Physics - Electrostatics Question with Solution | TestHub

$\because$At the axial point of a uniformly charged disc electric field is given by,

$E=\frac{\sigma }{2{ϵ}_0}\left(1-cos\theta \right)$



By superposition principle, when inner disc is removed , then, electric field due to remaining disc is,



$E=\frac{\sigma }{2{ϵ}_0} \left[\left(1-cos{\theta }_2\right)-\left(1-cos{\theta }_1\right)\right]$

$=\frac{\sigma }{2{ϵ}_0}\left[cos{\theta }_1-cos{\theta }_2\right]$

$=\frac{\sigma }{2{ϵ}_0}\left[\frac{h}{\sqrt{h^2+a^2 }}-\frac{h}{\sqrt{h^2+b^2}}\right]$

$=\frac{\sigma }{2{ϵ}_0}\left[\frac{h}{a\sqrt{1+\frac{h^2}{a^2}}}-\frac{h}{b\sqrt{1+\frac{h^2}{b^2} }}\right]$

$\because h \ll a$and b.

$\therefore E=\frac{\sigma }{2{ϵ}_0}\left[\frac{h}{a}-\frac{h}{b}\right]$

$=\frac{\sigma }{2{ϵ}_0}\left[\frac{h}{a}-\frac{h}{2a}\right]=\frac{\sigma h}{4{ϵ}_{0}a}$

$\Rightarrow C=\frac{\sigma }{4{aϵ}_0}$.