anonymous
  • anonymous
The capacitance of a parallel plate capacitor \[C=\frac{Q}{V}\] \[dV=-\int Edr\] \[E=\]
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
|dw:1362265815092:dw|
anonymous
  • anonymous
|dw:1362265934285:dw| should I use gauss's law to find the electric field?
experimentX
  • experimentX
yes .. use the gauss law. consider a single plate first.

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anonymous
  • 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
  • anonymous
since we have two plates do we use \(2\sigma\)?
experimentX
  • experimentX
yeah yeah ... that;s the way ... let me add few things ... |dw:1362266439292:dw|
experimentX
  • experimentX
how should the flux lines go? fit it geometrically
anonymous
  • anonymous
|dw:1362266583983:dw|
experimentX
  • experimentX
yep!! Now what is flux??
anonymous
  • anonymous
\[\sigma/\epsilon_0\]? or do I need to consider the cylinder?
experimentX
  • experimentX
|dw:1362266689639:dw| yes ... it's okay even if you consider square or rectangle.
anonymous
  • anonymous
I used L^2 for area
experimentX
  • 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
  • experimentX
looks like you can use that on gravity too ...
anonymous
  • 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
  • experimentX
No ... calculate the charge density.
anonymous
  • anonymous
\[\sigma=q/A= \frac{1.6\times10^{-19}}{.1^2}=1.6E-17 C/m^2\] \[E=\frac {\sigma}{}\]
experimentX
  • experimentX
looks like you are going in wrong direction ... can you state the original question
anonymous
  • 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
?
anonymous
  • anonymous
but what are we integration over....the distance between the plates?
experimentX
  • experimentX
Q <-- this should the be the total charge on the surface ... and A <-- this is the total area of the surface.
anonymous
  • anonymous
is this different from the gaussian surface? do I need different variable to use gauss's law?
anonymous
  • anonymous
yeah it looks like I haven't quite grasped the concept of Gauss's law
anonymous
  • anonymous
should we use the cylinder?
experimentX
  • experimentX
hmm ... what is your original question?
anonymous
  • anonymous
I have to find the capacitance of the parallel plate capacitor.
experimentX
  • experimentX
what information are given initially?
anonymous
  • anonymous
the plate is square and has length 10cm....the plates are separated by 1mm
experimentX
  • experimentX
ow ..
anonymous
  • anonymous
\[dV=\int\frac{\sigma}{\epsilon_0}dA\]
anonymous
  • anonymous
is that correct?
anonymous
  • anonymous
\[V=\frac{2\sigma A}{\epsilon_0}\]
experimentX
  • experimentX
Woops!! error
experimentX
  • experimentX
|dw:1362268232059:dw|
anonymous
  • anonymous
but would that still be \[V=\frac{\sigma A}{\epsilon_0}\]
experimentX
  • experimentX
|dw:1362268629160:dw|
anonymous
  • anonymous
Let me try ...One moment....
experimentX
  • experimentX
|dw:1362268884843:dw|
anonymous
  • 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
  • anonymous
so the charge density is that of the cylinder?
anonymous
  • anonymous
\[E=\frac{\rho L}{\epsilon_0}\]
anonymous
  • anonymous
I completely missed the point didn't I?
experimentX
  • experimentX
use sigma for surface charge density.
experimentX
  • experimentX
|dw:1362269453040:dw|
anonymous
  • 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
  • anonymous
yes?
anonymous
  • anonymous
\[\phi=\oint E\cdot dr=\frac{Q}{\epsilon_0}\]
experimentX
  • experimentX
No ... we are not integrating over the circle.
anonymous
  • anonymous
closed integral from a to b?
experimentX
  • experimentX
|dw:1362269839865:dw|
experimentX
  • experimentX
to understand Gauss law ... you should understand flux. Find the total flux .. this case |dw:1362269948919:dw|
anonymous
  • anonymous
is this a circular surface of charge +q ?
experimentX
  • experimentX
no ... not circular ... spherical!!
anonymous
  • anonymous
the flux would be |dw:1362270059079:dw| oh ok...I guess we'll work with a sphere instead...Let's see...
experimentX
  • experimentX
consider ... the charge in enclosed by sphere.
anonymous
  • anonymous
|dw:1362270254679:dw| in all directions out of the sphere...the Electric field is perpendicular to the surface everywhere
anonymous
  • anonymous
|dw:1362270435029:dw|
experimentX
  • experimentX
yes ...
anonymous
  • anonymous
Yay!!! sigh...
