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nj1202
 3 years 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
nj1202
 3 years 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

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akash123
 3 years ago
Best ResponseYou've already chosen the best response.0Take 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

akash123
 3 years ago
Best ResponseYou've already chosen the best response.0and its dirction will be radially outward..

akash123
 3 years ago
Best ResponseYou've already chosen the best response.0now break dE into tow components along x and y axis respectively...dE cos(theta) and dE sin(theta)

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0So 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?

akash123
 3 years ago
Best ResponseYou've already chosen the best response.0can you show me the solution of this...I think answer will be something different...but how it'll help you in this?

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0Let'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.

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0Not 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.

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0It'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.

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0I 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...

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0What does your k = ? 1/4piEo ?

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0Yea, 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?

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0But where did Eo disappear to in your last step?

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0OH! 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....

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0Ok.. hopefully that does the trick and now it's right xD

nj1202
 3 years ago
Best ResponseYou've already chosen the best response.0Just to let you know, what i wrote was right. it said correct online :) thank you so much for all the help !
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