Nonuniform Semicircle of Charge problem

[littlebilly91]

Homework Statement

A non-uniformly charged semicircle of radius R=28.0 cm lies in the xy plane, centered at the origin, as shown. The charge density varies as the angle θ (in radians) according to λ = + 5.8 θ, where λ has units of μC/m.
(picture included)
What is the y - component of the electric field at the origin?

Homework Equations

dE=\frac{dq}{4\pi\epsilon r^{2}}sin\theta

where \epsilon = 8.85*10^{-12}

\theta r = s

\lambda=\frac{q}{s}
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The Attempt at a Solution

First I derived (5.8*10^{-6})\theta^{2} r=q

and took the derivative 2(5.8*10^{-6}) r \theta d\theta=dq

substituted into equation 1
dE=\frac{2(5.8*10^{-6})}{4\pi\epsilon r}\theta sin(\theta)d\theta

Integrated from 0 to pi.
E=\frac{(5.8*10^{-6})}{2\pi\epsilon r}\int\theta sin(\theta)d\theta

The whole integral simplifies down to pi,which cancels out and leaves
E=\frac{(5.8*10^{-6})}{2\epsilon r}

I am getting 1.17*10^{6} N/C for an answer, but it's not correct.
Thanks for the help in advance.

 

[ideasrule]

First I derived (5.8*10^{-6})\theta^{2} r=q

It's supposed to be (5.8*10^{-6}/2)\theta^{2} r=q. A less roundabout way of doing this is to note that dq=λ*dL and dL=Rd(theta), so dq=λRd(theta).

 

[littlebilly91]

It's supposed to be (5.8*10^{-6}/2)\theta^{2} r=q. A less roundabout way of doing this is to note that dq=λ*dL and dL=Rd(theta), so dq=λRd(theta).

why is that? I'm not seeing where that 2 came from. Does the 2 come in before the derivative because it's only half a circle?

 

[ideasrule]

A better question would be how you got (5.8*10^{-6})\theta^{2} r=q in the first place. Didn't you use dq/dl=λ, dq=λ*dl=λ*R*dθ?

 

[littlebilly91]

I used \theta r = L and \lambda=\frac{q}{L} to get \lambda=\frac{q}{\theta r}. I substituted 5.8\theta for lambda and rearranged to solve for q. I didn't know how else to take the nonuniform charge into account.

Again, thanks for the help.

 

[ideasrule]

Lambda does not equal q/L because the charge density varies across the circle. Try using the method I described: dq=λ*dL and dL=Rd(theta), so dq=λRd(theta).

 

[littlebilly91]

Okay, I understand why I was wrong before. Is it correct to go from dq=λRd(theta) to dq =5.8(theta)Rd(theta)?

 

[ideasrule]

Yup, that's correct.

 

[littlebilly91]

Integrated from 0 to pi.
E=<br /> \frac{(5.8*10^{-6})}{4\pi\epsilon r}\int sin(\theta)d\theta<br />

=>(skipping a few steps) =>
E=<br /> \frac{(5.8*10^{-6})}{4\pi\epsilon r}(2)
I believe that would be correct.
And it should also be positive because it points towards the negative charge, right?

 

[ideasrule]

Yeah, it's positive.

 

[littlebilly91]

thanks so much for the help!

 

[kqian]

Integrated from 0 to pi.
E=<br /> \frac{(5.8*10^{-6})}{4\pi\epsilon r}\int sin(\theta)d\theta<br />

=>(skipping a few steps) =>
E=<br /> \frac{(5.8*10^{-6})}{4\pi\epsilon r}(2)
I believe that would be correct.
And it should also be positive because it points towards the negative charge, right?

How did you get rid of the
<br /> (\theta) from \int (\theta)sin(\theta) to \int sin(\theta)<br /> [\tex]