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(C) $1$. When the battery is disconnected,the charge $Q$ on the capacitor plates remains constant because there is no path for the charge to flow.$2$.

Vedclass · Accepted Answer

Charge remains constant,and both voltage and electrostatic potential energy decrease.

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(C) $1$. When the battery is disconnected,the charge $Q$ on the capacitor plates remains constant because there is no path for the charge to flow.$2$. When a dielectric slab of constant $K > 1$ is inserted,the capacitance $C$ increases according to the formula $C' = KC$.$3$. Since $Q = CV$,the new voltage $V' = Q / C'$. Since $C' > C$,the voltage $V'$ decreases $(V' = V/K)$.$4$. The electrostatic potential energy $U$ is given by $U = Q^2 / (2C)$. Since $C$ increases and $Q$ is constant,the potential energy $U$ decreases $(U' = U/K)$.

$A$ capacitor is charged and then the battery is disconnected. If a dielectric slab is inserted between the plates,choose the correct statement.

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