Q.Calculate potential on the axis of a disc of radius $R$ due to a charge $Q$, uniformly distributed on its surface.

Concept: Electric Potential — the work done per unit charge to bring a test charge from infinity to a point, computed by integrating contributions from each charge element.

Reasoning:

1. Consider a ring of radius $r$ and width $dr$ on the disc. Its area is $2\pi r\,dr$, so the charge on the ring is $dq = \sigma \cdot 2\pi r\,dr$, where $\sigma = Q/(\pi R^2)$.

2. Every point on this ring is at distance $\sqrt{r^2 + x^2}$ from a point on the axis at distance $x$ from the centre. The potential due to the ring is

3. Integrate from $r=0$ to $r=R$: …

Why potential, not field?

When asked for potential, we have a huge advantage: potential is a scalar. No vector components to resolve. For a continuous charge distribution, we simply sum (integrate) the scalar contributions $dV = \frac{1}{4\pi\varepsilon_0} \frac{dq}{r}$ from every tiny piece of charge.

The disc is symmetric about its axis. That symmetry lets us break it into rings — each ring is an equal-distance contour from any point on the axis. This is the classic approach: rings → integration over radius.

Step-by-step solution

1. Set up coordinates and charge density

Place the disc in the $y$-$z$ plane, centred at the origin. Its axis is the $x$-axis. We want potential at point $P$ on the axis at distance $x$ from the centre.

Total charge $Q$ is uniformly spread over area $\pi R^2$. Surface charge density:

2. Choose an infinitesimal ring element

Consider a ring of radius $r$ and thickness $dr$ (so $r$ runs from $0$ to $R$). The area of this ring is its circumference times thickness:

Charge on this ring:

3. Distance from ring to point P

Every point on this ring is at the same distance from $P$ on the axis. By Pythagoras:

4. Potential contribution from the ring

The potential due to a tiny charge $dq$ at distance $d$ is $dV = \frac{1}{4\pi\varepsilon_0} \frac{dq}{d}$. For the whole ring:

Substitute $dq$:

5. Integrate over all rings

The total potential is the sum of contributions from $r=0$ to $r=R$:

Simplify the constant:

6. Evaluate the integral

Let $u = x^2 + r^2$, so $du = 2r \, dr$, hence $r \, dr = du/2$. When $r=0$, $u = x^2$; when $r=R$, $u = x^2 + R^2$.

$= \frac{1}{2} \left[ 2u^{1/2} \right]_{x^2}^{x^2+R^2} = \sqrt{x^2 + R^2} - \sqrt{x^2}$ …

Method: Potential of a Continuous Charge Distribution by Elemental Integration

Use this whenever you need the electric potential due to a charge spread continuously over a line, surface, or volume (here, a uniformly charged disc), evaluated at a point with some symmetry axis.

Steps

Step 1: Exploit the symmetry to choose the right shape of charge element

Because potential is a scalar, you don't need to resolve components — you only need every point of your chosen element to be the same distance from the field point, so its contribution can be summed directly. For an axial point above a disc, every point on a thin ring (concentric with the disc's centre) is equidistant from the axial point, so a ring is the natural element. (For other geometries, the natural element differs — e.g. a strip for a rod, a shell for a sphere.)

Step 2: Express the charge on the element in terms of the density

For a uniformly charged disc of total charge $Q$ and radius $R$, the surface density is $\sigma = \dfrac{Q}{\pi R^2}$. A ring of radius $r$ and thickness $dr$ has area $dA = 2\pi r\,dr$, so it carries charge

Step 3: Write the potential from that element using the point-charge formula, then integrate

Treat $dq$ as a point charge at the (single, shared) distance to the field point, $d = \sqrt{x^2+r^2}$ for a point at axial distance $x$: …