In the figure below, a parallel-plate capacitor has square plates of edge length
ID: 1474657 • Letter: I
Question
In the figure below, a parallel-plate capacitor has square plates of edge length L = 1.0 m. A current of 3.6 A charges the capacitor, producing a uniform electric field E vector between the plates, with E vector perpendicular to the plates. What is the displacement current id through the region between the plates A What is dE/dt in this region V/m-s What is the displacement current encircled by the square dashed path of edge length d = 0.50 m A What is closed integral B vector-ds around this square dashed path T-mExplanation / Answer
We using the equations
The electric flux FE = E A = Q/ e0
The electric current I= dQ/dt
Displacement current Id =e0 dFE / dt
Frist part
We calculate the electric flow (FE)variation
d FE / dt = (dQ/dt) /e0 = I /e0 = 3.6 / 8.85 10-12 = 0.406 1012 V m/s
Id = e0 d FE / dt = I = 3.6 A
Second part
d FE / dt = d (EA)/dt =( dQ/dt) /e0 = I/e0
A= L d L=1 m d =0.5 m
dE/dt = I/(e0 A)
dE/dt = 3.6/ (8.85 10-12 1 0.5)= 0.814 1012 V/ m s
third part
Id = d(EA)/dt = A dE/dt = (0.5* 1) 0.814 1012
Id= 0.407 1012 A
Four part
B.ds = e0 m0 d FE / dt
A0 initial area L= 1 m A0=1 m2
A interest area d= 0.5 m
B .ds = e0 m0 (d EA / dt) = e0 m0 (I/ e0 Ao )A = m0 I A/A0 =4 10-7 3.6 (0.5)2
B .ds = 3.6 10-7 T m