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+1 vote
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in Electrostatics by (67 points)
imageA Sherical shell of radius \( R \) is given charge \( 3 q \) on its surface and a point charge \( q \) is placed at distarice \( R \) / 2 from its centre \( C \). Also there is a charge \( 2 q \) placed outside the shell at a distance of \( 2 R \) as shown. Then The magnitude of electric field at the centre \( C \) due to charge on the outer surface of shell is \( \frac{K q}{2 R^{2}} \) before closing the switch \( S \) The electric potential at the centre \( C \) due to charges on the outer surface of shell is \( \left(\frac{-K q}{R}\right) \) before closing the switch \( S \) The electric potential at the centre \( C \) due to charges on the outer surface of shell is \( \left(\frac{-K q}{R}\right) \) after closing the switch \( S \), Charge flown through the switch in to earth after closing the switch \( S \) is \( 5 q \)

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