# Electric potential at center of circular arc

1. Oct 12, 2006

### cyberstudent

An insulating rod of length l is bent into a circular arc of radius R that subtends an angle theta from the center of the circle. The rod has a charge Q ditributed uniformly along its length. Find the electric potential at the center of the circular arc.

Struggling with this problem.

I know that I have to divide the charge Q into many very small charges, essentially point charges, then sum them up (integration).

dV = dq/(4 Π Ε0 R)

Length of dq = ds
db = angle subtended by ds
dΒ=ds/R => ds = dBR

dq = λds => dq = λdBR

V = ∫ dq/(4 Π Ε0 R)

V = ∫ λdBR/(4 Π Ε0 R)

Now, this is where it all goes wrong for me.

I take out the constants V = λR/(4 Π Ε0 R) * ∫ dB
My Rs cancel out, which makes no sense.
The radius must be important in the calculation of the difference potential.

Notice also that I did not indicate the limits on the integration. In a similar problem which was done in a previous assignemnt to calculate the electric field at the center, the upper and lower limits were set to -B/2 and B/2, but I am not sure why.

Last edited: Oct 12, 2006
2. Oct 12, 2006

### OlderDan

It is. Define λ

The limits would be -theta/2 to +theta/2. You have to integrate over the angle subtended by the arc. You could actually use any pair of limits that differ by theta.

3. Oct 12, 2006

### cyberstudent

λ is the linear charge density.

That actually makes senses to me. Thanks. Of course, you would want to integrate over the angle subtended by the arc. Would 0 and theta also be valid limits then?

So, I now have my limits, but I still end up with the same absurd problem of losing my R. What is the mistake. Is the equation wrong? Am I not taking out constants?

Θ/2
V = ∫ λdBR/(4 Π Ε0 R)
-Θ/2

Θ/2
V = λR/(4 Π Ε0 R) ∫ dB
-Θ/2

I uploaded a diagram to clarify the problem and my thought process.

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4. Oct 12, 2006

### OlderDan

The limits of 0 and theta will work fine.
OK, you have defined λ. Now calculate λ.

5. Oct 12, 2006

### cyberstudent

I think I got it.

You must think I'm a bit slow. :)

λ = Q/ΘR (Charge/divided by length of the arc)

I haven't lost my R after all!

Θ/2
V = ∫ λdBR/(4 Π Ε0 R)
-Θ/2

Θ/2
V = λR/(4 Π Ε0 R) ∫ dB
-Θ/2

Θ/2
V = λ/(4 Π Ε0) B ׀
- Θ/2

V = (λ/(4 Π Ε0)) * Θ

V = QΘ/((4 Π Ε0)ΘR)

V = Q/(4 Π Ε0 R)

Interestingly, the equation for the potential difference at the center of the circular arc is the same regardless of Θ.
Have I solved the problem or did I make another dumb mistake?

6. Oct 13, 2006

### OlderDan

Why would I think you are slow? You did what I asked you to do, jut not what I had expected you would do

Your answer is interesting, and it is correct. It also makes sense. Regardless of the arc length, all the charge is the same distance from the center of the circle. Potential is a scalar inversely porportional to the distance from the charge. No matter how you distribute that charge along the arc the answer will be the same.

7. Mar 12, 2008

### chrisg8

sorry to dig up such an old post but i too had a problem like this and everything except how you went from

V = (λ/(4 Π Ε0)) * Θ

to

V = QΘ/((4 Π Ε0)ΘR)

makes sense... i realize that λ=Q/R but how come the Θ cancels out?

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