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The region between the parallel plates of capacitor is filled with parallel layers of air and paper (of dielectric constant 4). The space between the plates is $1 \mathrm{~mm}$ and the thickness of paper is $0.75 \mathrm{~mm}$. The ratio of the voltage across air and paper is
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Verified Answer
The correct answer is:
$\frac{4}{3}$
The given situation is shown below
$$
d=1 \mathrm{~mm}
$$
Thickness of paper, $t=0.75 \mathrm{~mm}$
According to diagram, it is clear that given capacitor is equivalent to a two capacitors connected in series. In series combination, charge on each capacitor is same.
Hence, $Q_{\text {air }}=Q_{\text {paper }}$
$\Rightarrow \quad C_{\text {air }} V_{\text {air }}=C_{\text {paper }} V_{\text {paper }}$
$$
\frac{V_{\text {air }}}{V_{\text {paper }}}=\frac{C_{\text {paper }}}{C_{\text {air }}}
$$
$$
\begin{aligned}
&=\frac{\frac{\varepsilon_{0} K A}{t}}{\frac{\varepsilon_{0} A}{d-t}} \\
&=\frac{K(d-t)}{t}=\frac{4(1-0.75)}{0.75}=\frac{4}{3}
\end{aligned}
$$
$$
d=1 \mathrm{~mm}
$$
Thickness of paper, $t=0.75 \mathrm{~mm}$
According to diagram, it is clear that given capacitor is equivalent to a two capacitors connected in series. In series combination, charge on each capacitor is same.
Hence, $Q_{\text {air }}=Q_{\text {paper }}$
$\Rightarrow \quad C_{\text {air }} V_{\text {air }}=C_{\text {paper }} V_{\text {paper }}$
$$
\frac{V_{\text {air }}}{V_{\text {paper }}}=\frac{C_{\text {paper }}}{C_{\text {air }}}
$$
$$
\begin{aligned}
&=\frac{\frac{\varepsilon_{0} K A}{t}}{\frac{\varepsilon_{0} A}{d-t}} \\
&=\frac{K(d-t)}{t}=\frac{4(1-0.75)}{0.75}=\frac{4}{3}
\end{aligned}
$$
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