Question

part a
calculate the displacement current ( i_d ) between the square plates, 6.4 cm on a side, of a capacitor if the electric field is changing at a rate of ( 1.8\times10^6v/mcdot s ).
express your answer to two significant figures and include the appropriate units.
( i_d=) value units

Explanation:

Step1: Calculate the area of the square plate

The side length of the square plate \(a = 6.4\space cm=0.064\space m\). The area of a square \(A=a^{2}\).

$$A=(0.064)^{2}\space m^{2}=4.096\times 10^{-3}\space m^{2}$$

Step2: Use the formula for displacement current

The formula for displacement current is \(I_{D}=\epsilon_{0}\frac{d\Phi_{E}}{dt}\), and since \(\Phi_{E} = EA\) (where \(E\) is the electric field and \(A\) is the area of the plate), then \(I_{D}=\epsilon_{0}A\frac{dE}{dt}\).
We know that \(\epsilon_{0} = 8.85\times 10^{-12}\space C^{2}/N\cdot m^{2}\) and \(\frac{dE}{dt}=1.8\times 10^{6}\space V/m\cdot s\) and \(A = 4.096\times 10^{-3}\space m^{2}\)

$$I_{D}=(8.85\times 10^{-12})\times(4.096\times 10^{-3})\times(1.8\times 10^{6})$$

$$I_{D}=8.85\times4.096\times1.8\times10^{-12 - 3+6}$$

$$I_{D}=8.85\times4.096\times1.8\times 10^{-9}$$

$$I_{D}\approx6.5\times 10^{-8}\space A = 65\space\mu A$$

Answer:

\(I_{D} = 65\space\mu A\)