Electrostatics Part 2

A circular disc is placed in xy-plane with its centre at the origin as shown below   Consider an elemental ring of thickness dr and radius r. Now, the area of the elemental ring can be given by $dA=2\pi rdr$ The charge stored in this elemental ring, $dq=\sigma dA$ Now, the electric field at the point on Z-axis at a distance of Z from origin can be given by $dE=\frac{kdqZ}{(r^{2}+Z^{2})^{3/2}}$ Substituting the value of dq and dA in above equation, we get $dE=\frac{kZ\sigma(2\pi rdr)}{(r^{2}+Z^{2})^{3/2}}$ $=\frac{\sigma Z}{2\varepsilon_{0}}\left(\frac{rdr}{(r^{2}+Z^{2})^{3/2}}\right)$ Calculating the total electric field by integrating the above expression from r = 0 to r = R, we get $E=\sigma Z 2\varepsilon_{0} \int\limits_{0}^{R} \frac{rdr}{(r^{2}+Z^{2})^{3/2}}$ Put $r^{2}+Z^{2}=u^{2}$ $\Rightarrow 2rdr = 2udu$ $\Rightarrow rdr = udu$ For lower limit, r = 0 ⇒ u = Z Upper limit, r = R $\Rightarrow u = \sqrt{R^{2}+Z^{2}}$ $E=\frac{\sigma Z}{2\varepsilon_{0}} \int\limits_{Z}^{\sqrt{R^{2}+Z^{2}}} \frac{udu}{u^{3}}$ $E=\frac{\sigma Z}{2\varepsilon_{0}} \int\limits_{Z}^{\sqrt{R^{2}+Z^{2}}} \frac{du}{u^{2}}$ $=\frac{\sigma Z}{2\varepsilon_{0}} \int\limits_{Z}^{\sqrt{R^{2}+Z^{2}}} u^{-2} du$ $=\frac{\sigma Z}{2\varepsilon_{0}} \left[ \frac{u^{-2+1}}{-2+1} \right]_{Z}^{\sqrt{Z^{2}+R^{2}}}$ $=\frac{\sigma Z}{2\varepsilon_{0}} \left[ -\frac{1}{u} \right]_{Z}^{\sqrt{Z^{2}+R^{2}}}$ $=\frac{\sigma Z}{2\varepsilon_{0}} \left[ -\frac{1}{\sqrt{Z^{2}+R^{2}}} + \frac{1}{Z} \right]$ $=\frac{\sigma Z}{2\varepsilon_{0}} \left[ \frac{1}{Z} - \frac{1}{\sqrt{Z^{2}+R^{2}}} \right]$ $=\frac{\sigma}{2\varepsilon_{0}} \left[ 1 - \frac{Z}{\sqrt{Z^{2}+R^{2}}} \right]$ Thus, the electric field at the point on Z-axis at a distance of Z from origin is $\frac{\sigma}{2\varepsilon_{0}} \left[ 1 - \frac{Z}{\sqrt{Z^{2}+R^{2}}} \right]$