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

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JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362265815092:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362265934285:dw should I use gauss's law to find the electric field?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yes .. use the gauss law. consider a single plate first.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\oint E dA=\frac{Q}{\epsilon_0}\] dw:1362266179361:dw \[Q=\sigma A\] \[EL^2=\frac{Q}{\epsilon_0}=\frac{\sigma L^2}{\epsilon_0}\] \[E=\frac{\sigma}{\epsilon_0}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
since we have two plates do we use \(2\sigma\)?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah yeah ... that;s the way ... let me add few things ... dw:1362266439292:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
how should the flux lines go? fit it geometrically
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362266583983:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yep!! Now what is flux??
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\sigma/\epsilon_0\]? or do I need to consider the cylinder?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362266689639:dw yes ... it's okay even if you consider square or rectangle.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I used L^2 for area
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
looks like you can use that on gravity too ...
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
No ... calculate the charge density.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\sigma=q/A= \frac{1.6\times10^{19}}{.1^2}=1.6E17 C/m^2\] \[E=\frac {\sigma}{}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
looks like you are going in wrong direction ... can you state the original question
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
but what are we integration over....the distance between the plates?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Q < this should the be the total charge on the surface ... and A < this is the total area of the surface.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
is this different from the gaussian surface? do I need different variable to use gauss's law?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
yeah it looks like I haven't quite grasped the concept of Gauss's law
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
should we use the cylinder?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
hmm ... what is your original question?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I have to find the capacitance of the parallel plate capacitor.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
what information are given initially?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
the plate is square and has length 10cm....the plates are separated by 1mm
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[dV=\int\frac{\sigma}{\epsilon_0}dA\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
is that correct?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[V=\frac{2\sigma A}{\epsilon_0}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362268232059:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
but would that still be \[V=\frac{\sigma A}{\epsilon_0}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362268629160:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
Let me try ...One moment....
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362268884843:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\oint EdA=\frac Q{\epsilon_0}\] \[E\pi r^2=\frac{Q}{\epsilon_0}\] \[\rho=\frac Q {\pi r^2L}\] \[E\pi r^2=\frac{\rho \pi r^2 L}{\epsilon_0}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
so the charge density is that of the cylinder?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[E=\frac{\rho L}{\epsilon_0}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I completely missed the point didn't I?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
use sigma for surface charge density.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362269453040:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\phi=\oint E\cdot dr=\frac{Q}{\epsilon_0}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
No ... we are not integrating over the circle.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
closed integral from a to b?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362269839865:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
to understand Gauss law ... you should understand flux. Find the total flux .. this case dw:1362269948919:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
is this a circular surface of charge +q ?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
no ... not circular ... spherical!!
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
the flux would be dw:1362270059079:dw oh ok...I guess we'll work with a sphere instead...Let's see...
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
consider ... the charge in enclosed by sphere.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362270254679:dw in all directions out of the sphere...the Electric field is perpendicular to the surface everywhere
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362270435029:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
what is the total flux?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
the sum of all of the electric field lines?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
sum of all the Electric field lines \[\phi=\oint E dr=\frac{Q}{\epsilon_0}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
You should understand what is flux first ... Flux is like ... dw:1362270773063:dw
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JenniferSmart1Best ResponseYou've already chosen the best response.0
the amount of water that flows through it is A times \(v\)
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JenniferSmart1Best ResponseYou've already chosen the best response.0
\[\frac{meter^2}{1}\times\frac{meters}{second}\] that doesn't seem right
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah .. that's correct. It's easy to visualize water ... dw:1362271042142:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362271173946:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
so the flux through the area would be the volume of the sphere times the electric field
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I tried to make your circle look more like a sphere :P
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
check your answer again!!
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh surface area of the sphere!
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yeah ... what is the total flux?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
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)
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
i meant surface area...not areas
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362271421314:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
what's s? surface area?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362271575577:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362271615713:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh are you saying that \[\int ds=s\] yeah I believe that but is s ...oh i see sphere
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
we're integrating with respect to the sphere, correct?
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
the surface area of the sphere
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yes ... but electric Field is constant .. since it's sphere.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
what would you limits be?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
and we are integrating surface ...not distance. lol ... the limits ... don't worry
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
so \[\oint \textrm{never has limits?}\]
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experimentXBest ResponseYou've already chosen the best response.1
dw:1362271874243:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
yeah I get that E is constant everywhere over the sphere
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362271927389:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
oh because were integrating \[\oint ds=s \;\;\;\textrm{we don't need to worry about the limits}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
since the answer is just s
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362272111497:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
yep , makes sense....
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
I finally understand flux.... :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Not complete yet!!
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
dw:1362272196631:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362272206784:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
i'm not gonna look
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
keep going...i'm sorry
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Now ... dw:1362272365512:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
this is called conservation of flux.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362272482209:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362272583908:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362272690179:dw
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JenniferSmart1Best ResponseYou've already chosen the best response.0
through both spheres? yes
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
This is why Gravity ... and Electrostatic force follow inverse square laws. Any shape ... but must be closed. and charge must be inside it.
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
interesting! yeah \(E\propto \frac 1{r^2}\) \[F_g\propto \frac 1{r^2}\]
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
or is it \[F_e\propto \frac 1{r^2}\] and \[F_g\propto \frac 1{r^2}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362273084528:dw
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362273139951:dw
 one year ago

JenniferSmart1Best ResponseYou've already chosen the best response.0
ohdw:1362273163066:dw flux is still the same
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
yes ... the flux is still the same.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1362273240830:dw
 one year ago