Two parallel plate air filled capacitors, each of capacitance $C$ , are joined in series to a battery of emf $V$ . The space between the plates of one of the capacitors is then completely filled up with a uniform dielectric having dielectric constant $K$ . The quantity of charge which flows through the battery is

\frac{C V}{2} (\frac{K - 1}{K + 1})

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\frac{C V}{2} (\frac{K + 1}{K - 1})

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\frac{2}{C V} (\frac{K - 1}{K + 1})

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\frac{2}{C V} (\frac{K + 1}{K - 1})

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Solution

The correct option is A $\frac{C V}{2} (\frac{K - 1}{K + 1})$
Inital capacitance $= \frac{C}{2}$
Final capacitance $= \frac{K C \times C}{K C + C} = \frac{K C}{K + 1}$
Capacitance changes from $\frac{C}{2}$ to $\frac{K C}{K + 1}$
Therefore, charge stored in system changes from $\frac{C V}{2}$ to $\frac{K C V}{K + 1}$
Quantity of charge flowing through the battery
$= \frac{K C V}{K + 1} - \frac{C V}{2} = \frac{C V}{2} [\frac{2 K}{K + 1} - 1]$
$= \frac{C V}{2} (\frac{K - 1}{K + 1})$

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The correct option is A CV2(K−1K+1)Inital capacitance =C2 Final capacitance =KC×CKC+C=KCK+1 Capacitance changes from C2 to KCK+1 Therefore, charge stored in system changes from CV2 to KCVK+1 Quantity of charge flowing through the battery =KCVK+1−CV2=CV2[2KK+1−1] =CV2(K−1K+1)