Animated Solution for Physics - Electrostatics: Three concentric metallic spherical shells of radii $R,2R$ and $3R$ are given charges $Q_1,Q_2$ and $Q_3$, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, $Q_1:Q_2:Q_3$, is

The problem of concentric metallic shells is a classic in electrostatics, testing your deep understanding of Gauss's Law and charge induction. Let's peel this problem layer by layer, just like an onion!

A Metallic Onion

Imagine three perfectly concentric metallic spherical shells. Their radii are $R$, $2R$, and $3R$. We are told that charges $Q_1$, $Q_2$, and $Q_3$ are given to these shells respectively.

The most crucial piece of information given is this: the surface charge densities on the outer surfaces of all three shells are equal. Let's call this uniform outer surface charge density $\sigma$.

Before we dive into the math, we must arm ourselves with the fundamental principles of conductors in electrostatics: 1. Charge resides on the surfaces: For a metallic shell, any charge given to it will distribute itself on its inner and outer surfaces. So, $Q_{\text{total}}=q_{\text{inner}}+q_{\text{outer}}$. 2. Gauss's Law for Conductors: The electric field inside the bulk of a conductor is always zero. If we draw a Gaussian surface inside the metal of a shell, the net charge enclosed must be zero. This means the inner surface of any shell will always induce a charge exactly equal and opposite to the total charge enclosed within its cavity. $q_{\text{inner}}=-Q_{\text{enclosed}}$.

Shell by Shell

Let's apply these rules starting from the innermost shell and moving outwards.

Shell 1 (Radius $R$): There is nothing inside this shell, so the enclosed charge is zero. The outer surface has a charge density $\sigma$. The area of this spherical surface is $4\pi R^2$. The total charge on the first shell is simply the sum of its surface charges:

Shell 2 (Radius $2R$): This shell encloses the first shell, which has a total charge of $Q_1$. To maintain a zero electric field inside the metal of Shell 2, its inner surface must induce a charge opposite to $Q_1$. The outer surface of Shell 2 also has a charge density $\sigma$. Its radius is $2R$, so its area is $4\pi (2R{)}^2=16\pi R^2$. The total charge given to the second shell is:

Shell 3 (Radius $3R$): This outermost shell encloses both Shell 1 and Shell 2. The total enclosed charge is $Q_1+Q_2$. Notice a beautiful pattern here: $Q_1+Q_2=4\pi \sigma R^2+12\pi \sigma R^2=16\pi \sigma R^2$, which is exactly equal to $q_{2,\text{outer}}$! The inner surface of Shell 3 will induce a charge opposite to this. The outer surface of Shell 3 has a charge density $\sigma$ and a radius of $3R$. Its area is $4\pi (3R{)}^2=36\pi R^2$. The total charge given to the third shell is:

The Ratio

We have successfully found the total charges given to each shell in terms of $\sigma$ and $R$: $Q_1=4\pi \sigma R^2$ $Q_2=12\pi \sigma R^2$ * $Q_3=20\pi \sigma R^2$

To find the ratio $Q_1:Q_2:Q_3$, we simply divide by the common factor $4\pi \sigma R^2$:

The beauty of this problem lies in the systematic application of Gauss's Law. By carefully tracking the induced charges layer by layer, a seemingly complex system simplifies into an elegant ratio.