Question

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 $\frac{\sigma}{2 \in_{0}} \cdot$ With respect to the field at the centre, the electric field along the axis at a distance $R$ from the centre of the disc :

• A. reduces by 70.7%
• B. reduces by 29.3%
• C. reduces by 9.7%
• D. reduces by 14.6%

Answer 1

Answer: (A)

reduces by 70.7%

Answer 2

Electric field intensity at the centre of the disc. $E=\frac{\sigma}{2 \in_{0}} \,$ (given)

Electric field along the axis at any distance x from the centre of the disc $E'=\frac{\sigma}{2 \in_{0}} \left(1-\frac{x}{\sqrt{x^{2}+R^{2}}}\right)$

From question, x=R (radius of disc)
$\therefore E' =\frac{\sigma}{2 \in_{0}} \left(1-\frac{R}{\sqrt{R^{2}+R^{2}}}\right)$ $=\frac{\sigma}{2 \in_{0}} \left(\frac{\sqrt{2}R-R}{\sqrt{2}R}\right)$ $=\frac{4}{14} E$

$\therefore$ % reduction in the value of electric field $=\frac{\left(E-\frac{4}{14}E\right)\times100}{E}=\frac{1000}{14} \% \approx70.7\%$

Hence, The correct answer is option (A): reduces by 70.7%