Half infinite cylinder uniform charge

In summary, the problem involves an infinite half cylinder with uniform charge density ρ. The goal is to find the force on a test charge at a specific position. The attempt at a solution involves dividing the cylinder into half circles and using cylindrical coordinates. However, there were errors in the calculations and questions were raised about the use of polar coordinates and the addition of infinite lines of charge. Ultimately, the correct expression for the electric force was found to be (ρqR^2ln(R))/(2ε) in the x-direction, where R is the radius of the cylinder.
  • #1
Kosta1234
46
1

Homework Statement


Hi. in my problem there is an infinite half cylinder uniformly charged with ρ.
What is the force is the cylinder is making on a test charge on a charge located on the position that is shown on the following picture:

Homework Equations


∫qdQ/(r2)

The Attempt at a Solution


Well first, if I know the charge density is ρ, I can know the total charge, but the problem is that because it is an infinite cylinder, so that the volume will be infinite as well and I cannot use the equation:
dQ = ρdV

In my attempt I tried to divide the half cylinder to half circles and to sum them from z=-inf to z=inf

Something is going wrong with my calculations, and I have few questions:
1. The force on the charged because of symmetry is normal to the part of the cylinder that is cutted (x axis)?
2. I tried to use cylindrical coordinates:

so dV= rdrd(theta)dz
so r [0,R]
theta [0, pi]
but z is [inf, -inf], doesn't it mean that the force will be endless?
 

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  • #2
Kosta1234 said:
but z is [inf, -inf], doesn't it mean that the force will be endless?
No, it doesn't mean that. You know that the electric field from an infinite line of charge at some distance ##r## from the line is finite. Have you considered adding infinite lines of charge to form the cylinder instead of adding half-disks?
 
  • #3
Thank you. Here is what I did I hope it is correct:

If I got volume charge density ##\rho## , and the area of half circle is ##\pi R^2/2## so I can say that the charge density of length is ##\lambda = \rho \pi R^2/2##

If I calculate the Electric Field to distance ##r## from one wire I will get:
##\int E dA = \lambda/\epsilon##
##dA = 2\pi r ##
## E = {\lambda }/{ 2\pi r \epsilon}##
## E = {\rho \pi R^2 }/{ 4\pi r \epsilon}##
so the Electric force is:
## F = {\rho \pi R^2 q }/{ 4\pi r \epsilon}##
If I will work on polar coordinates
##\theta = [-\pi/2,\pi/2 ] ##
## r = [0, R]##

## F\vec = \int cos \theta d\theta \int {\rho \pi R^2 q }/{ 4\pi r \epsilon} ##and I've used ##cos \theta ## because of the symmetry
 
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  • #4
Your area element ##dA=2\pi r## is not correct. That's the circumference of a circle of radius ##r##. What is your Gaussian surface? What is an element of area on the surface? Also, you need to write integrals correctly. The form is ##\int f(x)~dx##. If you are integrating over the radius, you need a ##dr## the presence of which you must justify. Hint: It's related to writing ##dA## correctly.
 
  • #5
My gauss surface is a thin cylinder of radius r.
maybe I need to include it's height? But it suppose to be infinity.

About the integral I've meant to write:
##F\vec = \int cos \theta d\theta \int \rho\pi R^2 q dr / 4 \pi r \epsilon ##
## \theta = [\pi / 2, - \pi / 2] ##
## r = [0, R] ##

Edit:
Maybe it should be
##\int E dA = \lambda L / \epsilon ##

while ## dA = 2\pi r L ##
But then the L goes away from both sides of the equation..
 
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  • #6
Kosta1234 said:
But then the L goes away from both sides of the equation..
There's nothing wrong with that. If you did have ##L## in your expression, you would run into trouble because the wire is infinitely long. So what is the electric field at the point of interest due to a single wire?
 
  • #7
Well in that case I am getting the same answer I get before..
if the length density is ## \lambda = \rho \pi R^2 /2 ##
and ## \int E dA = \lambda L / \epsilon ##
## \int 2\pi r L = \rho \pi R^2 L / 2 \epsilon ##
##\vec E(r) = \frac {\rho R^2 } { 4r \epsilon } ## to direction ## \hat r ##
when r is the distance between the wire to the point of interest..

