Can someone check my setup for finding E.dl?

[emhelp100]

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

[/B]

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

 

[BvU]

Apart from misssing brackets ?
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,

 

[emhelp100]

Apart from misssing brackets ?
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,

E1 is the line segment from (a/2,-a/2) to (a/2, a/2)

 

[BvU]

Of course, but there $\vec {dl} = (0, dy)$

 

[emhelp100]

Of course, but there $\vec {dl} = (0, dy)$

That is because x stays constant and y changes right?

 

[BvU]

Correct. And the expression for E₁ has a constant $x={a\over 2}$ and a varying $y$

 

[emhelp100]

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

 

[BvU]

You sound as if you hesitate. Have you not convinced yourself?

 

[emhelp100]

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?

 

[BvU]

Do you realize you have written two integrals ?

 

[emhelp100]

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

 

[BvU]

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 ?

 

[BvU]

Just to make sure: we see that we don't have to find a primitive to establish the value...

 

[emhelp100]

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 />

 

[kuruman]

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 90^o or 180^o or 270^o 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?

 

[emhelp100]

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 90^o or 180^o or 270^o 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...

 

[kuruman]

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?

 

[BvU]

Being lazier than @BvU

Impossible . Apart from that: We both want to exploit symmetry. Can be done from the math side and/or from the physics side