Q.Prove that a closed equipotential surface with no charge within itself must enclose an equipotential volume.

Concept: Electric Potential — Inside a charge-free region, the potential satisfies Laplace’s equation, and its behaviour on a closed surface determines the potential throughout the enclosed volume.

Reasoning:

1. On the closed equipotential surface, $V = V_0$ (constant). Inside, there is no charge, so $\nabla^2 V = 0$ (Laplace’s equation). …

Why This Must Be True

Imagine you have a closed surface — like a balloon — that is everywhere at the same electric potential. Inside this balloon, there is no electric charge. The question asks: can the potential inside be different from the potential on the surface? The answer is no, and here's why.

Electric potential is a continuous function in space (except at point charges). If the potential were higher or lower somewhere inside, there would be a potential difference between that point and the surface. A potential difference implies an electric field, since $\vec{E} = -\nabla V$. But an electric field inside a charge-free region must obey certain rules — and those rules forbid it from existing under these conditions.

Let's prove this properly.

Step-by-Step Proof

1. Set up the problem

We have a closed equipotential surface $S$ with potential $V = V_0$ (constant). The volume $V$ enclosed by $S$ contains no electric charge. We want to show that at every point inside $S$, the potential is also $V_0$.

2. Use the uniqueness theorem for Laplace's equation

Inside the volume, since there is no charge, the potential satisfies Laplace's equation:

The boundary condition is that on the surface $S$, $V = V_0$ (constant). The uniqueness theorem says: if a solution to Laplace's equation exists that satisfies the boundary conditions, it is the only solution.

3. Guess a solution and check it

Consider the constant function $V(x,y,z) = V_0$ everywhere inside the volume. Does it satisfy Laplace's equation? Yes — the Laplacian of a constant is zero. Does it match the boundary condition? Yes — on the surface, $V = V_0$. So $V = V_0$ everywhere inside is a valid solution.

By the uniqueness theorem, this must be the only solution. Therefore, the potential is $V_0$ at every interior point.

4. Alternative argument using Gauss's law (for intuition)

If you prefer a more physical argument: suppose there were a point inside where $V \neq V_0$. Then there would be an electric field $\vec{E} = -\nabla V$ in some region. Since the surface is equipotential, the field lines would have to start and end inside the volume (they can't cross the surface because that would require a potential difference on the surface). But field lines begin and end on charges. Since there are no charges inside, this is impossible. The only way out is that $\vec{E} = 0$ everywhere inside, which means $V$ is constant — and that constant must match the boundary value $V_0$. …

Method: Proving an Electrostatic Result via the Uniqueness Theorem

Use this method for "prove that..." questions about potential or field inside a charge-free region bounded by a surface with known potential (or known field) behaviour.

Steps

Step 1: Confirm the region is charge-free and state the governing equation

Inside a charge-free region, the potential obeys Laplace's equation:

Step 2: Write down the boundary condition given in the problem

State exactly what is specified on the boundary of the region — e.g. "$V$ is constant ($=V_0$) on the enclosing surface." This boundary condition is the second ingredient the uniqueness theorem needs.

Step 3: Invoke the uniqueness theorem

$\text{If } \nabla^2 V = 0 \text{ inside a region AND } V \text{ matches a given boundary condition, that solution is the ONLY one possible.}$ …