# A positively charged wire bent into a semi-circle

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1. Sep 24, 2015

### Callix

1. The problem statement, all variables and given/known data
A positively charged wire is bent into a semicircle of radius R, as shown in the figure below.
The total charge on the semicircle is Q . However, the charge per unit length along the semicircle is non-uniform and given by λ=λocos(θ)

What is the relationship between λo, R and Q?

2. Relevant equations
Q = ∫ λ ds
F = qE

3. The attempt at a solution
Q = ∫ λ ds = ∫(from θ=0 to π) λocos(θ) dθ.
However, this yields evaluating the sin function from 0 to π which is just 0. I'm having a hard time believing that there is no relationship between λ, R, and Q. I was wondering if someone would be able to check my work.

Any help would be greatly appreciated! :)

2. Sep 24, 2015

### ddd123

It completely depends on where $\theta = 0$ is on the semicircle. Obviously, if it's on one end, Q = 0 because for a quarter of a circle the charge is the negative of the other. But, from your drawing, it seems you should actually use $-\pi/2,\pi/2$ as boundaries, which changes everything (the charge has the same sign throughout).

3. Sep 24, 2015

### Titan97

[Deleted]

4. Sep 24, 2015

### SammyS

Staff Emeritus
Look at θ as it is shown in the figure.

It looks like θ goes from -π/2 to π/2 .

5. Sep 24, 2015

### Callix

I'm confused as to why θ goes from -π/2 to π/2

6. Sep 24, 2015

### TSny

According to the figure, Θ is measured from the y-axis (not the x-axis).

7. Sep 24, 2015

### Callix

Oh, I understand what you're saying now!
So now that I have the proper limits

Q = ∫(from θ=-π/2 to π/2) λcos(θ) ds = 2λs = 2λr.
Is this now correct?

8. Sep 24, 2015

### TSny

Why did you set s = r in the final step?

9. Sep 24, 2015

### Callix

Because I'm running off of 2 hours of sleep and making stupid mistakes like thinking that s is the distance when its not.

So after I have the integral,

∫(from θ=-π/2 to π/2) λcos(θ) ds

How can I implement R?

10. Sep 24, 2015

### Callix

Can I express it as the double integral: ∫∫ λcos(θ) r dr dθ?

11. Sep 24, 2015

### TSny

You need to express ds in terms of dθ in order to carry out the integration. (Yikes, only 2 hours of sleep? I feel for you.)

12. Sep 24, 2015

### Callix

But I think that's wrong because wouldn't that cover the whole circle as opposed to just the wire?

13. Sep 24, 2015

### TSny

Your setup of the single integral is fine. You do not want a double integral.

In the single integral you are integrating the function cosΘ where Θ is the variable. If you want to integrate with respect to θ, you need to express ds in terms of dθ.

Recall that if θ is measured in radians, then arc length on a circle of radius r is s = rθ.

14. Sep 24, 2015

### Callix

I just remembered the equation for arc length right when you posted that haha

Which means that ds/dθ = r θ → ds = r θ dθ

15. Sep 24, 2015

### TSny

No. You did not solve for ds correctly.

16. Sep 24, 2015

### Callix

Deleted

17. Sep 24, 2015

### Callix

s=rθ
ds=rdθ

18. Sep 24, 2015

### TSny

Yes.

19. Sep 24, 2015

### Callix

I swear I'm not normally this stupid :)
But this makes sense because for every little change in the arc length, there is a change in the angle, while of course R remains constant.
Thank you!

20. Sep 24, 2015

### TSny

OK, good. Better get some