A parallel plate capacitor with circular plates of radius R is being charged. The displacement current between the plates is equal to the conduction current in the wires. If the rate of change of electric field between the plates is dE/dt, what is the displacement current?

A. _0 R^2 dEdt — The displacement current is defined as the product of the permittivity of free space and the rate of change of electric flux. Since the electric flux is the product of the electric field and the area of the plates, pi*R^2, the result follows.

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$A parallel plate capacitor with circular plates of radius R is being charged. The displacement current I_{d} between the plates is given by I_{d} = ϵ_{0} \frac{d Φ _{E}}{d t} . If the rate of change of electric field between the plates is \frac{d E}{d t}, what is the displacement current?$

A. $ϵ_{0} π R^{2} \frac{d E}{d t}$
B. $ϵ_{0} R^{2} \frac{d E}{d t}$
C. $ϵ_{0} π R \frac{d E}{d t}$
D. $ϵ_{0} \frac{d E}{d t}$

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Correct answer: A — $ϵ_{0} π R^{2} \frac{d E}{d t}$

Solution

The displacement current is I_{d} = ϵ_{0} \frac{d}{d t} (E \cdot A) = ϵ_{0} A \frac{d E}{d t} . Substituting A = π R^{2}, we get I_{d} = ϵ_{0} π R^{2} \frac{d E}{d t} .

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A parallel plate capacitor with circular plates of radius R is being charged. The displacement current Id​ between the plates is given by Id​=ϵ0​dtdΦE​​. If the rate of change of electric field between the plates is dtdE​, what is the displacement current?

Solution

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$A parallel plate capacitor with circular plates of radius R is being charged. The displacement current I_{d} between the plates is given by I_{d} = ϵ_{0} \frac{d Φ _{E}}{d t} . If the rate of change of electric field between the plates is \frac{d E}{d t}, what is the displacement current?$