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

The time rate of change of the electric field in a parallel-plate capacitor with a current flowing is found by dividing the current by the vacuum permittivity and the area of the plates.

Step-by-step explanation:

The question asks to determine the time rate of change of the electric field between the plates of a parallel-plate capacitor while a current is present. Since current (I) is the rate of charge (Q) flow, I = dQ/dt, the time rate of change of the electric field can be related to the current in the capacitor using the formula for an electric field in a parallel-plate capacitor, which is E = σ/ε₀, where σ is the surface charge density (Q/A), and ε₀ is the vacuum permittivity. Given that the current is 5 A (∆Q/∆t), the area of the plates A can be calculated using the radius given and the formula for the area of a circle. The rate of the electric field change is then ∆E/∆t = I/ε₀A.

Answer 2

2 x 10^14 N/Cs

Step-by-step 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 = (\varepsilon _(0)A)/(d)* E* d`

`E/t = (5)/(\varepsilon _(0)A)`

`E/t = (5)/(8.854* 10^(-12)* 28.26* 10^(-4))`

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