Physics - Electrostatics Question with Solution | TestHub

Electric field intensity at the centre of the disc. $\mathrm{E}=\frac{\sigma }{2{ϵ}_0}\text{ (given) }$ Electric field along the axis at any distance $x$ from the centre of the disc ${\mathrm{E}}^{\prime }=\frac{\sigma }{2{ϵ}_0}\left(1-\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2}-{\mathrm{R}}^2}\right)$ From question, $\mathrm{x}=\mathrm{R}$ (radius of disc) $\begin{array}{rl}& \therefore E^{\prime }=\frac{\sigma }{2{ϵ}_0}\left(1-\frac{R}{\sqrt{R^2+R^2}}\right)\\ & =\frac{\sigma }{2{ϵ}_0}\left(\frac{\sqrt{2}R-R}{\sqrt{2}R}\right)\\ & =\frac{4}{14}E\end{array}$ $\therefore$ \% reduction in the value of electric field $=\frac{\left(\mathrm{E}-\frac{4}{14}\mathrm{E}\right)\times 100}{\mathrm{E}}=\frac{1000}{14}\%=70.7\%$