How is the Electric Field Rate Changing in a Charging Parallel-Plate Capacitor?

[jehan60188]

Homework Statement

A parallel-plate capacitor with circular plates of radius 1.7 m is being charged. Consider a circular loop centered on the central axis between the plates. The loop has a radius of 2.6 m and the displacement current through the loop is 2 A.

(a) At what rate is the electric field between the plates changing?

Homework Equations

maxwell's equations
1) electric flux through a closed surface = charge enclosed by the surface divided be e0
2) magnetic flux through an open surface = u0 times current through the surface
3) EMF in a closed path = derivative of magnetic flux through the path WRT time
4) currents cause magnetic fields

The Attempt at a Solution

there's an E-field in the capacitor
C = e0*A/d
q = C*V
E = q/(A*e0) SO dE/dt = dq/dt * 1/(A*e0)

I feel like I'm missing something.
Does a changing E-field induce a current in a closed path? is that current proportional to the electric flux through the path?

 

[tiny-tim]

hi jehan60188!

… The loop has a radius of 2.6 m and the displacement current through the loop is 2 A.
…
Does a changing E-field induce a current in a closed path? is that current proportional to the electric flux through the path?

"displacement current" isn't a current, it's ∂D/∂t, a current density (per area) see http://en.wikipedia.org/wiki/Displacement_current

(but it's measured in units of A/m², not A … perhaps they mean that 2A is the total current through the loop, ie the displacement current times the area? )

 

[jehan60188]

bump, googled all week, but still no idea what to do

 

[tiny-tim]

hi jehan60188!

A parallel-plate capacitor with circular plates of radius 1.7 m is being charged. Consider a circular loop centered on the central axis between the plates. The loop has a radius of 2.6 m and the displacement current through the loop is 2 A.

(a) At what rate is the electric field between the plates changing?

i was hoping you might find some more information (btw, what are parts (b) (c) etc of the question?), but if that's all there is, my guess is that the question means that ∂D/∂t times the area of the loop is 2 A

(D is in coulombs per m², so ∂D/∂t is amps per m², so the units are correct)

in that case, (a) is asking for ∂E/∂t, so i suppose all you need is an equation relating D and E

 

[jehan60188]

solution
phi = E*A
so
d(phi)/dt = d(E*A)/dt = dE/dt*(A)
so
e*d(phi)/dt = e* dE/dt*(A)
where e = 8.85e-12
so, we know displacement current = 2 (since capacitors only have displacement current)
divide that by the area of the plates to get dE/dt

 

[tiny-tim]

that's a rather confusing way of writing it

you're saying ∂E/∂t = (1/ε_o)∂D/∂t = 2/ε_o N/C.s ?

why do you want to divide by the area? (and why did you mention phi?)

 

[jehan60188]

phi is the electric flux through a surface
I_d = ε*d(phi)/dt is the definition of displacement current
I'm going to change over to using single-quote to represent time-derivatives (so dx/dt = x')

amperes law:

integral(B~ds) = u*(I + I_d)
where ~ is the dot product
or in words: a closed path integral in a magnetic field is equal to the permeability of free space times the sum of displacement current (I_d) and enclosed current

for uniform E-fields electric flux is E*A
since our A is constant, phi' = A*E'

so: ε*phi' = εAE'
between the plates of a capacitor, there is only displacement current
so: ε*phi' = εAE' = 2
so: E' = 2/(Aε)

 

[tiny-tim]

between the plates of a capacitor, there is only displacement current
so: ε*phi' = εAE' = 2
so: E' = 2/(Aε)

oh yes, i forgot , we decided that 2 amps was D' times area

so you're right, E' = D'/ε = D'A/Aε = 2/(Aε) N/C.s …

but you still don't need to mention phi, do you?​