Calculate the electric field at the center of the circle:

- anonymous

Calculate the electric field at the center of the circle:

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- schrodinger

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- anonymous

can the circle be the sphere? say @CShrix

- anonymous

@lall I'm not sure. It does not specify sphere. I would assume that the field inside a sphere is 0, given that the material is conductive.

- anonymous

read one of the statements in the middle, there is written that the 2 half circles are separated by thin insulator thus there is no charge conductance b/w the 2 half circles....
thus if the circles are charged then the charges shold lie on the boundary of the half circles so that the 2 half circles have an electric field uniformly on the centre...

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## More answers

- anonymous

could it be so?

- anonymous

i think u just need to apply the formula for electric field for the individual charges for half of the radius each

- anonymous

Wouldn't that just yield the normal equation:
\[E=\frac{kQ}{r^2}\]

- anonymous

|dw:1444331060490:dw|

- anonymous

a bit of confusion is the direction of electric field correct?

- anonymous

what do u think?

- anonymous

do u hav an ans book OR ans at back?

- anonymous

@CShrix

- anonymous

ar u there?

- anonymous

@thomaster

- anonymous

@IrishBoy123

- anonymous

what happened?

- anonymous

Yeah that looks to be the correct direction.

- anonymous

And I do not have the answer. These are questions that my professor created. I tried searching it online but to no avail.

- anonymous

i think it would form a dipole having net charge zero and u hav to find electric field in the centre of the dipole

- anonymous

this is the thing that the whole ques. wanted to ask i think

- anonymous

what is electric field at the centre of dipole

- anonymous

\[E=\frac{ kQ }{ (\frac{ D }{ 2 })^2 }\]
Where D is the distance from the positive side to the negative side, right?

- anonymous

Q=+ve Q=-ve

- anonymous

just go with the formulas of E at axial pnt. and E at equatorial pnt to get the E at the centre
if u dont get any thing, search on net for the the field at center of dipole...
if even not then the teacher will tell....

- anonymous

sorry need to go............bye..........

- IrishBoy123

my guess is that whether it is a circular dic or ring, it is conducting so the charge will aggregate around its edges, so it can be modelled as a ring
|dw:1444345240325:dw|
with a bit of calculus that is manageable.
that is how i would do it without further information

- IrishBoy123

|dw:1444345614703:dw|
and because of symmetry you only need to consider the x direction.....

- anonymous

Okay, so then how would i set up the integration? @IrishBoy123

- IrishBoy123

|dw:1444346032637:dw|
i'm doing this very quickly but this gives an idea
if we take a small element on the right side ring we can say \(dr = R d \phi\) where R is the radius of the ring
if the total charge +Q is distributed along its half circumference, it has linear chanrge density \(\sigma = { Q \over \pi r }\)
so the inward radial E field on a test charge +q at the centre due to small element is \(dE_r = k { dq \over R^2} = k {\sigma dr \over R^2} = k {Q d\phi \over \pi R^2}\) [\(k = \) Coulommb's constnat
the field from the equivalent -Q charge on the left hand ring act outward from the cemtre -- but has same magnitude if the charges have same magnitude so for that part we can say \(dE_r = -k {Q d\phi \over \pi R^2}\)
net net in in the x direction we therefore have \(dE_x = -2k {Q d\phi \over \pi R^2} \cos \phi\) because net field acts right to left
\(E_x = -2k {Q \over \pi R^2} \int\limits_{-\pi/2}^{\pi/2} \cos \phi \; d \phi \\ = -2k {Q \over \pi R^2} \left| \sin \phi\right|_{-\pi/2}^{\pi/2} \\= -4k {Q \over \pi R^2}\)
that could be wrong in terms of the answer but it shows the method
i can check it later, say tomorrow for silly errors but i am in a bit of a rush... soz
i hope you get the idea

- IrishBoy123

check it for silly errors against this, it is the same idea....
http://www.phys.uri.edu/gerhard/PHY204/tsl329.pdf

- anonymous

Did you get what you needed? @CShrix

- anonymous

@Mashy Kind of. If you could give me some more help that'd be appreciated too! IrishBoy123's link helped me to find the field along a conductive semicircular wire. Which helps me set up the equation for Ex. But I don't understand what I do when I have two of them. Do I simple add them? Wouldn't one be negative and one positive, so the net would be 0?

- anonymous

Both the electric field vectors at the center are in the same direction (away from positive and towards negative)
so you simply add them

- anonymous

Ah okay, I'll try that and see where I get

- anonymous

@Mashy
I still got 0..
|dw:1444369848115:dw|
\[E_x=0\]
\[E_y=\frac{ k \lambda }{ R }\int\limits_{0}^{\pi}\cos(\theta) d \theta=\frac{ -2k \lambda }{ R }\]
Since there is symmetry across the x-axis then we know the bottom one is going to be \[\frac{ 2k \lambda }{ R }\]
Adding them would still give me 0

- anonymous

I think my error is just a sign convention. Because it must be that the top one is pointing down and the bottom semi circle is point up, which ends up cancelling each other. But since the field flows from positive to negative, then both would flow downward. So maybe the field is really \[\frac{ -4k \lambda }{ R }\]?

- anonymous

@Mashy In my drawing, I assigned the top semi circle positive q and the bottom semi circle the negative q, just to clarify

- anonymous

@Mashy Oh wait maybe I solved it?
Ok so the result for the top semi-circle Ey is \[\frac{ -2k \lambda }{ R }\]
But the charge density lambda is
\[\lambda = \frac{ Q }{ \pi R }\] for the top semi circle and\[\lambda = \frac{-Q}{\pi R}\] for the bottom semi-circle.
Therefore subbing would give me\[E_{y_{upper}}=\frac{ -2k \frac{ Q }{ \pi R } }{ R }=-\frac{2 kQ }{ \pi }\]
and the bottom would also give me the same answer, so adding would give me\[\frac{ -4kQ }{ \pi }\]
But does this make sense how R is not a factor here? When Gauss' law clearly shows that an electric field is inversely related to its radius?

- anonymous

Your algebra is wrong :P
R doesn't cancel..
it gets squared in the denominator :D

- anonymous

Oh duhh! Geez..
Okay, so stupid mistakes aside.. lol.. is everything else correct?

- anonymous

lemme just quickly derive it myself.. and check the result..

- anonymous

Ok everything else is correct.
But the question is asking you to keep your final answer in terms of the positive, negative charges and Radius..
So i think its better if you call one charge as Q1 and another as Q2..
and just keep it that way.. but that is just a detail..

- anonymous

Okay, awesome!
Yeah that makes sense, I'll probably have to do that. Or just separate the answer into two terms, one +q and one -q.

- anonymous

sure :)

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