Animated Solution for Physics - Electrostatics: Concentric metallic hollow spheres of radii $R$ and $4R$ hold charges $Q_1$ and $Q_2$, respectively. Given that, surface charge densities of the concentric spheres are equal. The potential difference $V(R)-V(4R)$ is

Imagine you are looking at two concentric metallic hollow spheres. The inner sphere has a radius $R$ and carries a charge $Q_1$. Surrounding it is a larger outer sphere with a radius $4R$ and a charge $Q_2$.

The problem asks us to find the potential difference between the surfaces of these two spheres, specifically $V(R)-V(4R)$. Interestingly, the problem also mentions that the surface charge densities of both spheres are equal. Let's hold onto that piece of information and see if we actually need it.

To tackle this, we must rely on the fundamental principle of superposition and the behavior of electric potential for a charged spherical shell.

The Golden Rule: For a uniformly charged spherical shell of radius $R$ and charge $q$, the electric potential $V$ at a distance $r$ from the center is given by:

This means that outside the shell, it behaves exactly like a point charge located at its center. However, inside the shell, the electric field is zero, which implies the potential is constant everywhere inside and is equal to the potential at its surface.

Let's calculate the total potential at the surface of the inner sphere, $V(R)$. By the principle of superposition, this is the sum of the potentials created by $Q_1$ and $Q_2$ at a distance $R$ from the center.

For the inner charge $Q_1$, we are exactly at its surface, so its contribution is $\frac{k{Q}_1}{R}$. For the outer charge $Q_2$, we are inside the outer shell. Therefore, its contribution is constant and equals its surface potential, which is $\frac{k{Q}_2}{4R}$.

Next, let's find the total potential at the surface of the outer sphere, $V(4R)$.

At a distance $4R$, we are outside the inner shell, so it behaves like a point charge, contributing $\frac{k{Q}_1}{4R}$. For the outer shell itself, we are at its surface, so it contributes $\frac{k{Q}_2}{4R}$.

Now comes the beautiful part. We need to find the potential difference $V(R)-V(4R)$. Let's subtract the two equations we just derived:

Notice what happens to the terms involving $Q_2$? They perfectly cancel each other out!

This is a profound physical insight: The potential difference between two concentric conducting shells depends entirely on the charge of the inner shell. The outer shell raises or lowers the potential of the entire inner region uniformly, so it does not contribute to any difference in potential between the two shells.

Let's simplify the remaining expression by taking $\frac{k{Q}_1}{R}$ as a common factor:

Finally, we substitute the standard value for Coulomb's constant, $k=\frac{1}{4\pi {\epsilon }_0}$:

And what about the equal surface charge densities? It was a classic distractor! We arrived at the correct answer without ever needing to use it. Always trust the fundamental physics principles over the urge to use every single number given in a problem.