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A semicircle of radius r is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge Q is distributed uniformly around the right half of the semicircle. What is the magnitude of the net electric field at the origin produced by this distribution of charge? Express your answer in terms of the variables Q, r, and appropriate constants. Thanks in advance for any help in approaching this. Can't quite nab the hang of this one xD
 one year ago
 one year ago
A semicircle of radius r is in the first and second quadrants, with the center of curvature at the origin. Positive charge +Q is distributed uniformly around the left half of the semicircle, and negative charge Q is distributed uniformly around the right half of the semicircle. What is the magnitude of the net electric field at the origin produced by this distribution of charge? Express your answer in terms of the variables Q, r, and appropriate constants. Thanks in advance for any help in approaching this. Can't quite nab the hang of this one xD
 one year ago
 one year ago

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akash123Best ResponseYou've already chosen the best response.0
dw:1346993325948:dw
 one year ago

akash123Best ResponseYou've already chosen the best response.0
Take elementary charge dq in the ist quadrant at angle theta...electric field E due to this electric charge at O: dE= dq/ 4pi epsilon(0) r^2
 one year ago

akash123Best ResponseYou've already chosen the best response.0
and its dirction will be radially outward..
 one year ago

akash123Best ResponseYou've already chosen the best response.0
now break dE into tow components along x and y axis respectively...dE cos(theta) and dE sin(theta)
 one year ago

akash123Best ResponseYou've already chosen the best response.0
dw:1346993841419:dw
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
So I actually was doing an easier problem earlier where this semicircle was all positive Q. And I found: \[E=\frac{ Q }{ 2\pi ^{2} \epsilon _{0}r ^{2}}\] Can I use this at all to help me in this?
 one year ago

akash123Best ResponseYou've already chosen the best response.0
can you show me the solution of this...I think answer will be something different...but how it'll help you in this?
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
Let's simplify this by considering symmetry. Break up the electric field into x and y and see if any of them cancel. dw:1346994485662:dw Hey, what do you know, the ycoordinate cancels out and now you have a slightly easier problem to deal with.
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
Not to mention you have the exact same contribution from the electric field from one and the other going to the right, so you really only need to solve for the electric field of one and multiply that by 2 to get the magnitude right.
 one year ago

KainuiBest ResponseYou've already chosen the best response.0
It's tempting to say that the electric field is related to 2 times the integral of cos theta from 0 to pi/2, but it's not I think lol.
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
I typed in demitris's answer. It said that it was missing or had an incorrect numerical multiplier. Any idea what that could mean? he's probably very close to being right, the system is just being picky about what it has entered into it...
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
What does your k = ? 1/4piEo ?
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
Yea, okay. I just don't see how you went from the last part in the 2nd to last drawing to cancelling the 4s. Where is like Eo?
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
But where did Eo disappear to in your last step?
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
OH! cuz you had to double. so you wrote the constant, and then the numerator doubled. got it... but i still don't understand why Eo is missing....
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
Ok.. hopefully that does the trick and now it's right xD
 one year ago

nj1202Best ResponseYou've already chosen the best response.0
Just to let you know, what i wrote was right. it said correct online :) thank you so much for all the help !
 one year ago
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