Can someone check my setup for finding E.dl?

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AI Thread Summary
The discussion revolves around verifying the setup for calculating the line integral of the electric field around a square due to a point charge at the origin. Participants confirm that the integral should evaluate to zero based on symmetry, as the contributions from opposite sides of the square cancel each other out. There are clarifications on the expressions for the electric field components and the corresponding differential line elements. The conversation highlights the importance of recognizing symmetries in both the mathematical setup and the physical interpretation of the problem. Ultimately, the consensus is that the setup is correct, and the integral's evaluation confirms the expected result of zero.
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Homework Statement


A point charge +Q exists at the origin. Find \oint \vec{E} \cdot \vec{dl} around a square centered around the origin. I know the answer is 0, but can someone check my setup?

Homework Equations


E=\frac{Q(\vec{R}-\vec{R'})}{4\pi E_0 |\vec{R}-\vec{R'}|^3}

The Attempt at a Solution

upload_2018-4-15_0-41-44.png
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For E1 and E3, R = \hat{x}x+\hat{y}\frac{a}{2} and R' = 0
E1 = E3 = \frac{Q\hat{x}x+\hat{y}\frac{a}{2}}{4\pi E_0(x^2+\frac{a^2}{4})^{3/2}}
dl1=dl3 = \hat{x}dx

E2=E4=\frac{Q\hat{x}\frac{a}{2}+\hat{y}y}{4\pi E_0(y^2+\frac{a^2}{4})^{3/2}}
dl2=dl4 = \hat{y}dy
 

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Apart from misssing brackets :smile: ?
And asssuming the obvious integration limits ?
And assuming (from the picture) E1 is for (a/2, 0) to (a/2,a/2) -- in which case you mixed up 1,3 with 2,4 ?

However, not bad for a first thread! Hello EM, :welcome:
 
BvU said:
Apart from misssing brackets :smile: ?
And asssuming the obvious integration limits ?
And assuming (from the picture) E1 is for (a/2, 0) to (a/2,a/2) -- in which case you mixed up 1,3 with 2,4 ?

However, not bad for a first thread! Hello EM, :welcome:
E1 is the line segment from (a/2,-a/2) to (a/2, a/2)
 
Of course, but there ##\vec {dl} = (0, dy)##
 
BvU said:
Of course, but there ##\vec {dl} = (0, dy)##
That is because x stays constant and y changes right?
 
Correct. And the expression for E1 has a constant ##x={a\over 2}## and a varying ##y##
 
BvU said:
Correct. And the expression for E1 has a constant ##x={a\over 2}## and a varying ##y##
So would this setup be correct?
E1 = E3 = \frac{Q}{4\pi E_0} \frac{\hat{x}{x}+\hat{y}\frac{a}{2}}{(x^2+\frac{a^2}{4})^{3/2}}
dl1=dl3 = \hat{y}dy

E2 = E4 = \frac{Q}{4\pi E_0}\frac{\hat{x}\frac{a}{2}+\hat{y}{y}}{(y^2+\frac{a^2}{4})^{3/2}}
dl2=dl4 = \hat{x}dx
 
:wink: You sound as if you hesitate. Have you not convinced yourself?
 
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BvU said:
:wink: You sound as if you hesitate. Have you not convinced yourself?
So now that I have this:
\int_{\frac{-a}{2}}^{\frac{a}{2}} \frac{\hat{x}\frac{a}{2}+\hat{y}y}{(\frac{a^2}{4}+y^2)^{3/2}}dy
How do I integrate this?
 
  • #10
Do you realize you have written two integrals ?
 
  • #11
BvU said:
Do you realize you have written two integrals ?
\int_{\frac{-a}{2}}^{\frac{a}{2}}\frac{\hat{x}\frac{a}{2}}{(\frac{a^2}{4}+y^2)^{3/2}}dy+\int_{\frac{-a}{2}}^{\frac{a}{2}}\frac{\hat{y}y}{(\frac{a^2}{4}+y^2)^{3/2}}dy
 
  • #12
Yes, so one component in the ##\hat x## direction and one in the ##\hat y## direction.
I am too lazy to work out the integral, so I try to sneak my way out: what about the second one, in the ##\hat y## direction ?
 
  • #13
Just to make sure: we see that we don't have to find a primitive to establish the value...
 
  • #14
BvU said:
Just to make sure: we see that we don't have to find a primitive to establish the value...
What does this mean?
For E1 and E3, the x^ direction integral canceled out.

For E2 and E4, the y^ direction integral canceled out.I am left with:
<br /> \int_{\frac{-a}{2}}^{\frac{a}{2}}\frac{\hat{x}\frac{a}{2}}{(\frac{a^2}{4}+y^2)^{3/2}}dy - \int_{\frac{a}{2}}^{\frac{-a}{2}}\frac{\hat{x}\frac{a}{2}}{(\frac{a^2}{4}+y^2)^{3/2}}dy+<br /> \int_{\frac{a}{2}}^{\frac{-a}{2}}\frac{\hat{y}\frac{a}{2}}{(\frac{a^2}{4}+x^2)^{3/2}}dx-\int_{\frac{-a}{2}}^{\frac{a}{2}}\frac{\hat{y}\frac{a}{2}}{(\frac{a^2}{4}+x^2)^{3/2}}dx<br />
 
  • #15
Being lazier than @BvU, I would use symmetry and note that the four line integrals along each side must be equal because if one rotated the square by 90o or 180o or 270o while you were not looking at it, you wouldn't know the difference. So there is only one integral to do. But wait, there is another symmetry to consider. Do you see what it is?
 
  • #16
kuruman said:
Being lazier than @BvU, I would use symmetry and note that the four line integrals along each side must be equal because if one rotated the square by 90o or 180o or 270o while you were not looking at it, you wouldn't know the difference. So there is only one integral to do. But wait, there is another symmetry to consider. Do you see what it is?
I don't see it...
 
  • #17
You know that each of the four line integrals gives you the same number by symmetry. You also know that the result around the closed loop is zero. What does this say about the line integral along one of the sides? How can you get that number? Hint: Look at segment E2. Consider two points on it at ##+x## and ##-x##. What is ##\vec E \cdot d\vec l## at these points?
 
  • #18
kuruman said:
Being lazier than @BvU
Impossible :-p . Apart from that: We both want to exploit symmetry. Can be done from the math side and/or from the physics side
 
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