experimentX
  • experimentX
what is the total flux?
anonymous
  • anonymous
zero?
anonymous
  • anonymous
the sum of all of the electric field lines?
anonymous
  • anonymous
zero
anonymous
  • anonymous
final answer
experimentX
  • experimentX
No ...
anonymous
  • anonymous
sum of all the Electric field lines \[\phi=\oint E dr=\frac{Q}{\epsilon_0}\]
anonymous
  • anonymous
volume?
experimentX
  • experimentX
You should understand what is flux first ... Flux is like ... |dw:1362270773063:dw|
anonymous
  • anonymous
the amount of water that flows through it is A times \(v\)
anonymous
  • anonymous
\[\frac{meter^2}{1}\times\frac{meters}{second}\] that doesn't seem right
experimentX
  • experimentX
yeah .. that's correct. It's easy to visualize water ... |dw:1362271042142:dw|
anonymous
  • anonymous
yep
anonymous
  • anonymous
|dw:1362271173946:dw|
anonymous
  • anonymous
so the flux through the area would be the volume of the sphere times the electric field
anonymous
  • anonymous
I tried to make your circle look more like a sphere :P
experimentX
  • experimentX
check your answer again!!
anonymous
  • anonymous
oh surface area of the sphere!
experimentX
  • experimentX
yeah ... what is the total flux?
anonymous
  • 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
  • anonymous
i meant surface area...not areas
experimentX
  • experimentX
|dw:1362271421314:dw|
anonymous
  • anonymous
what's s? surface area?
anonymous
  • anonymous
ds...?
experimentX
  • experimentX
|dw:1362271575577:dw|
anonymous
  • anonymous
|dw:1362271615713:dw|
anonymous
  • anonymous
what's that?
experimentX
  • experimentX
sphere.
anonymous
  • anonymous
oh are you saying that \[\int ds=s\] yeah I believe that but is s ...oh i see sphere
anonymous
  • anonymous
we're integrating with respect to the sphere, correct?
anonymous
  • anonymous
the surface area of the sphere
experimentX
  • experimentX
yes ... but electric Field is constant .. since it's sphere.
anonymous
  • anonymous
yep
anonymous
  • anonymous
what would you limits be?
experimentX
  • experimentX
and we are integrating surface ...not distance. lol ... the limits ... don't worry
anonymous
  • anonymous
so \[\oint \textrm{never has limits?}\]
experimentX
  • experimentX
|dw:1362271874243:dw|
anonymous
  • anonymous
yeah I get that E is constant everywhere over the sphere
anonymous
  • anonymous
surface*
experimentX
  • experimentX
|dw:1362271927389:dw|
anonymous
  • anonymous
oh because were integrating \[\oint ds=s \;\;\;\textrm{we don't need to worry about the limits}\]
anonymous
  • anonymous
since the answer is just s
experimentX
  • experimentX
|dw:1362272111497:dw|
anonymous
  • anonymous
yep , makes sense....
anonymous
  • anonymous
I finally understand flux.... :)
experimentX
  • experimentX
Not complete yet!!
anonymous
  • anonymous
|dw:1362272196631:dw|
experimentX
  • experimentX
|dw:1362272206784:dw|
anonymous
  • anonymous
i'm not gonna look
anonymous
  • anonymous
not gonna look
anonymous
  • anonymous
not gonna look
anonymous
  • anonymous
ok fine lol
anonymous
  • anonymous
keep going...i'm sorry
experimentX
  • experimentX
Now ... |dw:1362272365512:dw|
experimentX
  • experimentX
this is called conservation of flux.
experimentX
  • experimentX
|dw:1362272482209:dw|
experimentX
  • experimentX
|dw:1362272583908:dw|
experimentX
  • experimentX
|dw:1362272690179:dw|
anonymous
  • anonymous
through both spheres? yes
experimentX
  • 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
  • anonymous
interesting! yeah \(E\propto \frac 1{r^2}\) \[F_g\propto \frac 1{r^2}\]
anonymous
  • anonymous
or is it \[F_e\propto \frac 1{r^2}\] and \[F_g\propto \frac 1{r^2}\]
experimentX
  • 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
  • experimentX
|dw:1362273084528:dw|
experimentX
  • experimentX
|dw:1362273139951:dw|
anonymous
  • anonymous
oh|dw:1362273163066:dw| flux is still the same
experimentX
  • experimentX
yes ... the flux is still the same.
experimentX
  • experimentX
|dw:1362273240830:dw|