In the adjoining figure,capacitors (1) and (2 | vedclass

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(D) Before inserting the dielectric,the capacitors are in series. The equivalent capacitance is $C_{eq} = \frac{C 	imes C}{C + C} = \frac{C}{2}$.The

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$\frac{(K - 1)CE}{2(K + 1)}$ from $C$ to $B$

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(D) Before inserting the dielectric,the capacitors are in series. The equivalent capacitance is $C_{eq} = \frac{C 	imes C}{C + C} = \frac{C}{2}$.The total charge stored is $Q_{eq} = C_{eq} E = \frac{CE}{2}$.Since the battery remains connected,the potential difference $E$ across the combination remains constant.After inserting the dielectric of constant $K$ into one capacitor,its capacitance becomes $KC$. The new equivalent capacitance is $C'_{eq} = \frac{(KC)C}{KC + C} = \frac{K}{K + 1} C$.The new total charge is $Q'_{eq} = C'_{eq} E = \frac{KCE}{K + 1}$.The charge flowing through the battery is $\Delta Q = Q'_{eq} - Q_{eq} = \frac{KCE}{K + 1} - \frac{CE}{2} = CE \left( \frac{2K - (K + 1)}{2(K + 1)} ight) = \frac{(K - 1)CE}{2(K + 1)}$.Since $Q'_{eq} > Q_{eq}$,the charge flows from the battery to the capacitors,which corresponds to the direction from $C$ to $B$ through the battery.

In the adjoining figure,capacitors $(1)$ and $(2)$ have a capacitance $C$ each. When a dielectric of dielectric constant $K$ is inserted between the plates of one of the capacitors,the total charge flowing through the battery is:

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