# Potential due to a curved rod

1. Sep 12, 2009

### exitwound

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

2. Relevant equations

V=kq/r

3. The attempt at a solution

V=kQ/r
dV=kdQ/r

$dQ=\lambda d\theta$ <---This is where I'll make a mistake if any.

$dV=k(\lambda d\Theta) /r$
$$V=\frac{k\lambda}{r}\int_0^{\phi}d\Theta$$
$\phi=2\pi/3$
$\lambda=Q/L= Q/(2\pi/3)$

$$V=k\frac{3Q}{2\pi r}\int_0^{\phi}d\Theta$$

...so far?

2. Sep 12, 2009

### RoyalCat

Yerr, that is a mistake. You forgot that the length of the rod is $$R\cdot \phi$$ and not just $$\phi$$.

$$dq=\lambda dl=\lambda R \cdot d\theta$$
$$\lambda \equiv \frac{Q}{R\cdot\phi}$$

$$dV=\frac{K}{R}\cdot dq$$

$$dV=\frac{K}{R}\cdot\lambda R \cdot d\theta$$

$$dV=K\lambda\cdot d\theta$$

$$dV=\frac{KQ}{R\cdot\phi}\cdot d\theta$$

And now it's just a question of taking the integral along the arc and you're done.

The surprising result I got is that the potential is just $$V=\frac{KQ}{R}$$

Last edited: Sep 12, 2009
3. Sep 12, 2009

### exitwound

See, that is where I knew I'd make a mistake. I can't understand why it's $$R\cdot \phi$$. No problems I've found online explain it either. Where does the R come from?

4. Sep 12, 2009

### RoyalCat

Just by definition. The length of a circular arc of radius $$R$$ resting on $$x$$ radians is $$R\cdotx$$

Just like how a circle of radius $$R$$ has a circumference of $$2\pi R$$, that's the same as saying you've got a circular arc of radius $$R$$ resting on $$2\pi$$ radians.