Two parallel plate capacitors with capacitanc | vedclass

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(C) Initially,the capacitors are connected in parallel to a battery of potential $V$. The total charge $Q_{net}$ stored in the system is:$Q_{net} = C_

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$\frac{3V}{K + 2}$

Vedclass · Answer

(C) Initially,the capacitors are connected in parallel to a battery of potential $V$. The total charge $Q_{net}$ stored in the system is:$Q_{net} = C_{eq}V = (C + 2C)V = 3CV$When the battery is disconnected,the total charge $Q_{net}$ remains constant.After inserting the dielectric of constant $K$ into the capacitor $C$,its new capacitance becomes $C' = KC$.The capacitors remain in parallel,so they share the same potential difference $V'$.The new total capacitance of the system is $C'_{eq} = C' + 2C = KC + 2C = (K + 2)C$.Since the total charge is conserved:$Q_{net} = C'_{eq}V'$$3CV = (K + 2)CV'$Solving for $V'$:$V' = \frac{3V}{K + 2}$

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