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  • A parallel plate capacitor with plate area  and separation between the plates <img alt="\mathrm{d}" src="https://entrancecorner.onc

A parallel plate capacitor with plate area \mathrm{A} and separation between the plates \mathrm{d}, is charged by a constant current \mathrm{T}. Consider a plane surface of area \mathrm{A / 2} parallel to the plates and drawn between the plates. The displacement current through this area is:
 

Option: 1

\mathrm{I}

 


Option: 2

\mathrm{I / 2}
 


Option: 3

\quad \mathrm{I} / 4
 


Option: 4

\quad \mathrm{I} / 8


Answers (1)

best_answer

Charge on capacitor plates at time \mathrm{t} is, \mathrm{q= It.}

Electric field between the plates at this instant is

\mathrm{ \mathrm{E}=\frac{\mathrm{q}}{\mathrm{A} \varepsilon_0}=\frac{\mathrm{It}}{\mathrm{A} \varepsilon_0} }                                 (i)

Electric flux through the given area \mathrm{ \mathrm{A} / 2 } is

\mathrm{ \phi_{\mathrm{E}}=\left(\frac{\mathrm{A}}{2}\right) \mathrm{E}=\frac{\mathrm{It}}{2 \varepsilon_0} }    (Using(i))       (ii)

Therefore, displacement current

\mathrm{ \mathrm{I}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}=\varepsilon_0 \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{It}}{2 \varepsilon_0}\right)=\frac{\mathrm{I}}{2} \quad(\text { Using (ii))} }

Hence option 2 is correct.
 

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