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  4. A uniformly charged non conducting disc with surface charge density 10 nC/m2 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.

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Q. 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.Physics Question Image

AIIMSAIIMS 2018Electric Charges and Fields

Solution:

$E=\frac{\sigma}{4\pi\varepsilon_{0}}\left(2\pi\right)\left[1-\frac{x}{\sqrt{R^{2}+x^{2}}}\right]$

$=9\times10^{9}\times10\times10^{-9}\times6.28\left[1-\frac{2}{\sqrt{4+9}}\right]$

$\therefore \,\, E=90\times6.28\left[1-\frac{2}{\sqrt{13}}\right]=251\,\,N/C$