# Nonuniformly Charged Semicircle

1. Sep 3, 2009

### saxyliz

1. The problem statement, all variables and given/known data
A thin glass rod is a semicircle of radius R, see the figure. A charge is nonuniformly distributed along the rod with a linear charge density given by $$\lambda=\lambda_{0} + sin(\theta)$$, where $$\lambda_{0}$$ is a positive constant. Point P is at the center of the semicircle.

Determine the acceleration (magnitude and direction) of an electron placed at point P, assuming R = 1.9 and $$\lambda_{0}$$ = 2.0 $$\mu C/m$$.

2. Relevant equations
Coulomb's Law $$F_{e}=kq_{1}q_{2}/r^{2}$$
Electric Field Equation $$E=F_{e}/q$$
Newton's 2nd Law

3. The attempt at a solution
I attempted to use Newton's 2nd Law to solve this problem. I know the mass of an electron, and since gravity can be ignored, I knew that all I needed to do was to find the electric force on the electron to find the acceleration. I know the the force points straight up do to the positive charge above and the negative charge below, but I'm having trouble quanitfying the forces. I attempted to use Coulomb's Law by plugging in the charges for just one electron and proton, but I know that's not right. Please help!
(Also, let me know if the image disappears, it should work out fine though...)

2. Sep 3, 2009

### tiny-tim

Welcome to PF!

Hi saxyliz ! Welcome to PF!

(have a theta: θ and a lambda: λ and a mu: µ )
Are you sure it isn't λ = λ0 times sinθ (that seems more consistent with the image)?

3. Sep 3, 2009

### saxyliz

Re: Welcome to PF!

Oh yes sorry! It's times! Silly me!

4. Sep 3, 2009

### tiny-tim

ok, then … yes, the force will point up, so you only need the up-components.

But you must slice the semicircle into bits of length dθ, and plug in the charge for that.

5. Sep 3, 2009

### saxyliz

So, dq = $$\lambda d \theta$$? Then I plug that into Coulomb's Law to solve for dE, but how do I solve for the components?

To be more specific, what I get is:

$$dE=k\lambda d \theta /R^{2}$$

Which is the same as

$$dE=k\lambda_{0} sin(\theta) d \theta /R^{2}$$

Do I use a sector for the components? Like polar coordinates?

Last edited: Sep 3, 2009
6. Sep 3, 2009

### tiny-tim

Hi saxyliz!

(what happened to that θ and λ i gave you? )
No, dq = λ sin θ dθ.

For components, you use cos of the angle as usual … what is worrying you about that?

7. Sep 3, 2009

### saxyliz

Sorry! I'm totally new to this and didn't get what you meant in the first place (about the θ, etc.)

Why isn't it λ_0 ? Wouldn't it be sin^2 then?

I think I get what you mean about the components, give me a second.

EDIT:
I got it!! Thanks so much for your help!

Last edited: Sep 3, 2009