# Electric field of a cylinder given the electric field of a ring

• notgoodatphysics
In summary: The symbol ##z## represents the position of the point on the ##z##-axis. It's fixed. You can't integrate over that variable. You need a different variable, namely ##z'##, which corresponds to the position of an infinitesimal ring. Then you integrate with respect to ##z'## to sum the contributions over the entire cylinder. When you break up a disk into a collection of rings, what's different about each ring? Similarly, when you divide the cylinder up into a collection of rings, what's different about those rings?
notgoodatphysics
Homework Statement
We’re given the equation for the electric field of a disk. From that the idea is to find the electric field of a cylinder.

I thought the best way would be to integrate the original equation over the surface area of a cylinder without the ends (2*pi*r*h). My attempt is similar to the solution except, the professor has introduced z’, and an R in the final solution has disappeared.

Why introduce z’ and have the dz’ above the origin instead of just using z like the original diagram?

And where did the R go in the numerator in final step of the solution?
Relevant Equations
The first pic is the question and my attempt, and the second pic is the solution.

notgoodatphysics said:
Why introduce z’ and have the dz’ above the origin instead of just using z like the original diagram?
The symbol ##z## represents the position of the point on the ##z##-axis. It's fixed. You can't integrate over that variable. You need a different variable, namely ##z'##, which corresponds to the position of an infinitesimal ring. Then you integrate with respect to ##z'## to sum the contributions over the entire cylinder.

notgoodatphysics said:
And where did the R go in the numerator in final step of the solution?
I think this is just a typo, and the ##R## should still be there. The final answer in your professor's solution isn't dimensionally correct.

notgoodatphysics and PeroK
vela said:
The symbol ##z## represents the position of the point on the ##z##-axis. It's fixed. You can't integrate over that variable. You need a different variable, namely ##z'##, which corresponds to the position of an infinitesimal ring. Then you integrate with respect to ##z'## to sum the contributions over the entire cylinder.I think this is just a typo, and the ##R## should still be there. The final answer in your professor's solution isn't dimensionally correct.

I’m not sure I’m totally understanding the need for z’ though. In the example of the electric field of due to a ring of charge with radius a, to find the electric charge due to a disk, the integral from 0 to R is calculated - isn’t this a similar case? Why isn’t it (a-R).

(From here: https://www.physics.udel.edu/~watson/phys208/exercises/kevan/efield1.html )

Also when we find the integral of (z-z’), why don’t we also take the direction of the electric field E(z) ##cos theta## because ##\cos \theta##is changing right?

notgoodatphysics said:
I’m not sure I’m totally understanding the need for z’ though. In the example of the electric field of due to a ring of charge with radius a, to find the electric charge due to a disk, the integral from 0 to R is calculated - isn’t this a similar case? Why isn’t it (a-R).
When you break up a disk into a collection of rings, what's different about each ring? Similarly, when you divide the cylinder up into a collection of rings, what's different about those rings?

## 1. How do you derive the electric field of a cylinder using the electric field of a ring?

To derive the electric field of a cylinder from the electric field of a ring, consider the cylinder as a stack of infinitesimally thin rings. The total electric field at a point can be found by integrating the contributions of all these rings along the length of the cylinder. This involves using the principle of superposition and integrating the electric field of each ring over the length of the cylinder.

## 2. What is the expression for the electric field of a ring?

The electric field at a point along the axis of a uniformly charged ring of radius R and total charge Q is given by: $E = \frac{1}{4 \pi \epsilon_0} \frac{Qz}{(z^2 + R^2)^{3/2}}$where $$z$$ is the distance from the center of the ring along its axis, and $$\epsilon_0$$ is the permittivity of free space.

## 3. How does the symmetry of the cylinder affect the electric field calculation?

The cylindrical symmetry simplifies the calculation because the electric field at any point along the axis of the cylinder only has a component along the axis. This means that contributions from opposite sides of the cylinder cancel out in the perpendicular direction, allowing us to focus on the axial component when integrating the electric field of each ring.

## 4. What are the boundary conditions for the electric field of a cylinder?

The boundary conditions for the electric field of a cylinder depend on whether the cylinder is finite or infinite. For a finite cylinder, the electric field must match the known field at the ends and outside the cylinder. For an infinite cylinder, the electric field far from the cylinder should approach zero, reflecting the fact that the field diminishes with distance.

## 5. Can the electric field inside a uniformly charged cylinder be zero?

No, the electric field inside a uniformly charged cylinder is not zero. Due to the linear charge distribution along the cylinder, there will be a non-zero electric field inside. The field inside a uniformly charged cylinder can be derived by integrating the contributions from each infinitesimal ring, and it generally increases linearly with distance from the axis up to the surface of the cylinder, then decreases outside the cylinder.

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