Animated Solution for Physics - Electrostatics: Three concentric spherical metallic shells, $A$, $B$ and $C$ of radii $a$, $b$ and $c$ ($a<b<c$) have surface charge densities $\sigma$, $-\sigma$ and $\sigma$ respectively. (a) Find the potential of the three shells $A$, $B$ and $C$. (b) If the shells $A$ and $C$ are at the same potential, obtain the relation between the radii $a$, $b$ and $c$.

The problem of concentric spherical shells is a classic in electrostatics, often appearing in JEE to test a student's grasp of the principle of superposition and Gauss's Law. Let's embark on a journey to unravel the electrical secrets hidden within these metallic layers.

Imagine three perfectly spherical, infinitely thin metallic shells nested inside one another, much like Russian Matryoshka dolls. We label them $A$, $B$, and $C$ from the inside out, with radii $a$, $b$, and $c$ respectively ($a<b<c$).

These shells are not electrically neutral. They possess surface charge densities of $\sigma$, $-\sigma$, and $\sigma$. Before we can calculate any potentials, we must determine the absolute charge residing on each shell. Since surface charge density is defined as charge per unit area ($\sigma =\frac{q}{A}$), the total charge is simply the density multiplied by the surface area of the sphere ($4\pi r^2$).

Thus, the charges are: $q_A=\sigma (4\pi a^2)$ $q_B=-\sigma (4\pi b^2)$ * $q_C=\sigma (4\pi c^2)$

To find the potential at any point, we rely on a fundamental property of charged spherical conductors.

Inside or on the surface of a charged shell of radius $R$, the electric field is zero, meaning no work is done moving a charge around. Consequently, the potential is constant and equals the potential at the surface:

Outside the shell, it behaves exactly as if all its charge were concentrated at a single point at its center. At a distance $r>R$:

Armed with these two rules and the principle of superposition, we can conquer any point in this system.

Let's find the potential $V_A$ right on the surface of the innermost shell $A$ (at distance $r=a$). We must sum the contributions from all three shells.

From Shell A: We are on its surface, so we use its radius $a$. From Shell B: We are inside shell B, so its contribution is constant and depends on its radius $b$. From Shell C: We are inside* shell C, so its contribution depends on its radius $c$.

Substituting the charges we found earlier:

Notice how the $4\pi$ terms cancel out beautifully, and one power of the radius in the numerator cancels with the denominator. Factoring out $\frac{\sigma }{{\epsilon }_0}$, we get a remarkably clean expression:

Now, let's move to the middle shell $B$ (at distance $r=b$).

From Shell A: We are outside shell A, so we use our current distance $b$. From Shell B: We are on its surface, so we use its radius $b$. From Shell C: We are inside* shell C, so we use its radius $c$.

Substituting the charges:

Simplifying this yields:

Finally, we evaluate the potential $V_C$ on the outermost shell $C$ (at distance $r=c$).

From Shell A: We are outside shell A, so we use distance $c$. From Shell B: We are outside shell B, so we use distance $c$. From Shell C:* We are on its surface, so we use its radius $c$.

Because we are at or outside all shells, the denominator for every term is simply $c$!

Substituting the charges:

This simplifies to our final potential expression:

Part (b) of the problem introduces a fascinating constraint: What if the potential of the innermost shell equals the potential of the outermost shell? Let's enforce $V_A=V_C$.

Immediately, the $\frac{\sigma }{{\epsilon }_0}$ terms cancel out. We can also split the fraction on the right side:

Subtracting $c$ from both sides leaves us with:

We recognize the numerator on the right as a difference of squares, which expands to $(a-b)(a+b)$:

Here is where we must be mathematically rigorous. Can we divide both sides by $(a-b)$? Yes, because the problem explicitly states that $a<b$. Therefore, $a-b$ is strictly negative and not equal to zero. Dividing by a non-zero quantity is perfectly valid.

Multiplying by $c$, we arrive at the final, elegant geometric relation:

$$a+b=c$$

This tells us that for this specific electrical state to exist, the radius of the outermost shell must be exactly equal to the sum of the radii of the two inner shells. It is a beautiful example of how complex electrostatic interactions can distill down to a simple, profound geometric truth.