Field of non-uniformly charged ring

[skyrolla]

Homework Statement

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.

Homework Equations

E=kq/r^2

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!

 

[facenian]

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

 

[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.

 

[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

 

[gabbagabbahey]

Desiring to integrate over Ø, my limits of integration will be 0 to pi.

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 $z=0$.<br /> <br /> If the ring is an entire circle, the limits on $\phi$ will be zero to $2\pi$<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Now dE=k*dq/r^2 with r=(R^2+z^2). <br /> <br /> Q/R=λo*sin(Ø) so dQ=R*λo*cos(Ø) dØ </div> </div> </blockquote><br /> Not quite. By definition of linear charge density, $dq=\lambda ds=\lambda_0 \sin\phi Rd\phi$<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> and dEz= k*R*λo*cos(Ø) dØ/(R^2+z^2) </div> </div> </blockquote><br /> No, not only should the cosine be a sine, but you haven't properly selected the z-component of $d\textbf{E}$.<br /> <br /> 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.<br /> <br /> 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 $E_y\neq0$.[/tex]

 

[skyrolla]

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