Q99 / 180 · JIPMER 2018 · Physics
Find the capacitance in shown figure
- Correct answerA$\frac{2 K A \varepsilon_0}{(K+1) d}$
- B$\frac{2 K A \varepsilon_0}{d}$
- C$\frac{(K+1) A \varepsilon_0}{2 d}$
- D$\frac{2 K A \varepsilon_0}{\left(K^2+1\right) d}$
Solution
This combination is same as the two capacitors are connected in series and distance between plates of each capacitor is $\mathrm{d} / 2$.
So, $\mathrm{C}_1=\frac{\mathrm{K} \varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}$ and $\mathrm{C}_2=\frac{\varepsilon_0 \mathrm{~A}}{\mathrm{~d} / 2}$
Hence, $C_{\text {net }}=\frac{C_1 C_2}{C_1+C_2}=\frac{\left(\frac{2 K \varepsilon_0 \mathrm{~A}}{d}\right)\left(\frac{2 \varepsilon_0 \mathrm{~A}}{d}\right)}{\left(\frac{2 K \varepsilon_0 \mathrm{~A}}{d}+\frac{2 \varepsilon_0 \mathrm{~A}}{d}\right)}$
$=\frac{2 \mathrm{KA} \varepsilon_0}{(\mathrm{~K}+\mathrm{l}) \mathrm{d}}$
Hence, the correct answer is option: 'A'.