The left plate of the capacitor shown in the figure carries a charge +Q while the right plate is uncharged. The total final charge on the right plate after closing the switch S will be

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Q2+Cε

Q2-Cε

-Q2

none of these

answer is B.

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Detailed Solution

Electric field between the capacitor platesE=Q+x2Aε0+x2Aε0=12ε0AQ+2xPotential difference =Ed=d2ε0AQ+2x=εε=Q+2x2C ⇒x=Q2-Cε

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