Animated Solution for Physics - Electrostatics: A thin spherical conducting shell of radius $R$ has a charge $q$. Another charge $Q$ is placed at the centre of the shell. The electrostatic potential at a point $P$ at a distance $R/2$ from the centre of the shell is

The problem of finding the electrostatic potential inside a charged conducting shell with a central point charge is a classic example of the elegance of the superposition principle. At first glance, the presence of multiple charges and a conducting boundary might seem intimidating, but by breaking the system down into its fundamental components, the solution becomes remarkably straightforward.

Imagine you are standing inside a thin spherical conducting shell of radius $R$. This shell carries a total charge $q$. Right at the center of this spherical room, there is a point charge $Q$. Our objective is to determine the total electrostatic potential at a specific point $P$, which is located halfway between the center and the shell, at a distance of $R/2$.

To tackle this, we rely on the Principle of Superposition. This principle states that the total electric potential at any point in space is simply the algebraic sum of the potentials produced by each individual charge configuration acting alone.

Let's first isolate the conducting shell. What is the potential at point $P$ due solely to the charge $q$ on the shell?

Here, we must recall a beautiful property of conductors in electrostatics: the electric field inside a hollow, charged conducting shell is exactly zero. Because the electric field is the spatial derivative of the potential ($E=-dV/dr$), a zero electric field implies that the potential does not change with position.

Therefore, the entire volume inside the shell is an equipotential region. The potential anywhere inside is exactly the same as the potential on the surface of the shell.

Notice that this value is completely independent of the distance $R/2$. Whether you are at the center, at $R/2$, or just inside the surface, the shell contributes this exact same constant potential.

Now, let's look at the point charge $Q$ located at the center. We need to find its contribution to the potential at point $P$.

Point $P$ is at a distance of $r=R/2$ from this central charge. Using the standard formula for the electric potential of a point charge, we get:

By bringing the $2$ to the numerator, this simplifies neatly to:

What About Induced Charges?

A sharp student might ask: "Wait, doesn't the charge $Q$ at the center induce charges on the conducting shell?"

Yes, it absolutely does! The positive charge $Q$ will attract electrons, inducing a charge of $-Q$ on the inner surface of the shell, and leaving a charge of $+Q$ on the outer surface.

However, here is the brilliant part: we don't need to worry about them for this specific calculation. Why? Because the potential at any point inside the cavity due to these induced charges perfectly cancels out. The uniform $-Q$ on the inner surface creates a constant potential of $-Q/(4\pi {\epsilon }_{0}R)$ everywhere inside. The uniform $+Q$ on the outer surface creates a constant potential of $+Q/(4\pi {\epsilon }_{0}R)$ everywhere inside.

When you add them up, their net contribution to the potential inside the shell is exactly zero! Thus, our initial superposition of just the original shell charge $q$ and the central charge $Q$ is perfectly valid.

Bringing it all together, we simply add the two contributions we calculated earlier:

This matches option (c) perfectly. The beauty of this problem lies in trusting the superposition principle and understanding the equipotential nature of conductors. Once you grasp these two concepts, even complex-looking electrostatic setups become a walk in the park.