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)

(a)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
$E^{\prime }=\frac{\sigma }{2{\in }_0}\left(1-\frac{x}{\sqrt{x^2+R^2}}\right)$
From question, x=R (radius of disc)
$\therefore E^{\prime }=\frac{\sigma }{2{\in }_0}\left(1-\frac{R}{\sqrt{R^2+R^2}}\right)$
$=\frac{\sigma }{2{\in }_0}\left(\frac{\sqrt{2}R-R}{\sqrt{2}R}\right)$
$=\frac{4}{14}E$
$\therefore$ % reduction in the value of electric field
$=\frac{\left(E-\frac{4}{14}E\right)\times 100}{E}=\frac{1000}{14}\%\approx 70.7\%$