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A parallel plate capacitor has a capacity $80 \times 10^{-6} \mathrm{~F}$, when air is present between the plates. The volume between the plates is then completely filled with a dielectric slab of dielectric constant 20. The capacitor is now connected to a battery of $30 \mathrm{~V}$ by wires. The dielectric slab is then removed. Then, the charge that passes now through the wire is
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$45.6 \times 10^{-3} \mathrm{C}$
Charge that through the wire
$\begin{aligned} \Delta q & =\Delta C V \\ & =\left(C^{\prime}-C\right) V \\ & =(k-1) \sigma V \\ & =(20-1)\left(80 \times 10^{-6}\right)(30) \\ & =4.56 \times 10^{-2} \\ & =45.6 \times 10^{-3} \mathrm{C}\end{aligned}$
$\begin{aligned} \Delta q & =\Delta C V \\ & =\left(C^{\prime}-C\right) V \\ & =(k-1) \sigma V \\ & =(20-1)\left(80 \times 10^{-6}\right)(30) \\ & =4.56 \times 10^{-2} \\ & =45.6 \times 10^{-3} \mathrm{C}\end{aligned}$
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