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two identical parallel plate capacitors of capacitance $C$ each are connected in series with abattery of emf, E as shown, If one of the capacitors is now filled with a dielectric of dielectric constant $\mathrm{k}$, the amount of charge which will flow through the battery is (neglect internal resistance of the battery)
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Verified Answer
The correct answer is:
$\frac{k-1}{2(k+1)} C E$
$\begin{array}{l}
Q=\frac{C}{2} E \\
C e q=\frac{C \cdot K C}{C+K C}=\frac{C^{2} K}{C(1+k)}=\frac{C K}{1+K} \\
Q=\frac{C K}{1+k} E \\
\Delta Q=\left(\frac{C K}{1+k}-\frac{C}{2}\right) E
\end{array}$
$\begin{aligned} \Delta Q &=\left(\frac{C K}{1+k}-\frac{C}{2}\right) E \\ &=\frac{2 c k-c(k-1)}{2(1+k)} \\ &=\frac{2 c k-c k-c}{2(1+k)}=\frac{c k-c}{2(1+k)} \end{aligned}$
Q=\frac{C}{2} E \\
C e q=\frac{C \cdot K C}{C+K C}=\frac{C^{2} K}{C(1+k)}=\frac{C K}{1+K} \\
Q=\frac{C K}{1+k} E \\
\Delta Q=\left(\frac{C K}{1+k}-\frac{C}{2}\right) E
\end{array}$
$\begin{aligned} \Delta Q &=\left(\frac{C K}{1+k}-\frac{C}{2}\right) E \\ &=\frac{2 c k-c(k-1)}{2(1+k)} \\ &=\frac{2 c k-c k-c}{2(1+k)}=\frac{c k-c}{2(1+k)} \end{aligned}$
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