Question

4.
The rate at which potential difference between the
plates of a parallel plate capacitor with a 2 uF
capacitance be changed to produce a
displacement current of 1 A is
(1) 0.5 MV/s
(2) 0.3 MV/s
(3) 0.2 MV/s
(4) 0.1 MV/s


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Answer 1

The correct answer is option 1) 0.5 MV/s

Given that displacement current = I(d) = 2 A

I(d) = ε.dΦ/dt

Φ - flux = E.A

dΦ/dt = change in flux

E is the electric field across the capacitor and A is the area of the plate of capacitor.

I(d) = ε.A dE/dt

E = V/d = potential difference / distance between the plates of the capacitor

=> I(d) = ε.A d(V/d)/dt

=> I(d) = (ε.A)/d × d(V)/dt

=> I(d) = C × dV/dt [ C = capacitance of the capacitor = (ε.A)/d ]

=> dV/dt = I(d)/C

=> dV/dt = 1/2 = 0.5 MV/s

The rate of change of potential difference of the capacitor = 0.5 MV/s