# Net electric field calc

1. Jan 30, 2017

### Jrlinton

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

2. Relevant equations
L=pi*r/2
λ=q/L
Enet=∑λ(2sinθ)*k

3. The attempt at a solution
So first to calculate L for each arc
L1=pi(.107m)/2
L2=pi(.214m)/2
L3=pi(.321m)/2
and lambda:
λ1=3.3E-6/L1=2E-5
λ2=-12E-6/L2=-3.9E-5
λ3=29.7E-6/L3=5.9E-5
So using θ=pi/2 and rearranging the equation found above with the distributive property i come up with:
k(2sin(π/2)*(2E-5-3.9E-5+5.9E-5)
=508551 N/C
This was incorrect and I am unsure of my mistake(s)

2. Jan 30, 2017

### TSny

This equation is incorrect. You can see that it doesn't have the correct dimensions for an electric field.
(When you use the correct formula, you will see that the electric fields from each arc have an interesting relation.)

3. Jan 30, 2017

### Cutter Ketch

I agree with TSny that the last equation is incorrect. I wonder if you were ever given an equation describing the electric field at the center of curvature of an arc. That seems too random. Do you think that perhaps you are supposed to integrate to get the net field of each arc? Has your class been doing similar integral problems?

4. Jan 31, 2017

### Jrlinton

We haven't done much integral problems besides calculating the number of excess electrons in a rod and things of that sort. I guess I am just without a formula for the net field and the center of an arc.

5. Jan 31, 2017

### Jrlinton

I suppose if i were to integrate it I could do so by setting the limits of integration so that the difference is θ but then I am unsure of what to include in the integration. As they are my only known information i have to assume that the radius, linear charge density and constant k is included. And then I would just add the integrations for each arc to find the net field.

6. Jan 31, 2017

### Staff: Mentor

Hint: Concentrate on finding the field at the radius center of a quarter circle arc. Redraw it to take best advantage of symmetry:

You can use angular charge density for the arc since it's natural to integrate over the enclosed angle.

7. Jan 31, 2017

### Jrlinton

So:
kλ/r∫cos(θ)dθ with the limits of integration being 0 to π/2 to get the x component of the field that it is asking for and to get the y component it would be identical but with the integration being ∫sin(θ)dθ?

8. Jan 31, 2017

### Staff: Mentor

Use symmetry! You should be able to see that one of the components is perfectly cancelled by symmetrically placed charge elements (if you orient the arc as I depicted).

9. Jan 31, 2017

### Jrlinton

Right, so the vertical component is absent....

10. Jan 31, 2017

### Staff: Mentor

Yes, for the shown orientation of the arc. You should be able to express the charge density in angular terms knowing the total angle enclosed by the arc and the total charge.