# Charge Distrubution evenly on Arc (Radius R)

1. Oct 10, 2011

### plat911

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

A charge Q is distributed evenly on a wire bent into an arc of radius R, as shown in
the figure below.What is the mathematical expression that describes the electric field
at the center of the arc (point P indicated) as a function of the angle θ? Sketch
a graph of the electric field as a function of θ for 0 < θ < 180.

I added the figure for the question as an attachment.

3. The attempt at a solution

lambda=Q/pi R

dE= kdQ/R^2
dE= (kdQ/R^2) cos θ

dQ=lambda dl
dl= Rd theta
dQ=lambda R dθ

dE=(k[lambda R d θ]/R^2)cos θ

E=(k lambda R cos θ/R^2)d θ(from pi/2 to -pi/2)
E=k lambda/R cos θ dθ
E=k lambda/Rsin θ
E=k lambda/R[sin(pi/2)-sin(-pi/2)]
E=k lambda/2R
E=k(Q/piR)/2R=2kQ/piR^2
E= 2kq/piR^2

Is this right? I used K for 1/4pi(E) to make it easier to type.
Also i am kind of lost on what is expected for the sketch.

#### Attached Files:

• ###### physics.jpg
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2. Oct 10, 2011

### lightgrav

looks good ... might make more sense if the E-field component was upward (to balance gravity, if you put a charged item in the center, say)

3. Oct 10, 2011

### SammyS

Staff Emeritus
Why are you integrating from -π/2 to π/2 ?

Integrate from -θ0 to θ0 or similar.

4. Oct 10, 2011

### plat911

i didn't notice that. i was integrating from the wrong limits. Thank you for pointing that out. I am still confused on the sketch of the electric field as a function of theta if anyone could point me in the right direction with this it would be much appreciated.

5. Oct 10, 2011

### SammyS

Staff Emeritus
Finish working out the solution before you can do the plot.