Two square shaped metal plates of side $1 \mathrm{~m}$, kept $0.01 \mathrm{~m}$ apart in air form a parallel…

Two square shaped metal plates of side $1 \mathrm{~m}$, kept $0.01 \mathrm{~m}$ apart in air form a parallel plate capacitor. It is connected to a battery of $500 \mathrm{~V}$. The plates of the capacitor are then immersed in an insulting oil by lowering the plates vertically with a speed of $0.001 \mathrm{~ms}^{-1}$. If the dielectric constant of the oil is 11 , then current drawn from the battery during this process is
  1. $4.425 \times 10^{-6} \mathrm{~A}$
  2. $4.425 \times 10^{-5} \mathrm{~A}$
  3. $4.425 \times 10^{-9} \mathrm{~A}$
  4. $4.425 \times 10^{-2} \mathrm{~A}$

Solution

$\begin{aligned} & \text { Capacitance, } C=\frac{(1-\mathrm{x} \cdot 1)}{\mathrm{d}}+\frac{\mathrm{K} \in_0 \mathrm{x}}{\mathrm{d}} \\ & =\frac{\in_0}{\mathrm{~d}}(1-\mathrm{x}+\mathrm{Kx}) \\ & \text { or, } \mathrm{C}=\frac{\in_0}{\mathrm{~d}}[1+(\mathrm{K}-1) \mathrm{x}]\end{aligned}$ $\begin{aligned} \frac{\mathrm{dC}}{\mathrm{dt}} & =\frac{\in_0}{\mathrm{~d}}(\mathrm{~K}-1) \mathrm{V}=\frac{8.85 \times 10^{-12}}{0.01} \times(11-1) \times 0.001 \\ & =8.85 \times 10^{-12} \\ & \because \mathrm{Q}=\mathrm{CV} \text { or, } \mathrm{I}=\mathrm{V}, \frac{\mathrm{dC}}{\mathrm{dt}}=500 \times 8.85 \times 10^{-12} \\ & =4.425 \mathrm{~A}\end{aligned}$

Asked in: AP EAMCET 2023 (15 May Shift 1)

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