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JEE Advanced 2023 · Paper 1, Physics question 5

A container has a base of 50 cm×5 cm50\ \text{cm}\times5\ \text{cm} and height 50 cm, as shown in the figure. It has two parallel electrically conducting walls each of area 50 cm×50 cm50\ \text{cm}\times50\ \text{cm}. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant 3 at a uniform rate of 250 cm3 s−1250\ \mathrm{cm^3\,s^{-1}}. What is the value of the capacitance of the container after 10 seconds?

[Given: Permittivity of free space ϵ0=9×10−12 C2N−1m−2\epsilon_0 = 9\times10^{-12}\ \mathrm{C^2N^{-1}m^{-2}}, the effects of the non-conducting walls on the capacitance are negligible]

A tall thin box container: the two large faces (50 cm × 50 cm) are conducting plates 5 cm apart, each connected to a terminal wire; the base is 50 cm × 5 cm and the height 50 cm. Liquid partly fills the bottom.
Choose one option
Answer and worked solution

Official answer: B

Liquid level. After 10 s the volume is 2500 cm32500\ \text{cm}^3. Over the 50×5 cm250\times5\ \text{cm}^2 base, this is a depth of 1010 cm.

Capacitor. The conducting walls are plates 50×5050\times50 cm, d=5d=5 cm apart. The filled and empty parts are side by side, so they act in parallel:

  • filled: A1=0.5×0.1=0.05 m2A_1=0.5\times0.1=0.05\ \text{m}^2 with K=3K=3;
  • empty: A2=0.5×0.4=0.2 m2A_2=0.5\times0.4=0.2\ \text{m}^2 in air.

C=ϵ0d(KA1+A2)=9×10−120.05(0.15+0.2)=63 pF.C=\frac{\epsilon_0}{d}(KA_1+A_2)=\frac{9\times10^{-12}}{0.05}(0.15+0.2)=63\ \text{pF}.

Topics:DielectricsSeries and parallel combinations
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