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|dw:1362265815092:dw|

|dw:1362265934285:dw|
should I use gauss's law to find the electric field?

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

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

|dw:1362266583983:dw|

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

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

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

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

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 ..

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

is that correct?

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

Woops!! error

|dw:1362268232059:dw|

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

|dw:1362268629160:dw|

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

|dw:1362268884843:dw|

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.

|dw:1362269453040:dw|

yes?

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

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

closed integral from a to b?

|dw:1362269839865:dw|

is this a circular surface of charge +q ?

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

consider ... the charge in enclosed by sphere.

|dw:1362270435029:dw|

yes ...

Yay!!! sigh...

what is the total flux?

zero?

the sum of all of the electric field lines?

zero

final answer

No ...

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

volume?

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|

yep

|dw:1362271173946: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?

i meant surface area...not areas

|dw:1362271421314:dw|

what's s? surface area?

ds...?

|dw:1362271575577:dw|

|dw:1362271615713:dw|

what's that?

sphere.

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.

yep

what would you limits be?

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

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

|dw:1362271874243:dw|

yeah I get that E is constant everywhere over the sphere

surface*

|dw:1362271927389:dw|

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

since the answer is just s

|dw:1362272111497:dw|

yep , makes sense....

I finally understand flux.... :)

Not complete yet!!

|dw:1362272196631:dw|

|dw:1362272206784:dw|

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.

|dw:1362272482209:dw|

|dw:1362272583908:dw|

|dw:1362272690179:dw|

through both spheres? yes

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
\[F_g\propto \frac 1{r^2}\]

|dw:1362273084528:dw|

|dw:1362273139951:dw|

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

yes ... the flux is still the same.

|dw:1362273240830:dw|