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  • Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, the displacement current through t

Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface is:

Option: 1

2 A


Option: 2

3 A


Option: 3

4 A


Option: 4

5 A


Answers (1)

best_answer

Given data :

Radius of the plates =R
Area of the parallel plate capacitor = A
Current in the wire connected to the plate =  \mathrm{I}_{\mathrm{c}}=5 \; \mathrm{Amp}

Electric field between the plates \mathrm{\mathrm{E}=\frac{Q}{\varepsilon_o A}}
The flux linked with the given area is \mathrm{\phi_E}

Hence, \mathrm{\phi_E=\frac{Q}{\varepsilon_o A} \times A=\frac{Q}{\varepsilon_o}}

The displacement current

\mathrm{\mathrm{I}_{\mathrm{d}}=\varepsilon_o \frac{d \phi_E}{d t}=\varepsilon_o \frac{d}{d t}\left(\frac{Q}{\varepsilon_o}\right)=\frac{d Q}{d t}}

\mathrm{\frac{dQ}{dt}} is rata of charge is carried to positive plate flow the connecting wire.

\mathrm{\therefore \mathrm{I}_{\mathrm{d}}=\mathrm{I}_{\mathrm{c}}=5 \mathrm{~A}}

Answer : \mathrm{\mathrm{Id}=\mathrm{Ic}=5 \mathrm{~A}}

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Rishi

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