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

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation:

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6