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Vaentus
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Homework Statement
Problem 2.4 from Griffiths Intro to Electro[/B]
Find the electric field a height z above the centre of a square loop with sides a and linear charge density λ.
height is given to be z and sides given to be a, ∴ distance from origin to side is given by a/2
Homework Equations
[tex] E = \frac{1}{4 \pi \epsilon_0} \int {\frac{1}{r^2} \hat {r}dq} [/tex]
[tex]dq = \lambda dl [/tex]
The Attempt at a Solution
Considering the side of the square perpendicular to the positive y axis
First using the position of point P (0, 0, z) and position on the side (x, a/2, z) where z and a/2 are constants the direction from origin to P is given by [itex]r=z\hat{z}[/itex] and direction from origin to side is given by [itex]r' = (a/2)\hat{y} + x\hat{x} [/itex].
∴direction of electric field is given by [itex] r = r - r' =z\hat{z}-(a/2)\hat{y}-x\hat{x} [/itex]
[tex] \hat{r}=\frac{r}{|r|}=\frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt{z^2+\frac{a^2}{4}+x^2}} [/tex]
considering the "right" side of the square loop y and z values are constant while the x goes from -a/2 to a/2
∴ [itex]dq=\lambda dx[/itex]
Plugging these equations into the integral we get
[tex] E = \frac{1}{4 \pi \epsilon_0} \int {\frac{1}{z^2+\frac{a^2}{4}+x^2} \frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt{z^2+\frac{a^2}{4}+x^2}}\lambda dx} [/tex]
which can be simplified to
[tex] E=\frac{\lambda}{4\pi\epsilon_0} \int{\frac{z\hat{z}-(a/2)\hat{y}-x\hat{x}}{\sqrt[3]{z^2+\frac{a^2}{4}+x^2}}} [/tex]
I know I can split the integral up into three different integrals w.r.t x y and z directions then group the y and z integrals because z and (a/2) are constants.while leaving the integral in the x direction due to its dependence on x which changes from -a/2 to a/2 but I am unsure whether that is the correct way to proceeed/what to do next.
Thanks for your help!