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A uniformly charged non conducting disc ...

A uniformly charged non conducting disc with surface charge density `10nC//m^(2)` having radius `R=3 cm`. Then find the value of electric field intensity at a point on the perpendicular bisector at a distance of `r=2cm`

A

`348.6N//C`

B

`305.6N//C`

C

`251.2N//C`

D

`116.8N//C`

Text Solution

Verified by Experts

The correct Answer is:
c
`E=(sigma2pi)/(4piepsilon_(0))[1-(x)/(sqrt(R^(2))+x^(2))]`
`E=9xx10^(9)xx10xx10^(-9)xx6.28[1-(2)/(sqrt(4)+9)]`
`E= 90xx6.28[1-(2)/(sqrt(13))]`
`E= 251.2N//C`
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