Electric potential of a conical surface

[davon806]

Homework Statement

(From Griffiths problem 2.26) :
A conical surface (an empty ice-cream cone) carries a uniform surface charge σ. The height of the cone is h, and the radius of the top is R. Find the potential difference between points a (the vertex) and b (the center of the top).

Homework Equations

The Attempt at a Solution

Solution to this problem: (EXAMPLE 4 of the page)
http://www.physicspages.com/2011/10/10/electric-potential-from-charges-examples-1/

Whenever we are talking about electric potential,we need to specify the reference point.
i.e.
V(r) \equiv -\int_O^\mathbf{r} \mathbf{E \cdot} d \mathbf{l}\

Where O is some standard reference point s.t. V = 0 at that point.In most circumstance we regard infinity as the reference point.

If you picture an inverted cone in your mind, and follow the solution in the above link, then for V(0),
you integrate from the base of cone to the tip of the cone.
While for V(R) ( the base of the cone), you integrate from the tip of the cone up to the base of the cone).

But this two directions are exactly opposite, so how can we have two different reference point when calculating the potential difference V(R) - V(0) ?

 

[BvU]

If you picture an inverted cone in your mind, and follow the solution in the above link, then for V(0),
you integrate from the base of cone to the tip of the cone.
While for V(R) ( the base of the cone), you integrate from the tip of the cone up to the base of the cone).

Not when I read it:
For z_v=0 they add all the contributions from the rings from z = 0 to z = R to find the potential V(0).
For z_v=R they add all the contributions from the rings from z = 0 to z = R to find the potential V(R).

 

[davon806]

Not when I read it:
For z_v=0 they add all the contributions from the rings from z = 0 to z = R to find the potential V(0).
For z_v=R they add all the contributions from the rings from z = 0 to z = R to find the potential V(R).

Yes, but since these two points are located at the limit of integration (z = 0 and z = R). For z = 0, the tip is at bottom,and all the contributions are above of the tip;
But for z = R, all the contributions are below of the centre of base . I am a bit confused,does this invert the upper and lower limit of integration? And using the definition of V = V(r) \equiv -\int_O^\mathbf{r} \mathbf{E \cdot} d \mathbf{l}\ , does the tip and centre of base have the same reference point ? I mean, because of "invert the upper and lower limit of integration" , I think the reference points in a and b are different as well. Sorry, I am very confused

 

[haruspex]

does this invert the upper and lower limit of integration?

No. The integral runs in the same direction in each case, from 0 to h.

does the tip and centre of base have the same reference point ?

Yes. The reference point in both cases is zero potential at infinity. This is built into equation (16). Consider what happens to the expression for dV there as z_v tends to infinity.