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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Physics
Chapter
Electromagnetic Waves

BITSAT 2024 · Physics · Electromagnetic Waves

A parallel plate capacitor consists of two circular plates of...

A parallel plate capacitor consists of two circular plates of radius R=0.1mR=0.1 \mathrm{ m}. They are separated by a short distance. If electric field between the capacitor plates changes as dEdt=6×1013Vm×s\frac{d E}{d t}=6 \times 10^{13} \frac{\mathrm{ V}}{\mathrm{ m} \times \mathrm{s}}, then the value of displacement current is

Options

  1. A15.25 A
  2. B6.25 A
  3. C16.67 A
  4. D4.69 A

Correct answer

C. 16.67 A

Step-by-step solution

Given, area of plates, A=πR2=3.14×(0.1)2m2A=\pi R^2=3.14 \times(0.1)^2 \mathrm{ m}^2 and dEdt=6×1013Vm×s\frac{d E}{d t}=6 \times 10^{13} \frac{\mathrm{ V}}{\mathrm{ m} \times \mathrm{s}} Displacement current between the plates, Id=ε0dϕdt⇒Id=ε0A(dEdt)[∵dϕdt=AdEdt]=8.85×10−12×3.14×(0.1)2×6×1013⇒Id=16.67A\begin{aligned} I_d & =\varepsilon_0 \frac{d \phi}{d t} \\ \Rightarrow \quad I_d & =\varepsilon_0 A\left(\frac{d E}{d t}\right) \quad\left[\because \frac{d \phi}{d t}=A \frac{d E}{d t}\right] \\ & =8.85 \times 10^{-12} \times 3.14 \times(0.1)^2 \times 6 \times 10^{13} \\ \Rightarrow \quad I_d & =16.67 \mathrm{ A} \end{aligned}
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