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Two circular plates of radius 9cm are separated in air by 2.0mm, forming a parallel

Two circular plates of radius 9cm are separated in air by 2.0mm, forming a parallel plate capacitor. a battery is connected across the plates. at a particular time, t1, the rate at which the charge is flowing through the battery from one plate to the other is 5a. (a)what is the time rate of change of the electric field between the plates at t1? (b)compute the displacement current between the plates at t1, and show it is equal to 5a.
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billy1123
billy1123
4,5(98 marks)

(a) 2.26\times 10^{13}\ N/C.s

(b) 5 A

Solution:

As per the question:

Radius of the circular plate, R = 9 cm = 0.09 m

Distance, d = 2.0 mm = 2.0\times 10^{- 3}\ m

At t_{1}, current, I = 5 A

Now,

Area, A = \pi R^{2} = \pi 0.09^{2} = 0.025

We know that the capacitance of the parallel plate capacitor, C = \frac{\epsilon_{o} A}{d}

Also,

q = CV

q = \frac{\epsilon A}{d}V

Also,

V = \frac{E}{d}

Now,

(a) The rate of change of electric field:

\frac{dE}{dt} = \frac{dq}{dt}(\frac{1}{A\epsilon_{o}})

where

I = \frac{dq}{dt} = 5\ A

\frac{dE}{dt} = 5\times (\frac{1}{0.025\times 8.85\times 10^{- 12}}) = 2.26\times 10^{13}\ N/C.s

(b) To calculate the displacement current:

I_{D} = epsilon_{o}\times \frac{d\phi}{dt}

where

\frac{d\phi}{dt} = Rate of change of flux

I_{D} = Aepsilon_{o}\times \frac{dE}{dt}

I_{D} = 0.025\times 8.85\times 10^{- 12}\times 2.26\times 10^{13} = 5\ A

probro1167
probro1167
4,5(66 marks)
Answer;

moving a magnet into a coil of wire in a closed circuit

Explanation;An electromagnet is a type of magnet that runs on electricity. It is created by introducing a magnetic material such as iron in a coil of wire  in a closed circuit. The strength of the electromagnet can be easily changed by changing the amount of electric current flowing in the closed circuit. This is different from the case of a permanent magnet whose strength is constant. Poles of an electromagnet may be reversed by changing the direction of electric current.

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