1. Set up the problem

A uniformly charged disc of radius $R$ lies in the $xy$-plane with surface charge density $\sigma$. We need the electric field at a point on the axis, i.e. at $(0,0,Z)$.

By symmetry, the electric field has only a $z$-component.

2. Take a ring element

Consider a thin ring of radius $r$ and thickness $dr$ on the disc.

Its charge is $dq = \sigma \cdot dA = \sigma (2\pi r\,dr).$

The field due to this ring at the point on the axis has magnitude along the axis: $dE = \frac{1}{4\pi\varepsilon_0}\frac{dq\,Z}{(Z^2+r^2)^{3/2}}.$

Substitute $dq$: $dE = \frac{1}{4\pi\varepsilon_0}\frac{\sigma (2\pi r\,dr) Z}{(Z^2+r^2)^{3/2}}.$

So, $dE = \frac{\sigma Z}{2\varepsilon_0}\frac{r\,dr}{(Z^2+r^2)^{3/2}}.$

3. Integrate from $r=0$ to $r=R$

$E = \frac{\sigma Z}{2\varepsilon_0}\int_0^R \frac{r\,dr}{(Z^2+r^2)^{3/2}}.$

Let $u = Z^2+r^2 \quad \Rightarrow \quad du = 2r\,dr.$

Then $\int \frac{r\,dr}{(Z^2+r^2)^{3/2}} = \frac{1}{2}\int u^{-3/2}du = -u^{-1/2}.$

Thus, $\int_0^R \frac{r\,dr}{(Z^2+r^2)^{3/2}} = \left[-\frac{1}{\sqrt{Z^2+r^2}}\right]_0^R.$

So, $= -\frac{1}{\sqrt{Z^2+R^2}} + \frac{1}{Z}$ (for $Z>0$; standard axis formula).

Therefore, $E = \frac{\sigma Z}{2\varepsilon_0}\left(\frac{1}{Z}-\frac{1}{\sqrt{Z^2+R^2}}\right).$

Simplifying, $E = \frac{\sigma}{2\varepsilon_0}\left(1-\frac{Z}{\sqrt{Z^2+R^2}}\right).$

4. Match with the options

This exactly matches: $\boxed{E = \frac{\sigma}{2\varepsilon_0}\left(1-\frac{Z}{\sqrt{Z^2+R^2}}\right)}$

So the correct option is A.

5. Check the other options briefly

• B has a plus sign, which is incorrect.
• C has wrong dimensions.
• D also has wrong form and wrong dimensions.

Hence only A is correct.

6. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They agree.