Animated Solution for Physics - Electrostatics: Three concentric metal shells $A,B$ and $C$ of respective radii $a,b$ and $c$ ($a<b<c$) have surface charge densities $+\sigma ,-\sigma$ and $+\sigma$, respectively. The potential of shell $B$ is

The problem of finding the electric potential of concentric spherical shells is a classic and beautiful application of the principle of superposition in electrostatics. It tests your fundamental understanding of how a charged spherical shell behaves both inside and outside its boundary.

Let's embark on a journey to decode this problem step-by-step, ensuring that the underlying physics becomes second nature to you.

Imagine three perfectly concentric, incredibly thin metal shells. Let's call them $A$, $B$, and $C$, moving from the innermost to the outermost. Their radii are $a$, $b$, and $c$ respectively, such that $a<b<c$.

We are given their surface charge densities: $+\sigma$ for shell $A$, $-\sigma$ for shell $B$, and $+\sigma$ for shell $C$.

Before we can calculate any potential, we need to know the actual amount of charge residing on each shell. The total charge $Q$ on a spherical shell is simply its surface charge density multiplied by its surface area ($4\pi r^2$).

Therefore, the charges on the three shells are:

Principle of Superposition

The electric potential is a scalar quantity. This is a massive relief because it means we don't have to worry about vector addition or complex angles. The total potential at any point in space is simply the algebraic sum of the potentials created by each individual charge distribution.

We need to find the potential exactly on the surface of shell $B$. Let's call this $V_B$. According to the principle of superposition:

Now, we must apply the golden rules of spherical shells: 1. Outside a shell ($r>R$): The shell behaves exactly as if all its charge were concentrated at its center. The potential is $V=\frac{kQ}{r}$. 2. On the surface ($r=R$): The potential is $V=\frac{kQ}{R}$. 3. Inside a shell ($r<R$): The electric field is zero, which means no work is done moving a charge around inside. Consequently, the potential everywhere inside is constant and equal to the potential on its surface. The potential is $V=\frac{kQ}{R}$.

Let's evaluate the contribution of each shell to the potential at the location of shell $B$ (which is at a distance $b$ from the center).

Contribution from Shell A: Shell $B$ is outside shell $A$ (since $b>a$). Therefore, shell $A$ acts like a point charge at the center.

Contribution from Shell B: We are calculating the potential exactly on the surface of shell $B$.

Contribution from Shell C: Shell $B$ is inside shell $C$ (since $b<c$). The potential anywhere inside shell $C$ is equal to the potential on its surface.

Putting it all together, the net potential at shell $B$ is:

Now, we substitute the expressions for the charges $Q_A$, $Q_B$, and $Q_C$ that we found earlier. We also substitute $k=\frac{1}{4\pi {\epsilon }_0}$.

Notice how beautifully the $4\pi$ terms cancel out across the entire equation. We can also factor out the surface charge density $\sigma$.

Simplifying the terms inside the bracket yields our final, elegant result:

This matches option (b). The beauty of this problem lies in carefully choosing the correct distance ($r$ or $R$) for each term based on whether the point of interest lies inside, on, or outside the respective shell.