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) $$\begin{array}{rl}\therefore {\mathrm{E}}^{\prime }& =\frac{\sigma }{2{ϵ}_0}\left(1-\frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\mathrm{R}}^2}}\right)\\ & =\frac{\sigma }{2{ϵ}_0}\left(\frac{\sqrt{2}\mathrm{R}-\mathrm{R}}{\sqrt{2}\mathrm{R}}\right)\end{array}$$ $$=\frac{4}{14}\mathrm{E}$$ $\therefore \%$ reduction in the value of electric field $$=\frac{\left(E-\frac{4}{14}E\right)\times 100}{E}=\frac{1000}{14}\%=70.7\%$$