AP EAMCET · PHYSICS · Capacitance
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
- A \(4.425 \times 10^{-6} \mathrm{~A}\)
- B \(4.425 \times 10^{-5} \mathrm{~A}\)
- C \(4.425 \times 10^{-9} \mathrm{~A}\)
- D \(4.425 \times 10^{-2} \mathrm{~A}\)
Answer & Solution
Correct Answer
(C) \(4.425 \times 10^{-9} \mathrm{~A}\)
Step-by-step Solution
Detailed explanation
\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)…
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