JEE MAIN - Physics (2023 - 25th January Morning Shift - No. 1)
A parallel plate capacitor has plate area 40 cm$$^2$$ and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is :
$$\mathrm{10\varepsilon_0~F}$$
$$\mathrm{24\varepsilon_0~F}$$
$$\mathrm{\frac{3}{10}\varepsilon_0~F}$$
$$\mathrm{\frac{10}{3}\varepsilon_0~F}$$
Explanation
_25th_January_Morning_Shift_en_1_1.png)
$$ \begin{aligned} & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{\mathrm{C}_1}+\frac{1}{\mathrm{C}_2} \\\\ & =\frac{1}{\frac{\mathrm{K} \in_0 \mathrm{~A}}{\mathrm{t}}}+\frac{1}{\frac{\in_0 \mathrm{~A}}{\mathrm{~d}-\mathrm{t}}} \end{aligned} $$
$$ \begin{aligned} & =\frac{\mathrm{t}}{\mathrm{K} \in_0 \mathrm{~A}}+\frac{\mathrm{d}-\mathrm{t}}{\epsilon_0 \mathrm{~A}} \\\\ & =\frac{1 \times 10^{-3}}{5 \in_0 \times 40 \times 10^{-4}}+\frac{1 \times 10^{-3}}{\in_0 40 \times 10^{-4}} \\\\ & \frac{1}{\mathrm{C}_{\mathrm{eq}}}=\frac{1}{20 \in_0}+\frac{1}{4 \in_0} \\\\ & \mathrm{C}_{\mathrm{eq}}=\frac{20 \times 4 \in_0}{24}=\frac{10 \in_0}{3} \mathrm{~F} \end{aligned} $$
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