Q172 / 180 · COMEDK 2014 · Physics
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
- A$\frac{1}{2}$
- B$\frac{3}{4}$
- Correct answerC$\frac{4}{3}$
- D$\frac{1}{3}$
Solution
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, $Q_{\text {air }}=Q_{\text {paper }}$ $\Rightarrow \quad C_{\text {air }} V_{\text {air }}=C_{\text {paper }} V_{\text {paper }}$ $$ \frac{V_{\text {air }}}{V_{\text {paper }}}=\frac{C_{\text {paper }}}{C_{\text {air }}} $$ $$ \begin{aligned} &=\frac{\frac{\varepsilon_{0} K A}{t}}{\frac{\varepsilon_{0} A}{d-t}} \\ &=\frac{K(d-t)}{t}=\frac{4(1-0.75)}{0.75}=\frac{4}{3} \end{aligned} $$
Hence, the correct answer is option: 'C'.