Question

at a given instant, a 3.6 - a current flows in the wires connected to a parallel - plate capacitor. part a what is the rate at which the electric field is changing between the plates if the square plates are 1.45 cm on a side? express your answer using two significant figures.

\\( \frac { \delta e } { \delta t } = \\)

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Explanation:

Step1: Calculate the area of the plate

The side length of the square plate \(s = 1.45\ cm=0.0145\ m\). The area of a square \(A = s^{2}\), so \(A=(0.0145)^{2}=2.1025\times 10^{-4}\ m^{2}\)

Step2: Use the formula for displacement current

The displacement current \(I_d=\epsilon_0\frac{\Delta\Phi_E}{\Delta t}=\epsilon_0A\frac{\Delta E}{\Delta t}\). Given \(I_d = 3.6\ A\) and \(\epsilon_0 = 8.85\times 10^{-12}\ C^{2}/N\cdot m^{2}\)

We can solve for \(\frac{\Delta E}{\Delta t}\): \(\frac{\Delta E}{\Delta t}=\frac{I_d}{\epsilon_0A}\)

Substitute the values: \(\frac{\Delta E}{\Delta t}=\frac{3.6}{8.85\times 10^{-12}\times2.1025\times 10^{-4}}\)

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Answer:

\(1.9\times 10^{15}\ V/(m\cdot s)\)