ is constant, its derivative is zero. This means that everywhere inside the sphere ($r<R_0$), the electric field is absolutely zero. There is no electrical push or pull inside this region.

Now, let's bring in Gauss's Law, one of the most powerful tools in electromagnetism. Gauss's Law tells us that the electric flux through any closed surface is proportional to the enclosed charge.

If we draw a spherical Gaussian surface anywhere inside $R_0$, the electric field on it is zero, meaning the flux is zero. Consequently, the enclosed charge must be zero. There is no charge hiding inside the volume of this sphere!

However, outside the sphere, the potential $V=\frac{q}{4\pi {\epsilon }_{0}r}$ tells us that the system behaves exactly as if a total charge $q$ were concentrated at the center. Since we just proved there is no charge inside, all of this charge $q$ must be crowded exactly on the boundary surface at $r=R_0$. This is the classic behavior of a charged conducting spherical shell.

Let's evaluate the given options based on our discoveries:

Option (a): Since all the charge $q$ is located at $r=R_0$, any larger sphere, such as one with radius $2R_0$, will enclose the entire charge $q$. This statement is correct.

Option (b): Electrostatic energy is stored in the electric field itself, with an energy density of $u=\frac{1}{2}{\epsilon }_0{E}^2$. Because the electric field $E$ is zero everywhere inside the shell ($r\le R_0$), the stored electrostatic energy in this region is perfectly zero. This statement is correct.

Option (c): Let's look at the electric field right at the boundary $r=R_0$. Just inside the shell, $E=0$. Just outside the shell, the field suddenly jumps to $E=\frac{q}{4\pi {\epsilon }_0{R}_0^2}$. This abrupt jump means the electric field is mathematically discontinuous at the surface. This statement is correct.

Option (d): As we deduced using Gauss's Law and the potential function, there is no charge inside the volume, and no charge outside. The entire charge $q$ is strictly confined to the surface $r=R_0$. This statement is correct.

In conclusion, all four options beautifully describe the physics of a uniformly charged spherical shell!