A uniformly charged disc of radius $$R$$ having surface charge density $$\sigma$$ is placed in the $$xy$$ plane with its center at the origin. Find the electric field intensity along the $$z$$-axis at a distance $$Z$$ from origin:

We have a thin, uniformly charged disc lying in the $$xy$$-plane with its centre at the origin. Its radius is $$R$$ and the surface charge density is $$\sigma$$. We wish to find the electric field intensity at a point located on the positive $$z$$-axis, a distance $$Z$$ from the origin. Because of circular symmetry, the field at this point must point purely along the $$z$$-direction; the transverse components cancel pair-wise.

To calculate the field, we break the disc into many thin concentric rings. Consider an elemental ring of radius $$r$$ and width $$dr$$. The area of this ring is

$$dA = 2\pi r\,dr.$$

Since the surface charge density is uniform, the charge on this ring is

$$dq = \sigma\,dA = \sigma\,(2\pi r\,dr).$$

Next we recall the standard formula for the magnitude of the electric field on the axis of a ring of radius $$r$$ carrying a total charge $$dq$$. For a point a distance $$Z$$ from the plane of the ring, the magnitude of the field contributed by that ring element is

$$dE = \frac{1}{4\pi\varepsilon_0}\;\frac{dq\,Z}{\left(Z^{2}+r^{2}\right)^{3/2}}.$$

Only the component along the axis survives; the horizontal components cancel by symmetry. Substituting the expression for $$dq$$ we obtain

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

To find the total field, we integrate $$r$$ from $$0$$ to $$R$$:

$$E = \int_{0}^{R} dE = \frac{\sigma Z}{2\varepsilon_0}\int_{0}^{R}\frac{r\,dr}{\left(Z^{2}+r^{2}\right)^{3/2}}.$$

We now evaluate the integral. Let

$$u = Z^{2}+r^{2}\;\;\Longrightarrow\;\;du = 2r\,dr\;\;\Longrightarrow\;\;r\,dr = \frac{du}{2}.$$

Substituting into the integral, we get

$$\int_{0}^{R}\frac{r\,dr}{\left(Z^{2}+r^{2}\right)^{3/2}} = \frac{1}{2}\int_{u=Z^{2}}^{u=Z^{2}+R^{2}} u^{-3/2}\;du.$$

Using the power-rule integral $$\int u^{n}\,du = \frac{u^{n+1}}{n+1}$$ (valid for $$n\neq-1$$), we first note that $$n=-\frac{3}{2}$$, so $$n+1=-\frac{1}{2}$$, and hence

$$\int u^{-3/2}\,du = \frac{u^{-1/2}}{-1/2} = -2\,u^{-1/2}.$$

Therefore,

$$\frac{1}{2}\int u^{-3/2}\,du = \frac{1}{2}\left[-2\,u^{-1/2}\right] = -u^{-1/2}.$$

Putting back the limits:

$$-u^{-1/2}\Big|_{Z^{2}}^{Z^{2}+R^{2}} = -\left[(Z^{2}+R^{2})^{-1/2} - (Z^{2})^{-1/2}\right] = \frac{1}{Z} - \frac{1}{\sqrt{Z^{2}+R^{2}}}.$$

Substituting this result into the expression for $$E$$, we have

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

Simplifying, the factor $$Z$$ in the numerator of the first term cancels with the $$Z$$ in the denominator:

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

This expression matches the option labelled C:

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

Hence, the correct answer is Option C.