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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
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Verified Answer
The correct answer is:
$45.6 \times 10^{-3} \mathrm{C}$
$C_{\text {air }}=80 \mu \mathrm{F}$
$$
C_{\text {dielectric }}=\varepsilon_r C_{\text {air }}=1600 \mu \mathrm{F}
$$
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
$$
\begin{aligned}
q & =q_d-q_{\text {air }} \\
& =(48-2.4) \times 10^{-3} \mathrm{C} \\
q & =45.6 \times 10^{-3} \mathrm{C}
\end{aligned}
$$
$$
C_{\text {dielectric }}=\varepsilon_r C_{\text {air }}=1600 \mu \mathrm{F}
$$
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
$$
\begin{aligned}
q & =q_d-q_{\text {air }} \\
& =(48-2.4) \times 10^{-3} \mathrm{C} \\
q & =45.6 \times 10^{-3} \mathrm{C}
\end{aligned}
$$
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