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The surface charge density of a conducting sphere is `8.85xx10^(-10)C//m^(2)` and the electric field intensity at a distance of 4m from the centre of the sphere is `10^(-2)V//m`. The radius of the sphere, assuming the sphere to be in vacuum isA. 3 cmB. 4 mmC. 4 cmD. 4 km |
Answer» Correct Answer - C `E=(sigmaR^(2))/(epsi_(0)kr^(2))` `therefore" "R^(2)=(Eepsi_(0)kr^(2))/sigma=(10^(-2)xx8.85xx10^(-12)xx1xx16)/(8.85xx10^(-10))` R = 4 cm |
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