Question

A parallel-plate capacitor in air has circular plates of radius 3.0 cm separated by 1.1 mm. Charge is flowing onto the upper plate and off the lower plate at a rate of 5 A. Find the time rate of change of the electric field between the plates.

Answer 1

Answer:

2 x 10^14 N/Cs

Explanation:

radius, r = 3 cm

Area , A = 3.14 x 3 x 3 = 28.26 cm^2 = 28.26 x 10^-4 m^2

d = 1.1 mm = 1.1 x 10^-3 m

i = 5 A

Let time be t and electric field strength is E.

Charge, q = i x t = 5 t

q = C V

q = C x E x d

$5 t = \frac{\varepsilon _{0}A}{d}\times E\times d$

$E/t = \frac{5}{\varepsilon _{0}A}$

$E/t = \frac{5}{8.854\times 10^{-12}\times 28.26\times 10^{-4}}$

E/t = 2 x 10^14 N/Cs