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

Question

A uniformly charged non conducting disc with surface charge density

10 \mathrm{nC} / \mathrm{m}^2

having radius

3 \mathrm{~cm}

. Then find the magnitude of electric field at a point on the perpendicular bisector at a distance

2 \mathrm{~cm}

from origin of the disc. A uniformly charged non conducting disc with surface charge density 10 nC / m^2 having radius 3 ~cm . Then find the

Quantrex Academy · Accepted Answer

Correct answer: A. 251 ~N / C

aligned & E=(σ)/(4 π _0)(2 π)[1- x R^2+x^2 ] & =9 × 10^9 × 10 × 10^ -9 × 6.28[1- 2 4+9 ] & E=90 × 6.28[1- 2 13 ]=251 N / C aligned

A uniformly charged non conducting disc with surface charge density...