Q.A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. A charge $Q$ is placed at the centre of the spherical cavity. What will be the surface charge density on

Concept: Gauss's law + electrostatic shielding in conductors.

When charge $Q$ sits at the center of the cavity, it induces charge on the conductor. Inside any conductor in electrostatic equilibrium, the electric field is zero. Applying Gauss's law to a surface just inside the metal shows that the inner surface must carry total charge $-Q$ to cancel the field from $Q$. Since the shell was initially neutral, the outer surface must carry $+Q$ to conserve charge.

Step 1 (Inner surface): The induced charge $-Q$ spreads uniformly over the inner surface of area $4\pi R_1^2$, so

$\sigma_{\text{inner}} = \frac{-Q}{4\pi R_1^2}.$ …

When a charge sits inside a conducting cavity, the conductor responds by rearranging its free electrons until equilibrium is reached. The key insight is that the electric field inside a conductor must be zero in electrostatic equilibrium. This constraint, combined with Gauss's law, tells us exactly how charge distributes on the surfaces.

The charge $Q$ at the center creates an electric field that would penetrate into the metal. But conductors won't allow that—electrons move until they cancel any internal field. This movement leaves one surface with a deficit of electrons (positive charge) and the other with an excess (negative charge).

Finding the charge distribution

1. Apply Gauss's law inside the conductor

Draw a Gaussian surface anywhere inside the metal itself (between $R_1$ and $R_2$). Since the electric field is zero throughout the conductor, the flux through this surface is zero:

By Gauss's law, this means the enclosed charge is zero:

2. Determine the charge on the inner surface

The Gaussian surface encloses both the central charge $Q$ and whatever charge $q_{\text{inner}}$ sits on the inner surface at radius $R_1$. For the total to be zero:

Therefore:

The inner surface must carry charge $-Q$ to neutralize the field inside the conductor.

3. Determine the charge on the outer surface

The spherical shell as a whole is electrically neutral (we started with an uncharged conductor). If the inner surface has $-Q$, and the total charge on the shell is zero, then:

The outer surface carries $+Q$.

Computing the surface charge densities

4. Inner surface density

The charge $-Q$ spreads uniformly over the inner spherical surface of area $4\pi R_1^2$: …

Method: Electrostatic Induction in a Conductor (Gauss's Law + Charge Conservation)

Concept: In a conductor in electrostatic equilibrium, the electric field inside the conductor material is zero. Any net charge resides only on the surfaces. When a charge is placed inside a cavity, it induces an equal and opposite charge on the inner surface.

Steps:

1. Identify the conductor region

   The metallic shell occupies $R_1 < r < R_2$. Inside the metal, $\vec{E} = 0$.

2. Apply Gauss's Law to the inner surface

   Draw a Gaussian sphere of radius $r$ such that $R_1 < r < R_2$ (inside the metal).

   • Flux through this surface: $\oint \vec{E} \cdot d\vec{A} = 0$ (since $E=0$ inside conductor).
   • By Gauss's Law: $Q_{\text{enclosed}} = 0$.
   • Enclosed charge = charge at centre ($+Q$) + charge on inner surface ($Q_{\text{inner}}$).
   • Therefore: $Q + Q_{\text{inner}} = 0 \implies Q_{\text{inner}} = -Q$.

3. Find inner surface charge density

   Inner surface area = $4\pi R_1^2$.

4. Apply charge conservation for the outer surface …

Here are the common mistakes students make on this classic electrostatics problem, along with how to avoid each.

Mistake 1: Forgetting that charge resides only on the surfaces of a conductor

Many students assume the charge $Q$ at the centre somehow "spreads" throughout the shell material.

Why this is wrong: In electrostatic equilibrium, net charge inside a conductor is zero. The field inside the conducting material must be zero.

How to avoid:

• Draw a Gaussian surface inside the conductor (between $R_1$ and $R_2$).
• Since $E = 0$ inside the conductor, the total enclosed charge must be zero.
• This forces the inner surface to acquire a charge of $-Q$ (to cancel the $+Q$ at the centre).

Mistake 2: Thinking the inner and outer surface charges are independent

Some students calculate the inner surface charge density correctly but then forget that the shell itself is neutral (unless stated otherwise).

Why this is wrong: The shell is uncharged overall. If the inner surface gets $-Q$, the outer surface must get $+Q$ to keep the net charge zero.

How to avoid:

• Always check conservation of charge for the conductor.
• If the shell has a net charge $Q_{\text{net}}$, then:

• In this problem, $Q_{\text{net}} = 0$, so $Q_{\text{outer}} = +Q$.

Mistake 3: Using the wrong area for surface charge density

Students often use the wrong radius when computing $\sigma = \frac{Q}{A}$.

Common errors:

• Using $R_2$ for the inner surface area.
• Using $R_1$ for the outer surface area.
• Forgetting that area is $4\pi r^2$, not $\pi r^2$.

How to avoid:

• Inner surface area = $4\pi R_1^2$
• Outer surface area = $4\pi R_2^2$
• Then:

Mistake 4: Confusing the sign of the induced charge

Some students write $\sigma_{\text{inner}}$ as positive, thinking the central $+Q$ "repels" positive charge to the inner surface.

Why this is wrong: The central $+Q$ attracts negative charge to the inner surface (opposites attract). The induced charge on the inner surface is negative.

How to avoid:

• Use the Gaussian surface argument: Inside conductor, $E = 0 \implies Q_{\text{enclosed}} = 0$ So $Q_{\text{inner}} + Q_{\text{centre}} = 0 \implies Q_{\text{inner}} = -Q$
• The sign follows automatically from this condition.

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