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1. Nov 15, 2016

### horsedeg

1. The problem statement, all variables and given/known data
A uniformly charged insulating rod of length 10.0 cm is bent into the shape of a semicircle as shown in the figure below. The rod has a total charge of -9.00 µC. Find the electric potential at O, the center of the semicircle.

2. Relevant equations
V=keQ/r

3. The attempt at a solution
I already found the answer. It is as simple as plugging in the radius and charge into the relevant equation I wrote. However, why? I don't really understand. At first I was thinking that I have to find the electric field or something and integrate using that one formula, but why is it this simple? Is direction completely irrelevant?

2. Nov 15, 2016

### cnh1995

Yes. Electric potential is a scalar quantity.

3. Nov 15, 2016

### horsedeg

I suppose that much should be obvious to me. Let me rephrase. Is there no "counteraction" between the top and bottom? Or am I having some sort of misunderstanding?

4. Nov 16, 2016

### cnh1995

The charge is uniformly distributed along the ring. I am not sure I understand what you mean by "counteraction". Electric "fields" from the top and bottom charges will cancel each other directly, if that's what you are thinking of.

5. Nov 16, 2016

### horsedeg

Yes this is what I'm thinking of. I was thinking the top and bottom would "cancel" but I guess that's just a misunderstanding of how voltage works?

6. Nov 16, 2016

### cnh1995

Electric "fields" will cancel because their directions are opposite. Potentials will not cancel as they have no direction. They will add up (or subtract in case of opposite charges).

7. Nov 16, 2016

### Delta²

To be more accurate, the electric field in O will cancel only in the vertical direction . In the horizontal direction there will be electric field in O.

But we are interested in potential. The potential from an element $dl$ of the circumference will be will be $K\frac{\rho dl}{r}$, where $\rho=\frac{-9}{10}$ charge density and $\pi r=10$ the radius r, and to find the total potential we just sum all the potentials due to all $dl$s (since direction is irrelevant as you have pointed out cause potential is scalar, we don't take into account the various angles of the various $dl$s with the horizontal in order to put sine or cosine terms).