Q.Can the potential function have a maximum or minimum in free space?

Concept: Electric Potential — In free space (no charges), the potential $V$ satisfies Laplace's equation $\nabla^2 V = 0$.

Reasoning:

1. Laplace's equation states that the sum of the second partial derivatives of $V$ is zero everywhere in free space.
2. A maximum or minimum of $V$ would require all second derivatives to have the same sign (all negative for a maximum, all positive for a minimum). …

Why This Question Matters

This isn’t just a yes/no puzzle — it gets at the very heart of electrostatics. The electric potential in free space is not arbitrary; it’s governed by a deep mathematical constraint. Understanding why a maximum or minimum is impossible reveals how charges shape the field around them, and why a test charge placed in empty space can never be in stable equilibrium without external forces.

The key is Laplace’s equation.

The Concept: What “Free Space” Means

“Free space” means a region with no electric charges — no point charges, no dipoles, no charge density anywhere inside that region. In such a region, Gauss’s law tells us:

Since the electric field is the negative gradient of the potential ($\mathbf{E} = -\nabla V$), this becomes:

This single equation is the entire reason for the answer. Now let’s unpack what it means for maxima and minima.

Step-by-Step Reasoning

1. What does $\nabla^2 V = 0$ say about curvature?

The Laplacian $\nabla^2 V$ is the sum of the second partial derivatives in all directions. In one dimension, $\frac{d^2 V}{dx^2} = 0$ means the graph is a straight line — no curvature, so no maximum or minimum (unless the line is flat, which is a degenerate case). In three dimensions, the condition is more subtle: the average curvature is zero.

At a point where $V$ has a local maximum, the potential curves downward in every direction — all second derivatives are negative. Their sum would then be negative. Similarly, at a local minimum, all second derivatives are positive, so their sum would be positive. But Laplace’s equation demands that sum be exactly zero. Therefore, a true local maximum or minimum cannot exist.

2. The only possibility: saddle points

If the sum of curvatures is zero, some directions must curve upward and others downward, balancing out. This is the signature of a saddle point — like the centre of a Pringle chip, where one direction is a maximum and the perpendicular direction is a minimum. The potential can have such points, but never a pure peak or valley.

3. The mathematical proof (mean value property)

There’s a beautiful theorem that makes this crystal clear: any solution of Laplace’s equation satisfies the mean value property. For any point in free space, the value of $V$ at that point equals the average of $V$ over any sphere centred there (provided the sphere lies entirely in free space). …

Method: Testing a Potential Function for a Maximum or Minimum

Use this method whenever you're asked whether the electrostatic potential can have a local maximum, minimum, or other extremum at some point in a given region.

Steps

Step 1: Identify whether the region is charge-free

Check whether the point in question lies in a region containing no charge (free space/vacuum, no source charges nearby). This matters because the governing equation for $V$ changes depending on whether charge is present.

Step 2: Apply the governing equation

In a charge-free region, $V$ must satisfy Laplace's equation:

This says the sum of the second derivatives of $V$ along every direction is exactly zero — the "curvatures" in different directions must cancel.

Step 3: Reason about what a true maximum/minimum would require …