Question

Now setting up another problem regarding calculating the displacement current of a parallel-plate capacitor, not sure if I'll need help yet, will be posting attempted workings beneath.

Answer 1

"A parallel-plate capacitor has square plates 10 cm on a side and 0.50 cm apart. The voltage across the plates is increasing at 220 V/ms. What is the displacement current across the capacitor?"

Answer 2

Alright, so what I ultimately need to be able to find is the Electric Flux. I'm guessing I can use the information given about the capacitor to calculate its electric field strength and then calculate its electric flux from that, and then go from flux to displacement current.

Answer 3

Alright, I found a formula for the electric field strength between two idealized, large parallel, charged plates.
\[E=\frac{\sigma}{\epsilon}\]Sigma is charge density, and epsilon is permittivity.

Found another equation expressing the voltage across the two plates to be equal to the electric field times the distance in between them,
\[V = Ed.\]Since I have both the voltage and the distance between the two plates, I can solve for electric field strength:

\[E = \frac{V}{d}=\frac{}{}\]

Wait, no, the voltage isn't constant...

Answer 4

@blurbendy or @hba , I'm not sure if you're any good with Physics stuff, but I'm linking you in case you can help me; any help would be appreciated if you can.

Answer 5

Displacement current is always equal to the current that is charging the capacitor, apparently.

Answer 6

@ganeshie8 , are you any good with Electricity & Magnetism?

Answer 7

According to the Wiki page for Capacitors, http://upload.wikimedia.org/math/7/5/6/756d1aa6542ef0266a7aed62154c1c76.png

Answer 8

@dan815

Answer 9

\[I(t)=\frac{dQ(t)}{dt}=C\frac{dV(t)}{dt}\]

Answer 10

Alright, so it looks like we're given the rate of change of voltage with respect to time, so I'm guessing that's \[\frac{dV(t)}{dt}=\frac{220 V}{1 \ ms}\]

(Speaking of, hey Ganeshie, do you know how I put text upright in the editor? I know there's a command like \textup in normal editors, but I can never get it to work on OS.)

Answer 11

\[C = \frac{\varepsilon A}{D}=\frac{k \varepsilon_{0} A}{D}\]

Answer 12

(But yeah, Ganeshie, do you know how I can put text upright in the editor?)

Answer 13

use `\rm` ?

Answer 14

\[C = \frac{(8.854 \cdot 10^{-12}F/ \rm m)(1)}{[(0.1 \ \rm m ) \cdot (0.1 \ \rm m)]}=\]

Answer 15

Thanks! I've been wanting to know how to do that for a long time, heh.

Answer 16

that doesn't look right

Answer 17

A is in numerator and d is in denominator

Answer 18

What doesn't look right about the calculation for C? epsilon naught and k are given values, the permittivity of free space, together, divided by-OH, yeah

Answer 19

Yep, you're right, denominator is distance, not area, thank you.

Answer 20

k=1 for air i think

Answer 21

So then it's just, \[\frac{8.854\cdot10^{-12}}{0.1\ \rm m}=8.854\cdot10^{-11}\]

Answer 22

`A parallel-plate capacitor has square plates 10 cm on a side and 0.50 cm apart.`

A = 0.1^2
d = 0.005

Answer 23

UGUUUH thank you, I keep getting ahead of myself (and actually excited I appeared to have figured this out on my own) and am making mistakes everywhere...)

Answer 24

Nice :) I don't seem to remember much of these anymore :O

Answer 25

Yeah, alright, so trying to do this without making mistakes, heh, here we go:

Answer 26

Due to the fact that for a capacitor,\[I_{D}=I,\]or displacement current is equivalent to the current running *through* a capacitor, \[I_{D}=I=C \frac{dV(t)}{dt}=\frac{\varepsilon A}{D}\frac{dV(t)}{dt}\]

Answer 27

\[\frac{\varepsilon A}{D} \frac{dV(t)}{dt}=\frac{8.854\cdot10^{-12}F/ \rm m}{0.5\cdot10^{-2}\ \rm m}\bigg(220\cdot 10^{-3} \rm V/\rm ms\Bigg)\]

Answer 28

Whoops, the units in the dV/dt should be just volts per second due to the times ten to the negative three.

Answer 29

\[=3.89576\cdot10^{-10}\] is what I got, my algebra might be off. I hope not.

Answer 30

\[\frac{\varepsilon A}{D} \frac{dV(t)}{dt}=\frac{(8.854\cdot10^{-12}F/ \rm m )(0.1 )^2}{0.5\cdot10^{-2}\ \rm m}\bigg(220~ \rm V/\rm ms\Bigg) \]

Answer 31

http://www.wolframalpha.com/input/?i=8.854*10%5E%28-12%29*0.1%5E2%2F0.005*220

Answer 32

3.896 nA

Answer 33

Whoah, lol, I just....totally forgot a term in there. Thank you!

Answer 34

np:) the currentlooks very small hmm

Answer 35

(Uh oh?)