Question

A parallel-plate capacitor has circular plates and no dielectric between the plates. Each plate has a radius equal to 2.8 cm and the plates are separated by 1.1 mm. Charge is flowing onto the upper plate (and off the lower plate) at a rate of 4.1 A. Find the time rate of change of the electric field between the plates.

Answer 1

Answer:
`1.88* 10^(14) V/m-s`

Step-by-step explanation:

Given

radius of capacitor Plate
`r=2.8 cm`

Area
`A=\pi r^2=\pi * 2.8^* 10^(-4) m^2`

`A=24.63* 10^(-4) m^2`

current
`I=4.1 A`

separation
`d=1.1 mm`

Electric Field strength is given by

`E=(Q)/(\epsilon _0A)`

`E=(I\cdot t)/(\epsilon _0A)`

`\frac{\mathrm{d} E}{\mathrm{d} t}=(I)/(\epsilon _0A)`

`\frac{\mathrm{d} E}{\mathrm{d} t}=(4.1)/(8.85* 10^(-12)* 24.63* 10^(-4))`

`\frac{\mathrm{d} E}{\mathrm{d} t}=1.88* 10^(14) V/m-s`