Aparallel-plate capacitor with circular plates of radius 0.10 m is being discharged. a circular loop of radius 0.20 m is concentric with the capacitor and halfway between the plates. the displacement current through the loop is 2.0 a. at what rate is the electric field between the plates changing?

7.1934 x 10^12 V/m.s

Explanation:

In order to do this exercise, you need to use the correct formula. Besides that, we need to identify our data.

First we have the radius of the plates which are circular, and it's 0.1 m. The current of the loop (I) is 2.0 A, and the radius of the loop is 0.2 m.

Now with this data, we use the next formula:

I = dE/dt Eo A

Where:

dE/dt = rate of electric field

Eo = constant of permittivity of free space

A = Area of circle

Solving for dE/dT:

dE/dt = I / Eo*A

Now, the area of the circle is A = πr²

A = 3.1416 * (0.1)² = 0.031416 m²

Now solving the electric field:

dE/dt = 2 / (8.85x10^-12 * 0.031416)

dE/dt = 7.1934 x 10^12 V/m.s

force tends to be low and long diustance masses are high