Confusion about the potential reference point

[jegues]

Homework Statement

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

Homework Equations

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!

 

[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

 

[jegues]

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

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

Just think, how would you substitute infinity for z into the potential function here?

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?

Can you please clarify?

 

[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

 

[jegues]

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

This was the missing piece I was looking for!

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

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

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

 

[ehild]

And what is this potential at infinity ?

ehild