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The surface charge density of a thin charged disc of radius R is $\sigma .$ The value of the electric field at the centre of the disc is $\frac{\sigma }{2\epsilon _{0}}$ . With respect to the field at the centre, the electric field along the axis at a distance R from the centre of the disc :
Option: 1
reduces by 70.7 %

Option: 2
reduces by 29.3%

Option: 3
reduces by 9.7%

Option: 4
reduces by 14.6%

Answers (1)

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$Electric \ field \ intensity \ at \ the \ centre\ of \ the \ disc \\\ \Rightarrow \mathrm{E}=\frac{\sigma}{2 \epsilon_{0}}(\text { given })$ \\ Electric field along the axis at any distance $x$ from the centre of the disc $\mathrm{E}^{'}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}-\mathrm{R}^{2}}}\right)$ \\ From question, $x=R$ (radius of disc) \\ $\therefore \mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{\mathrm{R}}{\sqrt{\mathrm{R}^{2}+\mathrm{R}^{2}}}\right)$ $=\frac{\sigma}{2 \epsilon_{0}}\left(\frac{\sqrt{2} \mathrm{R}-\mathrm{R}}{\sqrt{2} \mathrm{R}}\right)$ $=\frac{4}{14} \mathrm{E}$ \\ So reduction in the value of electric field \\ $=\frac{\left(\mathrm{E}-\frac{4}{14} \mathrm{E}\right) \times 100}{\mathrm{E}}=\frac{1000}{14} \%=70.7 \%$

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