# Homework Help: Nonuniform Semicircle of Charge problem

1. Jan 15, 2010

### littlebilly91

1. The problem statement, all variables and given/known data
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?

2. Relevant 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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3. 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.

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Last edited: Jan 15, 2010
2. Jan 15, 2010

### ideasrule

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).

3. Jan 15, 2010

### littlebilly91

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?

4. Jan 15, 2010

### 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?

5. Jan 15, 2010

### 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.

6. Jan 15, 2010

### 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).

7. Jan 15, 2010

### 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)?

8. Jan 15, 2010

### ideasrule

Yup, that's correct.

9. Jan 15, 2010

### littlebilly91

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

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

10. Jan 15, 2010

### ideasrule

Yeah, it's positive.

11. Jan 15, 2010

### littlebilly91

thanks so much for the help!

12. Jan 25, 2010

### kqian

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