How Is Electric Potential Calculated for a Curved Rod?

[exitwound]

Homework Statement

Homework Equations

V=kq/r

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?

 

[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}

 

[exitwound]

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

 

[RoyalCat]

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

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.