The capacitance of a parallel plate capacitor
\[C=\frac{Q}{V}\]
\[dV=-\int Edr\]
\[E=\]

- anonymous

The capacitance of a parallel plate capacitor
\[C=\frac{Q}{V}\]
\[dV=-\int Edr\]
\[E=\]

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- anonymous

|dw:1362265815092:dw|

- anonymous

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should I use gauss's law to find the electric field?

- experimentX

yes .. use the gauss law. consider a single plate first.

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## More answers

- anonymous

\[\oint E dA=\frac{Q}{\epsilon_0}\]
|dw:1362266179361:dw|
\[Q=\sigma A\]
\[EL^2=\frac{Q}{\epsilon_0}=\frac{\sigma L^2}{\epsilon_0}\]
\[E=\frac{\sigma}{\epsilon_0}\]

- anonymous

since we have two plates do we use \(2\sigma\)?

- experimentX

yeah yeah ... that;s the way ...
let me add few things ... |dw:1362266439292:dw|

- experimentX

how should the flux lines go? fit it geometrically

- anonymous

|dw:1362266583983:dw|

- experimentX

yep!!
Now what is flux??

- anonymous

\[\sigma/\epsilon_0\]?
or do I need to consider the cylinder?

- experimentX

|dw:1362266689639:dw|
yes ... it's okay even if you consider square or rectangle.

- anonymous

I used L^2 for area

- experimentX

everything is okay ... the answer is interesting ... If you have infinite charged sheet ... not matter how far you fly, the electric field will never decrease.

- experimentX

looks like you can use that on gravity too ...

- anonymous

should I try putting in values? Let's say 10cm=L and the plate seperation is 1mm.
wait, so the electric field does not depend on the dimensions? would I need to substitute for sigma and epsilon to find the electric field if I'm only given geometric dimensions?

- experimentX

No ... calculate the charge density.

- anonymous

\[\sigma=q/A= \frac{1.6\times10^{-19}}{.1^2}=1.6E-17 C/m^2\]
\[E=\frac {\sigma}{}\]

- experimentX

looks like you are going in wrong direction ... can you state the original question

- anonymous

I think we're on the right track...once we have E we can solve for V and so on....should I leave E in variable form for now.

- anonymous

?

- anonymous

but what are we integration over....the distance between the plates?

- experimentX

Q <-- this should the be the total charge on the surface ... and A <-- this is the total area of the surface.

- anonymous

is this different from the gaussian surface? do I need different variable to use gauss's law?

- anonymous

yeah it looks like I haven't quite grasped the concept of Gauss's law

- anonymous

should we use the cylinder?

- experimentX

hmm ... what is your original question?

- anonymous

I have to find the capacitance of the parallel plate capacitor.

- experimentX

what information are given initially?

- anonymous

the plate is square and has length 10cm....the plates are separated by 1mm

- experimentX

ow ..

- anonymous

\[dV=\int\frac{\sigma}{\epsilon_0}dA\]

- anonymous

is that correct?

- anonymous

\[V=\frac{2\sigma A}{\epsilon_0}\]

- experimentX

Woops!! error

- experimentX

|dw:1362268232059:dw|

- anonymous

but would that still be
\[V=\frac{\sigma A}{\epsilon_0}\]

- experimentX

|dw:1362268629160:dw|

- anonymous

Let me try ...One moment....

- experimentX

|dw:1362268884843:dw|

- anonymous

\[\oint EdA=\frac Q{\epsilon_0}\]
\[E\pi r^2=\frac{Q}{\epsilon_0}\]
\[\rho=\frac Q {\pi r^2L}\]
\[E\pi r^2=\frac{\rho \pi r^2 L}{\epsilon_0}\]

- anonymous

so the charge density is that of the cylinder?

- anonymous

\[E=\frac{\rho L}{\epsilon_0}\]

- anonymous

I completely missed the point didn't I?

- experimentX

use sigma for surface charge density.

- experimentX

|dw:1362269453040:dw|

- anonymous

oh so the flux is only though the circle...
\(\phi\) is of that surface....because we're looking for the flux through the plate, and we're using a cylinder to model that....I think I got it now

- anonymous

yes?

- anonymous

\[\phi=\oint E\cdot dr=\frac{Q}{\epsilon_0}\]

- experimentX

No ... we are not integrating over the circle.

- anonymous

closed integral from a to b?

- experimentX

|dw:1362269839865:dw|

- experimentX

to understand Gauss law ... you should understand flux.
Find the total flux .. this case
|dw:1362269948919:dw|

- anonymous

is this a circular surface of charge +q ?

