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Three metallic spheres of radii R, 2R and 3R are far apart from each other. First sphere is initially given a charge Q and other two are uncharged. Third sphere (radius = 3R) is connected with first sphere (radius R) by a conducting wire. Now connecting wire is removed and then it is connected by wire with second sphere (radius 2R) and finally the wire is removed. Ratio of electric field intensity on surfaces of first and second sphere is:

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3 : 2

2 : 3

27 : 4

10 : 3

answer is D.

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Detailed Solution

${(Q_{Final})}_{1} = (\frac{Q_{1} + Q_{2}}{R_{1} + R_{2}}) R_{1} {(Q_{Final})}_{2} = (\frac{Q_{1} + Q_{2}}{R_{1} + R_{2}}) R_{2}$

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