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Two identical parallel plate capacitors of capacitance $C$ each are connected in series with a battery of emf, $E$ as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant $k$, then the amount of charge which will flow through the battery is (neglect internal resistance of the battery)
$\frac{k+1}{2(k-1)} \cdot C E$
$\frac{k-1}{2(k+1)} \cdot C E$
$\frac{k-2}{k+2} \cdot C E$
$\frac{k+2}{k-2} \cdot C E$
Solution
(b)
Initially, $\quad C_{ eq }=\frac{C}{2}, V=E$
So, charge that is delivered by cell is
$Q_1=C_{ eq } E=\frac{C E}{2}$
In series charge remains same for both capacitors.
When one of capacitors is filled with a dielectric then,
$C_{ eq }=\frac{k C^2}{(C+k C)}=\left(\frac{k}{1+k}\right) C$
As battery remains connected, $V=E$. So, charge of combination
$Q_2=C_{ eq } V=\left(\frac{k}{1+k}\right) C E$
So, extra charge given by cell after insertion of dielectric is
$\Delta Q=Q_2-Q_1=\left(\frac{k}{1+k}-\frac{1}{2}\right) C E$
$=\frac{k-1}{2(k+1)} \cdot C E$
