Q.A regular hexagon of side $10\ \text{cm}$ has a charge $5\ \mu\text{C}$ at each of its vertices. Calculate the potential at the centre of the hexagon.

Concept: Electric Potential — the potential at a point due to a system of point charges is the scalar sum of the potentials due to each charge.

Step 1: For a regular hexagon, the centre is equidistant from all vertices. The distance from the centre to any vertex equals the side length.

Given side $a = 10\ \text{cm} = 0.1\ \text{m}$, so $r = 0.1\ \text{m}$.

Step 2: Potential due to a single charge $q = 5\ \mu\text{C} = 5 \times 10^{-6}\ \text{C}$ at distance $r$ is

Step 3: Since potential is scalar, the net potential at the centre is the sum from all 6 vertices:

Why potential, not field?

The question asks for potential at the centre — a scalar quantity. This is much simpler than finding the electric field. Potential adds as numbers (with sign), so we don't need to worry about directions or vector components. Each charge contributes independently, and we just sum them up.

The key insight: in a regular hexagon, all six vertices are at the same distance from the centre. And all charges are identical ($+5\ \mu\text{C}$ each). So the potential at the centre is simply six times the potential due to one charge.

Step-by-step

1. Find the distance from centre to any vertex

For a regular hexagon of side $a$, the distance from the centre to any vertex equals the side length $a$. Why? A regular hexagon can be divided into six equilateral triangles, each of side $a$. The centre is the common vertex of all six triangles, and the distance from centre to any outer vertex is exactly the side of that equilateral triangle.

So here, $r = 10\ \text{cm} = 0.1\ \text{m}$.

2. Potential due to a single point charge

The electric potential at a distance $r$ from a point charge $q$ is:

where $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9\ \text{N m}^2/\text{C}^2$.

3. Potential due to one vertex charge

4. Total potential at centre

Since potential is a scalar, we simply add the contributions from all six vertices:

5. Check the units

$9 \times 10^9 \times \frac{5 \times 10^{-6}}{0.1}$ gives units of $\frac{\text{N m}^2/\text{C}^2 \cdot \text{C}}{\text{m}} = \text{N m}/\text{C} = \text{J}/\text{C} = \text{V}$. All good.

Method: Potential at a Point Due to Several Point Charges Arranged Symmetrically

This method applies whenever multiple point charges sit at the vertices of a symmetric shape (polygon, ring, etc.) and you need the potential at a special point such as the centre.

Steps

Step 1: Confirm potential is a scalar, so no vector components are needed

Unlike an electric-field superposition problem, you never resolve into components or worry about cancellation by direction — every contribution simply adds algebraically, with its own sign.

Step 2: Use the geometry to find the common distance

For many symmetric shapes, the distance from the centre (or point of interest) to every vertex is the same — e.g. in a regular hexagon the centre-to-vertex distance equals the side length, because the hexagon divides into six equilateral triangles meeting at the centre. Work out this common distance $r$ from the given geometry before doing anything else.

Step 3: Compute the potential due to a single charge

Step 4: Multiply by the number of identical, equidistant charges

If the charges are not all identical or not all equidistant, this shortcut fails — instead sum $\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i}$ individually for each charge, respecting its own sign.