A parallel plate capacitor has a capacity $80 \times 10^{-6} \mathrm{~F}$, when air is present between its…
A parallel plate capacitor has a capacity $80 \times 10^{-6} \mathrm{~F}$, when air is present between its plates. The space between the plates is 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 passing through the wire is
$12 \times 10^{-3} \mathrm{C}$
$25.3 \times 10^{-3} \mathrm{C}$
$120 \times 10^{-3} \mathrm{C}$
$45.6 \times 10^{-3} \mathrm{C}$
Solution
$C_{\text {air }}=80 \mu \mathrm{F}$
$
C_{\text {dielectric }}=\varepsilon_r C_{\text {air }}=1600 \mu \mathrm{F}
$
Charge stored in the presence of air,
$
q_{\text {air }}=C_{\text {air }} \times V=80 \times 30 \times 10^{-6}=2400 \mu \mathrm{C}
$
Charge stored in presence of dielectric medium,
$
\begin{aligned}
q_d=C_{\text {dielectric }} \times V & =1600 \times 30 \times 10^{-6} \\
& =48000 \mu \mathrm{C}
\end{aligned}
$
When dielectric is removed, effective charge remained is
$
\begin{aligned}
q & =q_d-q_{\text {air }} \\
& =(48-2.4) \times 10^{-3} \mathrm{C} \\
q & =45.6 \times 10^{-3} \mathrm{C}
\end{aligned}
$