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A parallel- plate capacitor with plate a...

A parallel- plate capacitor with plate area A and separation between the plates d, is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn summetrically between the plates. Find the displacement current through this area.

Text Solution

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Solution:
Suppose the charge on the capacitor at time t is Q. The
electric field between the plates of the capacitor is
` E=(Q/ (varepsilon_0)A). The flux through the area considered is `
` Phi_E = (Q/(varepsilon_0)A).(A/2)=(Q/2(varepsilon_0)).`
The displacement current is
` (i_d)=(varepsilon_0)((d Phi_E)/dt)= (varepsilon_0)(1/2(varepsilon_0)) (dQ/dt) = i/2.`
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Knowledge Check

  • Area of a parallel plate capacitor of capacitance 2F and separation between the plates 0.5cm will be

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