Electric field at the center of an arc

[tag16]

Homework Statement

A charge Q is arranged evenly on a wire bent into an arc or radius R as shown in the
Figure. What is the electric field at the center of the arc as a function of the opening
angle theta? Sketch a graph of the electric field as a function of theta for 0<theta<180 degrees.

Homework Equations

E= KQ/R^2 cos theta

The Attempt at a Solution

Is the above equation correct? If not, I'm not really sure what to do, also I'm not sure how the graph should look.

 

[ApexOfDE]

where is the figure?

 

[MaxL]

Where'd you get that equation?

Also, if you're worried about how the graph should look, start by figuring it out for some easy points. What's the E field going to be at the center if theta is zero (so all the charge is concentrated in a single point)? How bout if Theta is 360 degrees (so a full circle)? It's a bit harder to figure out for 180 degrees, or 90 or 270 degrees, but you should be able to make a very rough estimate. That will give you some insight into the shape of the graph.

 

[tag16]

sorry this is the best I can do: http://i848.photobucket.com/albums/ab41/tag16/problem1.jpg?t=1259862072

The equation I found in my physics book except the cos theta part, I was just guessing there. Though it would probably be integral cos theta or sin theta, if it's suppose to be in there at all.

 

[tag16]

[tex]\lambda[/tex]=Q/[tex]\pi[/tex]R

dE= kdQ/R^2
dE= (kdQ/R^2) cos[tex]\theta[/tex]

dQ=[tex]\lambda[/tex]dl
dl= Rd[tex]\theta[/tex]
dQ=[tex]\lambda[/tex]Rd[tex]\theta[/tex]

dE=(k[[tex]\lambda[/tex]Rd[tex]\theta[/tex]]/R^2)cos[tex]\theta[/tex]


E=[tex]\int[/tex](k[tex]\lambda[/tex]Rcos[tex]\theta[/tex]/R^2)d[tex]\theta[/tex](from [tex]\pi[/tex]/2 to -[tex]\pi[/tex]/2)
E=k[tex]\lambda[/tex]/R[tex]\int[/tex] cos[tex]\theta[/tex]d[tex]\theta[/tex]
E=k[tex]\lambda[/tex]/R[tex]\int[/tex]sin[tex]\theta[/tex]
E=k[tex]\lambda[/tex]/R[sin([tex]\pi[/tex]/2)-sin(-[tex]\pi[/tex]/2)]
E=k[tex]\lambda[/tex]/2R
E=k(Q/[tex]\pi[/tex]R)/2R=2kQ/[tex]\pi[/tex]R^2
E= 2kq/[tex]\pi[/tex]R^2

Is this right?