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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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a
Q2+Cε
b
Q2-Cε
c
-Q2
d
none of these

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detailed solution

Correct option is B

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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In the circuit shown in figure, calculate the charge on 2 μF capacitor in steady state (in μC). 


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