Question

1. Conside
consider a parallel plate capacitor
whose plates are closely spaced. LetRbe
the radius of the plates and the current
in the wire connected to the plates is
5 A, calculate 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.​

Answer 1

Answer:

Rate of change in electric flux is given as

$\frac{d\phi}{dt} = 5.65 \times 10^{11} Vm/s$

Explanation:

Let the charge at any moment of time on the plate is given as q

so the electric field between the plates is given as

$E = \frac{q}{\pi R^2 \epsilon_0}$

now we have

$q = \pi R^2 \epsilon_0 E$

now we have

$i = \frac{dq}{dt}$

so we have

$i = \epsilon_0 \frac{d\phi}{dt}$

Now we have

$i = 5A$

so we have

$5 = (8.85 \times 10^{-12})\frac{d\phi}{dt}$

so we have

$\frac{d\phi}{dt} = \frac{5}{8.85 \times 10^{-12}}$

$\frac{d\phi}{dt} = 5.65 \times 10^{11} Vm/s$

#Learn

Topic : Displacement Current

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