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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`

Text Solution

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`E=k6.2pi[1-(x)/(sqrt(R^(2)+x^(2)))]`
`E=9xx10^(9)xx10xx10^(-9)xx6.28[1-(x)/(sqrt(4+9))]`
`E=90xx6.28[1-(2)/(sqrt13)]`
`E=251.2N//C`
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