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 $\frac{\sigma }{2{ϵ}_0}$. With respect to the field at the centre, the electric field along the axis at a distance $R$ from the centre of the disc.

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

Answer 1

The correct option is A Reduces by $70.7$%

Electric field intensity at the centre of the disc.
$E=\frac{\sigma }{2{ϵ}_0}\left(given\right)$
Electric field along the axis at any distance $x$ from the centre of the disc
$E^{\prime }=\frac{\sigma }{2{ϵ}_0}(1-\frac{x}{\sqrt{x^2}-R^2})$
From question, $x=R$ (radius of disc)
$\therefore E^{\prime }=\frac{\sigma }{2{ϵ}_0}(1-\frac{R}{\sqrt{R^2+R^2}})$
$=\frac{\sigma }{2{ϵ}_0}\left(\frac{\sqrt{2}R-R}{\sqrt{2}R}\right)$
$=\frac{4}{14}E$
$\therefore$ % reduction in the value of electric field
$=\frac{(E-\frac{4}{14}E)\times 100}{E}=\frac{1000}{14}$% $=70.7$%.