Question

A uniformly charged non conducting disc with surface charge density $10\, nC/m^2$ having radius $3\,cm$. Then find the magnitude of electric field at a point on the perpendicular bisector at a distance $2\,cm$ from origin of the disc.

• A. 251 N/C
• B. 126 N/C
• C. 502 N/C
• D. 72 N/C

Answer 1

Answer: (A)

251 N/C

Answer 2

$E=\frac{\sigma}{4\pi\varepsilon_{0}}\left(2\pi\right)\left[1-\frac{x}{\sqrt{R^{2}+x^{2}}}\right]$ $=9\times10^{9}\times10\times10^{-9}\times6.28\left[1-\frac{2}{\sqrt{4+9}}\right]$ $\therefore\,\, E=90\times6.28\left[1-\frac{2}{\sqrt{13}}\right]=251\,\,N/C$