Animated Solution for Physics - Electrostatics: A charge $Q$ is distributed over two concentric conducting thin spherical shells radii $r$ and $R$ ($R>r$). If the surface charge densities on the two shells are equal, the electric potential at the common centre is

Imagine you are standing at the center of two massive, concentric conducting spheres. The inner sphere has a radius $r$, and the outer sphere has a larger radius $R$. A total charge $Q$ has been sprinkled over these two shells.

Our first task is to figure out exactly how much of this total charge $Q$ resides on the inner shell (let's call it $Q_1$) and how much is on the outer shell (let's call it $Q_2$). We know that charge is conserved, so we can immediately write down our first fundamental equation:

Equal Charge Densities

The problem gives us a beautiful constraint: the surface charge densities on both shells are perfectly equal. Let's denote this surface charge density as $\sigma$.

Recall that surface charge density is simply the total charge on a surface divided by the area of that surface. For a sphere, the surface area is $4\pi \times {\text{radius}}^2$. Therefore, we can express the charge densities for both shells and equate them:

This equation is the key to unlocking the problem. By canceling out the common $4\pi$ terms from the denominators, we get a clean, direct relationship between the charges and their respective radii:

From this, we can easily express $Q_2$ in terms of $Q_1$:

Now, let's bring this relationship back to our very first equation, $Q_1+Q_2=Q$. By substituting our expression for $Q_2$, we get an equation entirely in terms of $Q_1$:

Let's factor out $Q_1$ to isolate it:

Solving for $Q_1$, we find the exact amount of charge on the inner shell:

Because the geometry is perfectly symmetric, we don't even need to do the algebra again to find $Q_2$. We can simply swap the $r^2$ in the numerator for an $R^2$:

We are finally ready to find the electric potential at the common center of the shells. According to the principle of superposition, the total potential at the center is simply the scalar sum of the potentials created by each individual shell.

Here is a crucial piece of physics intuition: the electric potential everywhere inside a charged conducting spherical shell is constant and equal to the potential on its surface. Therefore, the potential at the center due to the inner shell is $\frac{1}{4\pi {\epsilon }_0}\frac{Q_1}{r}$, and the potential due to the outer shell is $\frac{1}{4\pi {\epsilon }_0}\frac{Q_2}{R}$.

Adding them together gives us our master equation for the potential $V$:

Let's substitute the expressions for $Q_1$ and $Q_2$ that we worked so hard to find:

Notice the beautiful cancellation that happens next. In the first term, the $r$ in the denominator cancels one power of $r$ in the numerator. In the second term, the $R$ in the denominator cancels one power of $R$ in the numerator:

Finally, we can factor out the common terms $\frac{Q}{R^2+r^2}$ to arrive at our elegant final answer:

This matches option (d) perfectly. The physics here is a wonderful dance between geometry (the surface areas) and the fundamental laws of electrostatics (superposition and potential).