Question 20: Electrostats & Capacitance

Question

A parallel plate capacitor consists of two circular plates of radius $ R=0.1 \mathrm{~m} $. They are separated by a short distance. If electric field between the capacitor plates changes as $ \frac{d E}{d t}=6 	imes 10^{13} \frac{\mathrm{~V}}{\mathrm{~m} 	imes \mathrm{s}} $, then the value of displacement current is

Accepted Answer

Given the radius of the capacitor plates $ R = 0.1 \, \mathrm{m} $, we can calculate the area of the plates as:
$ A = \pi R^2 = 3.14 	imes (0.1)^2 \, \mathrm{m}^2  $
We are also given that the rate of change of the electric field between the capacitor plates is:
$ \frac{dE}{dt} = 6 	imes 10^{13} \, \frac{\mathrm{V}}{\mathrm{m} \cdot \mathrm{s}} $
The displacement current $ I_d $ between the plates is calculated using the formula:
$ I_d = \varepsilon_0 \frac{d\phi}{dt} $
where $ \frac{d\phi}{dt} = A \frac{dE}{dt} $. Hence, the formula becomes:
$ I_d = \varepsilon_0 A \left(\frac{dE}{dt}ight) $
Substituting the values, where $ \varepsilon_0 = 8.85 	imes 10^{-12} \, \mathrm{F/m} $, we have:
$ I_d = 8.85 	imes 10^{-12} 	imes 3.14 	imes (0.1)^2 	imes 6 	imes 10^{13} $
Calculating the above expression yields:
$ I_d = 16.67 \, \mathrm{A} $
Therefore, the value of the displacement current is $ 16.67 \, \mathrm{A} $.

Answer

15.25 A

Answer

6.25 A

Answer

16.67 A

Answer

4.69 A

Electrostats & Capacitance

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