Question

A physics problem dealing with the ambiguity in Ampere's Law, posted below momentarily.

Answer 1

"A uniform electric field is increasing at 1.5(V/m) microseconds. FInd the displacement current through a 1 cm^2 area through the field."

Answer 2

I have pretty much.......no clue how to solve this.

Answer 3

Alright well, do you know of any formulas that contain "displacement current" in them? That will no doubt be your best place to begin, start with making sure you have the formula that gives you the result and then start to piece your way back to what you're given. The other bit of advice to give is when you read, try to understand and you will understand.

Answer 4

"try to understand and you will understand."

That sounded borderline insulting, but sure, I'll try to do that. I definitely try to understand.

Dealing with the formula, I don't remember, we've been covering optics for some time now, so I've forgotten most of this. I'm rereading the chapter now, but don't want to get lost and sink time into rereading and for it to potentially come up wasted.

Answer 5

(I'm guessing this has to do with Ampere's Law, but despite trying, my application of both that and Biot-Savart is shaky at best despite putting a good amount of time into both.)

Answer 6

Anyways, just going to go read, I guess.

Answer 7

I think Ampere's Law with Maxwell's Modification is related, the quantity in it called "Displacement Current".

We literally never covered this stuff in the form of a test. \[\int\limits_{}^{}B\cdot dr=\mu_{0}I+\mu_{0}e_{0}\frac{d \phi_{E}}{dt}\]
Dealing with the right hand term.

Answer 8

Alright, now not sure what I'm supposed to do. Think I maybe have to take the derivative of something, but-again-i'm not sure.

Answer 9

@Kainui , I'm really not sure how to proceed from here or how I'm supposed to do this.

Answer 10

@jim_thompson5910 , are you at all good with Physics? I could use some help on this.

Answer 11

With some physics yes, but the basics really. This kind of stuff seems very foreign to me.

Answer 12

I have some quantity like Electric Flux given in the prompt if I'm not mistaken(?), or is it something near it?

No problem, thanks nonetheless for responding.

Answer 13

try asking if anyone is an electrical engineer. That seems like the stuff they study (maybe look in the EE or CS section of OS?)

EE = electrical engineering
CS = computer science

Answer 14

\[\phi_{E}\]Is electric flux, and I have to find \[\frac{d \phi_{E}}{dt}\]

Answer 15

Yeah, will do. Thank you. @perl , @wio , are you decent with Electricity & Magnetism/Maxwell's Equations related stuff?

Answer 16

It's been a while

Answer 17

Heh, no problem, I'm having to re-read everything.

Answer 18

(But yeah, if you're not sure, don't hold your breath, I legitimately do not want to waste your time if you don't know, lol.)

Answer 19

It seems like a flux integral.

Answer 20

Since it is a vector field acting on an area.

Answer 21

From the Wikipedia page for Electric Flux:

"If the electric field is uniform, the electric flux passing through a surface of vector area S is

\[\Phi_E = \mathbf{E} \cdot \mathbf{S} = ES \cos \theta"\]

Answer 22

So I'm guessing that might be useful? But the statement in the prompt that it's perpendicular would make me think that it's zero. \[ES \cos(\pi/2)\]

Answer 23

Hah! I've seen this make appearances on the physics GRE.

It's just multiplication.

Area * Change in Field * epsilon naught

Answer 24

^Yeah, but I'm not sure how you got there/what formulas you used to get there. I'm finding, poking around the internet, uses of formulas dealing with specific physical things like a capacitor, but not just an increasing electric field.

The most common example I've seen is with a capacitor, where they first, based on a voltage, calculate the electric field of the capacitor, and then calculate the electric flux through the capacitor, and then calculate the displacement current from that; I'm not sure how to approach it from dealing with *just* a field.

Answer 25

It derives from the concept of a capacitor, would you like me to show the derivation?

Answer 26

(V/m) is units of electric density

Answer 27

"A uniform electric field is increasing at 1.5(V/m) microseconds."

Electric field is increasing, what term might this represent in your equation?

" Find displacement current"

You have identified this term, so now you are trying to work with everything else, good job so far.

"through a 1 cm^2 area through the field."

Area and amount through a field, which term is this in your equation?

Is it possible that any of the terms in your equation are representing things that aren't here, and you can safely set equal to zero? If you see a term you do not know what it is, look at its units. How is this unit measured or derived from other units?

---

I'm not being insulting; I would be if I had said, "try to understand, and you will fail." =)

Answer 28

intensity^

Answer 29

(Is anybody else not getting notifications sometimes?)

Just realized people posted a bunch of stuff, playing catchup at the moment.

Answer 30

V/m is units of electric density, or electric intensity? Or electric field strength, or are all three synonymous.

Answer 31

intensity

Answer 32

(But yeah, I'd like the derivation of how they got there very much.)

Answer 33

"Electric field is increasing, what term might this represent in your equation?"

I'm guessing just the magnitude of the electric field, and this should be put into the EScos(theta) formula? The thing is, *I don't just want to throw random numbers everywhere it looks convenient, and am not doing that.*

Answer 34

And that's the thing, too, I'm thrown off by that "perpendicular to the field" thing for using the formula I last posted for Electric Flux; Theta is 90 degrees, isn't it, so you'd have cos(pi/2), and thus have the whole thing become zero?

Answer 35

We can use the following:
\[E = \frac{ Q }{ \epsilon _{0} A }\]\[\phi _{E} = EA\]
Therefore:\[\phi _{E} = \frac{Q}{\epsilon _{0}}\]
Rate of change of charge is the definition of current:\[\frac{d \phi _{E}}{dt}= \frac{1}{\epsilon _{0}} \frac{dQ}{dt}= \frac {I}{\epsilon _{0}}\]
This means that the displacement current would be equal to:\[I = \epsilon _{0} \frac{d \phi _{E}}{dt}\]

You're given rate of change of electric field and an area.
The rate of change times the area gives you rate of change of flux.

\[I = \epsilon _{0} \frac{d (EA)}{dt}\]

Answer 36

The meters^2 of area cancels V/m and F/m

A F*V is a coulomb C

You're left with time in the denominator.

C/s = Ampere A

Answer 37

Just saw this, taking a look at it now. Thanks so much.

Answer 38

Alright, cool, this totally makes sense. Thanks so much. Where exactly did the original formula come from, though? (I mean the very first one in your derivation-had I known or remembered that, I'd probably gotten along just fine.)

I've honestly been neglecting the living heck out of Physics and been putting all of my time towards ODE and Calc III, to the point that I've forgotten fundamentals badly.

\[E = \frac{Q }{\epsilon_{0}A}\]Came from what?

Answer 39

That's the equation of electric field of large parallel plates.

Gauss's Law: \[\int\limits Eda = \frac{Q}{\epsilon _{o}}\]

Answer 40

If you take the area vector that is normal to the surface of a plate

Answer 41

Alright, cool, thank you. dA is the area element differential thing?

Answer 42

yes

Answer 43

should be dot product, but cosine(0) = 1

Answer 44

area vector and e-field are the same direction on the plate (away from plate)

Answer 45

Alright, awesome. Thank you very much, had I looked in the proper places and found Gauss' Law in the first place, I wouldn't have had to ask, heh. Thanks again, man.