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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main · Physics · Electromagnetic Waves

Consider a parallel plate capacitor which is maintained at...

Consider a parallel plate capacitor which is maintained at potential 200 V200 \mathrm{~V}. The separation distance between the plates of capacitor and the area of plates are 1 mm1 \mathrm{~mm} and 20 cm220 \mathrm{~cm}^2, respectively. Calculate displacement current in 1μs1 \mu \mathrm{s}.

Options

  1. A3.5 mA3.5 \mathrm{~mA}
  2. B2.5 mA2.5 \mathrm{~mA}
  3. C1.5 mA1.5 \mathrm{~mA}
  4. D0.5 mA0.5 \mathrm{~mA}

Correct answer

A. 3.5 mA3.5 \mathrm{~mA}

Step-by-step solution

Displacement current, Id=ε0dϕdt=ε0ΔϕΔt=ε0ΔEAΔt(∵ϕ=E⋅A)\begin{aligned} I_d & =\varepsilon_0 \frac{d \phi}{d t}=\varepsilon_0 \frac{\Delta \phi}{\Delta t} \\ & =\varepsilon_0 \frac{\Delta E A}{\Delta t} \quad(\because \phi=E \cdot A)\end{aligned}=\frac{\varepsilon_0 \cdot \Delta V \cdot A}{\Delta t \cdot d} \quad \left(\becauseE=E=\frac{V}{d}\right) Here, \DeltaV=200V=200\mathrm{~V}, \Deltat=1μs=10−6 s,d=1 mm=1×10−3 m and \quadA=20 cm2=20×10−4 m2\begin{aligned}\Deltat & =1 \mu \mathrm{s}=10^{-6} \mathrm{~s},\\ d & =1 \mathrm{~mm}=1 \times10^{-3} \mathrm{~m}\\ \text { and } \quadA & =20 \mathrm{~cm}^2=20 \times10^{-4} \mathrm{~m}^2\end{aligned}∴\quadId and =8.85×10−12×200×20×10−410−6×10−3 and =3.5×10−3 A=3.5 mA\begin{aligned} \therefore \quadI_d \text{ and } =\frac{8.85 \times 10^{-12} \times 200 \times 20 \times 10^{-4}}{10^{-6} \times 10^{-3}}\\ \text{ and } =3.5 \times10^{-3} \mathrm{~A}=3.5 \mathrm{~mA}\end{aligned}
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