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

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|dw:1362265934285:dw| should I use gauss's law to find the electric field?
yes .. use the gauss law. consider a single plate first.

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\[\oint E dA=\frac{Q}{\epsilon_0}\] |dw:1362266179361:dw| \[Q=\sigma A\] \[EL^2=\frac{Q}{\epsilon_0}=\frac{\sigma L^2}{\epsilon_0}\] \[E=\frac{\sigma}{\epsilon_0}\]
since we have two plates do we use \(2\sigma\)?
yeah yeah ... that;s the way ... let me add few things ... |dw:1362266439292:dw|
how should the flux lines go? fit it geometrically
yep!! Now what is flux??
\[\sigma/\epsilon_0\]? or do I need to consider the cylinder?
|dw:1362266689639:dw| yes ... it's okay even if you consider square or rectangle.
I used L^2 for area
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.
looks like you can use that on gravity too ...
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?
No ... calculate the charge density.
\[\sigma=q/A= \frac{1.6\times10^{-19}}{.1^2}=1.6E-17 C/m^2\] \[E=\frac {\sigma}{}\]
looks like you are going in wrong direction ... can you state the original question
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.
but what are we integration over....the distance between the plates?
Q <-- this should the be the total charge on the surface ... and A <-- this is the total area of the surface.
is this different from the gaussian surface? do I need different variable to use gauss's law?
yeah it looks like I haven't quite grasped the concept of Gauss's law
should we use the cylinder?
hmm ... what is your original question?
I have to find the capacitance of the parallel plate capacitor.
what information are given initially?
the plate is square and has length 10cm....the plates are separated by 1mm
ow ..
is that correct?
\[V=\frac{2\sigma A}{\epsilon_0}\]
Woops!! error
but would that still be \[V=\frac{\sigma A}{\epsilon_0}\]
Let me try ...One moment....
\[\oint EdA=\frac Q{\epsilon_0}\] \[E\pi r^2=\frac{Q}{\epsilon_0}\] \[\rho=\frac Q {\pi r^2L}\] \[E\pi r^2=\frac{\rho \pi r^2 L}{\epsilon_0}\]
so the charge density is that of the cylinder?
\[E=\frac{\rho L}{\epsilon_0}\]
I completely missed the point didn't I?
use sigma for surface charge density.
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
\[\phi=\oint E\cdot dr=\frac{Q}{\epsilon_0}\]
No ... we are not integrating over the circle.
closed integral from a to b?
to understand Gauss law ... you should understand flux. Find the total flux .. this case |dw:1362269948919:dw|
is this a circular surface of charge +q ?
no ... not circular ... spherical!!
the flux would be |dw:1362270059079:dw| oh ok...I guess we'll work with a sphere instead...Let's see...
consider ... the charge in enclosed by sphere.
|dw:1362270254679:dw| in all directions out of the sphere...the Electric field is perpendicular to the surface everywhere
yes ...
Yay!!! sigh...
what is the total flux?
the sum of all of the electric field lines?
final answer
No ...
sum of all the Electric field lines \[\phi=\oint E dr=\frac{Q}{\epsilon_0}\]
You should understand what is flux first ... Flux is like ... |dw:1362270773063:dw|
the amount of water that flows through it is A times \(v\)
\[\frac{meter^2}{1}\times\frac{meters}{second}\] that doesn't seem right
yeah .. that's correct. It's easy to visualize water ... |dw:1362271042142:dw|
so the flux through the area would be the volume of the sphere times the electric field
I tried to make your circle look more like a sphere :P
check your answer again!!
oh surface area of the sphere!
yeah ... what is the total flux?
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)
i meant surface area...not areas
what's s? surface area?
what's that?
oh are you saying that \[\int ds=s\] yeah I believe that but is s ...oh i see sphere
we're integrating with respect to the sphere, correct?
the surface area of the sphere
yes ... but electric Field is constant .. since it's sphere.
what would you limits be?
and we are integrating surface ...not distance. lol ... the limits ... don't worry
so \[\oint \textrm{never has limits?}\]
yeah I get that E is constant everywhere over the sphere
oh because were integrating \[\oint ds=s \;\;\;\textrm{we don't need to worry about the limits}\]
since the answer is just s
yep , makes sense....
I finally understand flux.... :)
Not complete yet!!
i'm not gonna look
not gonna look
not gonna look
ok fine lol
keep going...i'm sorry
Now ... |dw:1362272365512:dw|
this is called conservation of flux.
through both spheres? yes
This is why Gravity ... and Electrostatic force follow inverse square laws. Any shape ... but must be closed. and charge must be inside it.
interesting! yeah \(E\propto \frac 1{r^2}\) \[F_g\propto \frac 1{r^2}\]
or is it \[F_e\propto \frac 1{r^2}\] and \[F_g\propto \frac 1{r^2}\]
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.
oh|dw:1362273163066:dw| flux is still the same
yes ... the flux is still the same.