Animated Solution for Physics - Electrostatics: The given graph shows variation (with distance $r$ from centre) of

Field vs. Potential

When you encounter a graphical question in electrostatics, the secret to cracking it lies in reading the mathematical story hidden in the curves. Let's break down this classic JEE problem step by step.

Look closely at the given graph. It is divided into two distinct regions based on the distance $r$ from the center: 1. Inside the boundary ($r<r_0$): The graph is a perfectly horizontal line. This means the physical quantity is constant and, importantly, non-zero. 2. Outside the boundary ($r>r_0$): The graph curves downwards, indicating that the quantity decreases as the distance $r$ increases.

Our job is to match this visual fingerprint with the correct physical scenario from the options.

Let's test the first suspect: the electric field of a uniformly charged spherical shell. By Gauss's Law, we know that the charge enclosed by any Gaussian surface inside the shell is zero. Therefore, the electric field inside is exactly zero ($E=0$ for $r<r_0$). Since our graph shows a non-zero constant value inside, we can immediately rule out the electric field.

Now, let's examine the electric potential of the same spherical shell. The relationship between electric field and potential is given by $E=-\frac{dV}{dr}$. Since $E=0$ inside the shell, the derivative of the potential must be zero. This means the potential $V$ must be a constant everywhere inside the shell, and its value is equal to the potential at the surface:

Outside the shell ($r>r_0$), the shell behaves exactly like a point charge concentrated at its center. The potential falls off inversely with distance:

This perfectly matches our graph! A horizontal line inside, and a $1/r$ curve outside.

To build a rock-solid intuition, let's ask: what if the options were about a uniformly charged solid sphere?

For a solid sphere, charge is distributed throughout its volume. The electric field inside grows linearly with distance ($E\propto r$), which would look like a straight line passing through the origin.

The potential inside a solid sphere is even more interesting. It follows a parabolic curve given by:

This would look like an inverted parabola starting from a maximum value at the center and smoothly joining the $1/r$ curve at the surface. Since our graph is flat inside, it definitely represents a hollow shell, not a solid sphere.

By systematically analyzing the behavior of the function inside and outside the boundary, we can confidently conclude that the graph represents the potential of a uniformly charged spherical shell.