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

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362265815092:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362265934285:dw should I use gauss's law to find the electric field?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yes .. use the gauss law. consider a single plate first.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since we have two plates do we use \(2\sigma\)?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yeah yeah ... that;s the way ... let me add few things ... dw:1362266439292:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1how should the flux lines go? fit it geometrically

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362266583983:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yep!! Now what is flux??

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sigma/\epsilon_0\]? or do I need to consider the cylinder?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362266689639:dw yes ... it's okay even if you consider square or rectangle.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1everything 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
 3 years ago
Best ResponseYou've already chosen the best response.1looks like you can use that on gravity too ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should 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
 3 years ago
Best ResponseYou've already chosen the best response.1No ... calculate the charge density.

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1looks like you are going in wrong direction ... can you state the original question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0but what are we integration over....the distance between the plates?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Q < this should the be the total charge on the surface ... and A < this is the total area of the surface.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is this different from the gaussian surface? do I need different variable to use gauss's law?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah it looks like I haven't quite grasped the concept of Gauss's law

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0should we use the cylinder?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1hmm ... what is your original question?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I have to find the capacitance of the parallel plate capacitor.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1what information are given initially?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the plate is square and has length 10cm....the plates are separated by 1mm

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[dV=\int\frac{\sigma}{\epsilon_0}dA\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[V=\frac{2\sigma A}{\epsilon_0}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362268232059:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but would that still be \[V=\frac{\sigma A}{\epsilon_0}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362268629160:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let me try ...One moment....

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362268884843:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the charge density is that of the cylinder?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[E=\frac{\rho L}{\epsilon_0}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I completely missed the point didn't I?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1use sigma for surface charge density.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362269453040:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[\phi=\oint E\cdot dr=\frac{Q}{\epsilon_0}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1No ... we are not integrating over the circle.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0closed integral from a to b?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362269839865:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1to understand Gauss law ... you should understand flux. Find the total flux .. this case dw:1362269948919:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0is this a circular surface of charge +q ?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1no ... not circular ... spherical!!

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1consider ... the charge in enclosed by sphere.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362270254679:dw in all directions out of the sphere...the Electric field is perpendicular to the surface everywhere

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362270435029:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1what is the total flux?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the sum of all of the electric field lines?

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1You should understand what is flux first ... Flux is like ... dw:1362270773063:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the amount of water that flows through it is A times \(v\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{meter^2}{1}\times\frac{meters}{second}\] that doesn't seem right

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yeah .. that's correct. It's easy to visualize water ... dw:1362271042142:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362271173946:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so the flux through the area would be the volume of the sphere times the electric field

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I tried to make your circle look more like a sphere :P

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1check your answer again!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh surface area of the sphere!

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yeah ... what is the total flux?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0because 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
 3 years ago
Best ResponseYou've already chosen the best response.0i meant surface area...not areas

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362271421314:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what's s? surface area?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362271575577:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362271615713:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we're integrating with respect to the sphere, correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the surface area of the sphere

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yes ... but electric Field is constant .. since it's sphere.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what would you limits be?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1and we are integrating surface ...not distance. lol ... the limits ... don't worry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so \[\oint \textrm{never has limits?}\]

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362271874243:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah I get that E is constant everywhere over the sphere

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362271927389:dw

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since the answer is just s

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362272111497:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep , makes sense....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I finally understand flux.... :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1362272196631:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362272206784:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0keep going...i'm sorry

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Now ... dw:1362272365512:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1this is called conservation of flux.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362272482209:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362272583908:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362272690179:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0through both spheres? yes

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

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

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

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1just 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
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362273084528:dw

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362273139951:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ohdw:1362273163066:dw flux is still the same

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1yes ... the flux is still the same.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1362273240830:dw