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(A) The capacitors $C_1$ and $C_2$ are in series across a voltage source $V$. The initial charge on $C_1$ is given by $q_1 = \frac{C_1 C_2 V}{C_1 + C_

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(A) The capacitors $C_1$ and $C_2$ are in series across a voltage source $V$. The initial charge on $C_1$ is given by $q_1 = \frac{C_1 C_2 V}{C_1 + C_2}$.When a dielectric of constant $K$ is inserted into $C_2$,its new capacitance becomes $C_2^{\prime} = K C_2$.The new charge on $C_1$ becomes $q_1^{\prime} = \frac{C_1 C_2^{\prime} V}{C_1 + C_2^{\prime}} = \frac{K C_1 C_2 V}{C_1 + K C_2}$.Dividing the numerator and denominator by $K$,we get $q_1^{\prime} = \frac{C_1 C_2 V}{(C_1/K) + C_2}$.Since $(C_1/K) + C_2  1$,the denominator decreases,which means the charge $q_1^{\prime}$ increases compared to $q_1$.

In the circuit below,if a dielectric is inserted into $C_2$,then the charge on $C_1$ will:

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