Animated Solution for Physics - Electrostatics: A charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R$ ($>r$) such that the surface densities are equal. Find the potential at the common centre.

Welcome to a beautiful problem from JEE Advanced 1981. This question elegantly combines the concepts of surface charge density, charge distribution, and the principle of superposition to find the electric potential at the center of a system of concentric spheres.

Imagine you are standing at the exact center of two massive, concentric hollow spheres. The inner sphere has a radius $r$, and the outer sphere has a larger radius $R$. A total charge $Q$ is sprinkled over this entire system.

Let's assume the inner sphere takes a portion of this charge, say $q_1$, and the outer sphere takes the rest, $q_2$. Naturally, the sum of these individual charges must equal the total charge given to the system:

Surface Charge Density

The problem gives us a golden key to unlock the charge distribution: the surface charge densities of both spheres are exactly equal.

We know that surface charge density, denoted by $\sigma$, 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, equating the surface charge densities ${\sigma }_1$ and ${\sigma }_2$, we get:

From the equation above, we can easily find the ratio of the charges on the two spheres. By rearranging the terms, we see that the charges are distributed in the ratio of the squares of their radii:

Now, we need to express $q_1$ and $q_2$ individually in terms of the total known charge $Q$. Using the ratio we just found and the fact that $q_1+q_2=Q$, we can write the individual charges as fractions of the total charge:

Potential at the Center

Now for the grand finale: finding the electric potential at the common center, $O$.

Here is a crucial concept in electrostatics: The electric potential at any point inside a charged hollow conducting sphere is constant and is exactly equal to the potential on its surface.

Therefore, the potential at the center due to the inner sphere is simply $V_1=\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r}$, and the potential due to the outer sphere is $V_2=\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{R}$.

Because electric potential is a scalar quantity, we can use the principle of superposition and simply add these two potentials algebraically to find the total potential $V$ at the center:

Let's substitute our expressions for $q_1$ and $q_2$ into the potential equation:

Notice the beautiful cancellation that happens here! One $r$ cancels out in the first term, and one $R$ cancels out in the second term.

Finally, by factoring out the common terms $\frac{Q}{4\pi {\epsilon }_0(r^2+R^2)}$, we arrive at our elegant final answer:

This result beautifully demonstrates how symmetry and fundamental electrostatic principles can simplify seemingly complex charge distributions.