- experimentX

no ... not circular ... spherical!!

- anonymous

the flux would be
|dw:1362270059079:dw|
oh ok...I guess we'll work with a sphere instead...Let's see...

- experimentX

consider ... the charge in enclosed by sphere.

- anonymous

|dw:1362270254679:dw|
in all directions out of the sphere...the Electric field is perpendicular to the surface everywhere

- anonymous

|dw:1362270435029:dw|

- experimentX

yes ...

- anonymous

Yay!!! sigh...

- experimentX

what is the total flux?

- anonymous

zero?

- anonymous

the sum of all of the electric field lines?

- anonymous

zero

- anonymous

final answer

- experimentX

No ...

- anonymous

sum of all the Electric field lines
\[\phi=\oint E dr=\frac{Q}{\epsilon_0}\]

- anonymous

volume?

- experimentX

You should understand what is flux first ...
Flux is like ... |dw:1362270773063:dw|

- anonymous

the amount of water that flows through it is A times \(v\)

- anonymous

\[\frac{meter^2}{1}\times\frac{meters}{second}\]
that doesn't seem right

- experimentX

yeah .. that's correct. It's easy to visualize water ... |dw:1362271042142:dw|

- anonymous

yep

- anonymous

|dw:1362271173946:dw|

- anonymous

so the flux through the area would be the volume of the sphere times the electric field

- anonymous

I tried to make your circle look more like a sphere :P

- experimentX

check your answer again!!

- anonymous

oh surface area of the sphere!

- experimentX

yeah ... what is the total flux?

- anonymous

because the flux is through the surface of the sphere :D
The total flux is the sum of all of (the surface areas multiplied by the electric field)

- anonymous

i meant surface area...not areas

- experimentX

|dw:1362271421314:dw|

- anonymous

what's s? surface area?

- anonymous

ds...?

- experimentX

|dw:1362271575577:dw|

- anonymous

|dw:1362271615713:dw|

- anonymous

what's that?

- experimentX

sphere.

- anonymous

oh are you saying that
\[\int ds=s\] yeah I believe that
but is s ...oh i see sphere

- anonymous

we're integrating with respect to the sphere, correct?

- anonymous

the surface area of the sphere

- experimentX

yes ... but electric Field is constant .. since it's sphere.

- anonymous

yep

- anonymous

what would you limits be?

- experimentX

and we are integrating surface ...not distance.
lol ... the limits ... don't worry

- anonymous

so \[\oint \textrm{never has limits?}\]

- experimentX

|dw:1362271874243:dw|

- anonymous

yeah I get that E is constant everywhere over the sphere

- anonymous

surface*

- experimentX

|dw:1362271927389:dw|

- anonymous

oh because were integrating
\[\oint ds=s \;\;\;\textrm{we don't need to worry about the limits}\]

- anonymous

since the answer is just s

- experimentX

|dw:1362272111497:dw|

- anonymous

yep , makes sense....

- anonymous

I finally understand flux.... :)

- experimentX

Not complete yet!!

- anonymous

|dw:1362272196631:dw|

- experimentX

|dw:1362272206784:dw|

- anonymous

i'm not gonna look

- anonymous

not gonna look

- anonymous

not gonna look

- anonymous

ok fine lol

- anonymous

keep going...i'm sorry

- experimentX

Now ... |dw:1362272365512:dw|

- experimentX

this is called conservation of flux.

- experimentX

|dw:1362272482209:dw|

- experimentX

|dw:1362272583908:dw|

- experimentX

|dw:1362272690179:dw|

- anonymous

through both spheres? yes

- experimentX

This is why Gravity ... and Electrostatic force follow inverse square laws.
Any shape ... but must be closed. and charge must be inside it.

- anonymous

interesting! yeah \(E\propto \frac 1{r^2}\)
\[F_g\propto \frac 1{r^2}\]

- anonymous

or is it
\[F_e\propto \frac 1{r^2}\]
and
\[F_g\propto \frac 1{r^2}\]

- experimentX

just think that ... charge is mouth of that pipe on that water example ... and outer envelopes is the other hose water comes out thorugh.
It is independent of spape of pipe.

- experimentX

|dw:1362273084528:dw|

- experimentX

|dw:1362273139951:dw|

- anonymous

oh|dw:1362273163066:dw|
flux is still the same

- experimentX

yes ... the flux is still the same.

- experimentX

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