91Ó°ÊÓ

Considering that disk as an infinite number of very thin loops,

Thus, the law of potential is

$V=\frac{kQ}{\sqrt{a^2+x^2}}$

Here (a) is the radius of the loop and (x) is the displacement on the x-axis.

Assuming Q spread on disk, now the charge density is given by

$\mathrm{σ}=\frac{Q}{A}=\frac{Q}{{\mathrm{Ï€²¹}}^2}$

Now differential charge per differential area is

$$\begin{gathered} dQ=\mathrm{σ}dA=\mathrm{σ}d\left({\mathrm{Ï€²¹}}^2\right) \\ =2\mathrm{Ï€²¹Ïƒda} \end{gathered}$$

As each differential charge contribute a potential at (x),

Therefore, the equation is:

$dV=\frac{kdQ}{\sqrt{x^2+r^2}}$

By integration;

$$\begin{gathered} V={∫}_0^{V}dV \\ =2k\mathrm{πσ}{∫}_0^{\mathrm{a}}\frac{\mathrm{rdr}}{\sqrt{{\mathrm{r}}^2+{\mathrm{x}}^2}} \\ =2\mathrm{°ìπσ}{\left[\sqrt{{\mathrm{r}}^2+{\mathrm{x}}^2}\right]}_0^{\mathrm{a}} \\ =2\mathrm{°ìπσ}\left[\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}-\mathrm{x}\right] \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\left[\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}-\mathrm{x}\right] \end{gathered}$$

Hence, the electric potential is $V=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\left[\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}-\mathrm{x}\right]$.

As the electric field is the gradient of the potential;

$$\begin{gathered} E_x=-\frac{∂V}{∂x} \\ =-\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\frac{∂\left[\sqrt{a^2+x^2}-x\right]}{∂x} \\ =-\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\left[\frac{x}{\sqrt{a^2+x^2}}-1\right] \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\left[\frac{1}{x}-\frac{1}{\sqrt{a^2+x^2}}\right] \end{gathered}$$

Hence, the expression is Hence, the expression is $E_x=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_0}\left[\frac{1}{x}-\frac{1}{\sqrt{a^2+x^2}}\right]$..