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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 $\frac{\sigma }{{2\,{ \in _0}}}$. With respect to the field at the centre, the electric field along the axis at a distance $R$ from the centre of the disc
reduces by $70. 7\%$
reduces by $29.3\%$
reduces by $9.7\%$
reduces by $14.6\%$
Solution
Electric field intensity at the centre of the disc.
$E=\frac{\sigma}{2 \epsilon_{0}} \quad(\text { given })$
Electric field along the axis at any distance
$x$ from the centre of the disc
$E^{\prime}=\frac{\sigma}{2 \epsilon_{0}}\left(1-\frac{x}{\sqrt{x^{2}+R^{2}}}\right)$
From question, $x=R(\text { radius of disc })$
$\therefore \mathrm{E}^{\prime}=$ $ \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) $
$=\frac{4}{14} \mathrm{E}$
$\therefore$ $\%$ reduction in the value of electric field
$=\frac{\left(\mathrm{E}-\frac{4}{14} \mathrm{E}\right) \times 100}{\mathrm{E}}=\frac{1000}{14} \%=70.7 \%$
