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  • Consider a parallel plate capacitor which is maintained at potential of 200 V. If the separation distance between the plates of the capacitor and area of the plates are 1 and <img alt="20 \; \text{

Consider a parallel plate capacitor which is maintained at potential of 200 V. If the separation distance between the plates of the capacitor and area of the plates are 1 and 20 \; \text{cm}^{2}.Then the displacement current for the time in \mathrm{\mu \mathrm{s}}.

 

Option: 1

1.5 mA


Option: 2

2.5 mA


Option: 3

3.5 mA


Option: 4

4.5 mA


Answers (1)

best_answer

Potential difference between the plates of the capacitor, V = 200 V 
The distance between the plates,

\mathrm{d}=1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m}

Area of the plates of the capacitor,

\mathrm{A}=20 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2

Time is given in micro-second, \mu \mathrm{s}=10^{-6} \mathrm{~s}

Displacement current

\mathrm{I_d=\varepsilon_{\mathrm{o}} \frac{d \Phi_B}{d t} \Rightarrow I_d=\varepsilon_{\mathrm{o}} \frac{E A}{t}}

But electric field, E = V/d

Therefore,

\mathrm{I=\frac{V}{d} I_d=\varepsilon_{\mathrm{a}} \frac{V A}{t d}=8.85 \times 10^{-12} \times \frac{200 \times 20 \times 10^{-4}}{10^{-6} \times 1 \times 10^{-3}}}

\mathrm{=35400 \times 10^{-7}=3.5 \mathrm{~mA}}

Posted by

Pankaj

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