# Confusion about the potential reference point

1. Oct 12, 2011

### jegues

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

See figure attached for problem statement, solution provided and my solution.

2. Relevant equations

3. The attempt at a solution

I'm getting confused about the reference point for the potential.

First find the potential at point z with reference at infinity,

$$\phi(z) - \phi(\infty) = \alpha$$

Now find the voltage at point z=0 with reference at infinity,

$$\phi(0) - \phi(\infty) = \beta$$

Now the potential at z with the reference at z=0, which is the same as the voltage difference between the individual voltages at z and z=0 with their reference point at infinity. (Is this statement true?)

$$\alpha - \beta = \phi(z) - \phi(0)$$

The question wants me to solve for,

$$\phi(z)$$

but it seems as though I've given them,

$$\phi(z) - \phi(0)$$

What am I interpreting differently? What am I confusing myself with?

Thanks again!

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Last edited: Oct 12, 2011
2. Oct 12, 2011

### ehild

The reference (zero) point of the potential can be chosen anywhere. Here it is at the origin instead of infinity: Φ(0)=0. Just think, how would you substitute infinity for z into the potential function here? In practice, only the potential difference counts.

When calculating the definite integral ∫dΦ = -∫Edz', integrate from z'=0 to z, then the left hand side becomes Φ(z) - Φ(0). No sense to use infinity.

ehild

Last edited: Oct 12, 2011
3. Oct 13, 2011

### jegues

I don't understand what you're trying to say with this line,

I don't have a potential function yet, in order to get it I had to calculate the potential at z with the reference at infinity. Are you making the assumption that I did this already?

4. Oct 13, 2011

### ehild

OK, I misunderstood the problem as i could not read the attached picture and your handwriting. I thought you integrated the electric field to get the potential. It is much better if you type in your work.

So the potential of the ring was calculated by integrating the Coulomb potential for the charge distribution. The Coulomb potential U=kQ/r has its zero value at infinity, and the resultant potential of the charged ring is also zero at infinity. But you can add an arbitrary constant to any potential function, so

$$V(z)=\frac{\rho_s}{2\epsilon_0}(\sqrt{a^2+z^2}-\sqrt{b^2+z^2})+C$$

is also a potential function of the same ring. Choose C so as V(0)=0.

ehild

Last edited: Oct 13, 2011
5. Oct 13, 2011

### jegues

This was the missing piece I was looking for!

So, $$C =\frac{\rho_s}{2\epsilon_0}(b-a)$$.

Now completing the expression for $$V(z)$$,

$$V(z) = \frac{\rho_s}{2\epsilon_0}(\sqrt{a^2+z^2}-\sqrt{b^2+z^2}) + \frac{\rho_s}{2\epsilon_0}(b-a)$$

6. Oct 13, 2011

### ehild

And what is this potential at infinity ?

ehild