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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` |
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Answer» Correct Answer - 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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