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:$1.88\times 10^{14} V/m-s$

Explanation:

Given

radius of capacitor Plate $r=2.8 cm$

Area $A=\pi r^2=\pi \times 2.8^\times 10^{-4} m^2$

$A=24.63\times 10^{-4} m^2$

current $I=4.1 A$

separation $d=1.1 mm$

Electric Field strength is given by

$E=\frac{Q}{\epsilon _0A}$

$E=\frac{I\cdot t}{\epsilon _0A}$

$\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{I}{\epsilon _0A}$

$\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{4.1}{8.85\times 10^{-12}\times 24.63\times 10^{-4}}$

$\frac{\mathrm{d} E}{\mathrm{d} t}=1.88\times 10^{14} V/m-s$