Field of non-uniformly charged ring

1. Sep 21, 2009

skyrolla

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

A thin rubber ring of radius R lies in the xy-plane and is centered at the origin. The ring carries line charge density λ=λo*sin(Ø), with a constant λo with units C/m. define tan(Ø)=y/x. Calculate the magnitude and direction of the electric field E(z) created by the ring on points on the positive z axis.

2. Relevant equations

E=kq/r^2

3. The attempt at a solution

Symmetry implies only the component Ez will be nonzero along the z axis.
Desiring to integrate over Ø, my limits of integration will be 0 to pi.

Now dE=k*dq/r^2 with r=(R^2+z^2).

Q/R=λo*sin(Ø) so dQ=R*λo*cos(Ø) dØ

and dEz= k*R*λo*cos(Ø) dØ/(R^2+z^2)

however, integrating this over Ø results in multiplying everything by sin(pi)-sin(0) = 0, so clearly I've made at least one mistake here.

All help is appreciated!

Last edited: Sep 21, 2009
2. Sep 21, 2009

facenian

the result must be zero for the sake of symmetri. Your result is correct (I think)

3. Sep 21, 2009

skyrolla

Symmetry in the x and y directions makes sense, as an equal amount of ring and angle phi are distributed to both the positive and negative halves. However the field would certainly have a Z component at points along the Z axis, as the full ring exerts charge in either a positive or negative z direction.

4. Sep 21, 2009

facenian

the charge is ditributed in iqual amount in each semi-circle but with defferent signs, so symmetry implies that the z component must be zero. However I now realize that there must be a component in the x-y plane

5. Sep 21, 2009

gabbagabbahey

I think you should draw a picture...the angle [tex]\phi[/itex] is measured entirely in the xy-plane, and any point on the ring is given by $x=R\cos\phi$, $y=R\sin\phi$ and $z=0$.

If the ring is an entire circle, the limits on $\phi$ will be zero to $2\pi$

Not quite. By definition of linear charge density, $dq=\lambda ds=\lambda_0 \sin\phi Rd\phi$

No, not only should the cosine be a sine, but you haven't properly selected the z-component of $d\textbf{E}$.

Again, draw a picture...the field $d\textbf{E}$ from a tiny piece of the ring points from that piece, to the point on the z-axis at which it is measured...you want to find the z-component of that vector.

Even after you do the calculation properly, you will still find that $E_z=0$, however, if you also calculate the other components you should find that $E_y\neq0$.

6. Sep 22, 2009

skyrolla

Thanks. After figuring out I had the wrong component, I was able to solve the problem.