Question

In the figure, a parallel-plate capacitor is being discharged by a current ? = 5.0 ?. The plates are square with edge length
? = 8.0 ??.

1. (a) What is the displacement current ?! and displacement current density ?! (current per unit of area) in the air space between the plates?

2. (b) What is the rate at which the electric field between the plates is changing?

3. (c) What is the value of ? ∙ ?? around the dashed path, where ? = 2.0 ?? and ? =3.0 ???

Answer 1

(a)What is the displacement current ?! and displacement current density ?! (current per unit of area) in the air space between the plates?

Given conduction current i_c=5.0 ?

So the displacement current ?­_d = conduction current i_{c =}5.0 ?

and current density J= i_d/A

where A=L²=8²=64mm=64*10^-3m

Current density J= 5/(64*10^-3)=78.125 A/m²

(b) What is the rate at which the electric field between the plates is changing?

The rate at which the electric field between the plates is changing

is rate of electric field

= 78.125/ (8.854*10^-12)

                                                            = 8.823**10¹² A/Fm

(c) What is the value of ? ∙ ?? around the dashed path, where ? = 2.0 ?? and ? =3.0 ???

Answer: No diagram is given in picture