# Electric field and potential at the origin

1. May 18, 2015

### cdummie

1. The problem statement, all variables and given/known data
I have to find a electric field vector and a potential of the point in the origin of the x0y coordinate system (0,0) due to a longitudinal charge placed on a half circle as shown in the picture, with radius a.

2. Relevant equations
Coulombs law.

3. The attempt at a solution

Using Coulombs law:

Vector E has two components:

Which means E=Ey.

If someone notices any mistakes up here, please point it out, since i haven't noticed any.

Now, for potential:

dl=dy because E has only y component, now when i integrate this i have V=∞, which seems wrong to me, so if someone knows how to solve this please post solution here.

2. May 18, 2015

### Staff: Mentor

Why do you integrate from a to infinity?
Also, E will depend on y. There is an easier way to calculate the potential, without electric fields.

3. May 19, 2015

### cdummie

Well i thought, since distance from point whose potential i want to know and charge is a i put first limit a, and since reference point is at infinity (i think i haven't pointed that out in the statement). I think i know what do you mean by easier way its probably dV=dQ/4πξ0a and then since dQ=Q'dl when i integrate it i get V=Q'/4ξ0. Is this correct? If it is then i should get the same thing doing it the way i did up there, but obviously i won't get the same thing as here.

4. May 19, 2015

### Staff: Mentor

Yes that is correct.

The approach with the electric field would need a path from the reference point to the position where you want to know the potential. And you would have to calculate the electric field everywhere along this path. The second part is quite messy.

5. May 19, 2015

### cdummie

That means that second approach is better for finding potential (i mean it's easier), i've done it that that way, but i thought if i solve it using electric field it could confirm that i did it correctly since the second approach seemed simple and i thought, if it's that simple, then probably i did something wrong. Anyway, thank you for help. :)

6. May 19, 2015

### Staff: Mentor

Solving it via electric fields is much more work. It would be a cross-check, but too tricky for such a homework problem.