The surface charge density of a thin charged disc of radius $\mathrm{R}$ is $\sigma$. The value of the…
The surface charge density of a thin charged disc of radius $\mathrm{R}$ 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 $\mathrm{R}$ 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.
$
\mathrm{E}=\frac{\sigma}{2 \epsilon_0} \quad \text { (given) }
$
Electric field along the axis at any distance $x$ from the centre of the disc
$
\mathrm{E}^{\prime}=\frac{\sigma}{2 \epsilon_0}\left(1-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+\mathrm{R}^2}}\right)
$
From question, $\mathrm{x}=\mathrm{R}$ (radius of disc)
$
\begin{aligned}
\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)
\end{aligned}
$
$
=\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 \%
$