Is that correct so far?

now to find the electric force I will use:
## Fq = E(r) ##
and this is due to a single wire..For the whole half cylinder I will integrate to "create" half circle with the polar coordinates so
## r = [R,0] ##
## \theta = [\pi/2, -\pi/2] ##
and istead of the direction## \hat r ## I can write ##\hat x cos \theta ## because of the symmetry in the problem..

## F\vec (r) = \frac { \rho q R^2 } {4 \epsilon } \int cos \theta \, d\theta \int {dr}/{r} ##

and after integrating:

## \vec F = \frac {\rho q R^2 *(sin(\pi/2)-(sin(-pi/2)) * ln(R) }{ 4 \epsilon } \hat x ##
## \vec F = \frac {\rho q R^2 ln(R) }{ 2 \epsilon } \hat x ##
 
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  • #8
Kosta1234 said:
Well in that case I am getting the same answer I get before..
if the length density is ## \lambda = \rho \pi R^2 /2 ##

This can't be correct. In any problem like this, the linear density must have infinitesimal dimensions. For example, in Cartesian coordinates, the linear density in the z-direction would be:

##\lambda = \rho dxdy##

If you integrate that over a rectangular volume of length ##L##, you get:

##Q = L \int_0^a \int_0^b \rho dxdy = \rho Lab = \rho V##

As expected, the charge in the rectangular volume is the density times the volume.

In your problem you must take a similar approach but with cylindrical coordinates.
 
  • #9
But in this case if I will work with the polar coordinates and I will integrate:
## \lambda = \rho d\theta r dr ##
So ## \lambda = \rho \int_0^\pi d\theta \int_0^R rdr = \rho \frac {\pi R^2}{2} ##
so it's the same thing, am I wrong?By the way so much thanks that you making me think and not immediately giving me the answer. Appreciate that.
 
  • #10
Kosta1234 said:
But in this case if I will work with the polar coordinates and I will integrate:
## \lambda = \rho d\theta r dr ##
So ## \lambda = \rho \int_0^\pi d\theta \int_0^R rdr = \rho \frac {\pi R^2}{2} ##
so it's the same thing, am I wrong?

You can't integrate ##\lambda## at that stage. Instead, you need to use ##\lambda## and the known field of an infinite wire.
 
  • #11
Do you mean to calculate without gauss surface, I mean directly for the Electric force?
In that case:

## \vec F = \int_{-\pi/2}^{\pi/2} \int_0^R \frac {\rho q k r dr d\theta}{r^2} cos \theta \hat x ##

## \vec F = kq \rho \int_{-\pi/2}^{\pi/2} cos \theta d\theta \int_0^R \frac {1}{r} ,dr #### \vec F = 2kq\rho \cdot ln(R) \hat x ##
 
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  • #12
Kosta1234 said:
Do you mean to calculate without gauss surface, I mean directly for the Electric force?
In that case:

## \vec F = \int_0^\pi \int_0^R \frac {\rho q r dr d\theta}{r^2} cos \theta \hat x ##

## \vec F = q \rho \int_0^\pi cos \theta d\theta \int_0^R \frac {1}{r} ,dr ##
isn't it the same?

I assumed you were going to use the field for an infinite wire. If you don't have that already, then the next step is to calculate that (for a general ##\lambda##).

Your idea is not the same. You could make it work mathematically, but you'd have to be very careful about how you are modeling the half-cylinder.

The whole point, I thought, of splitting the cylinder into infinitesimal wires was to use the field for a wire, which you either already know or can be calculated.

If you have a different approach, you'd better explain it!
 
  • #13
Your calculation in post #3 for the electric field due to an infinite wire correctly gives ##E=\frac{\lambda}{2\pi \epsilon_0 r}##. The question is what do you use for ##\lambda## if you know ##\rho##? Remember that in cylindrical coordinates a charge element is ##dq=\rho \ dV=\rho~ rdr~d\theta~dz##. The linear charge density is ##\lambda = \frac{dq}{dz}##.
So ...
 
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  • #14
thanks.
I think I got it
if ## \vec E = \frac {\lambda}{\epsilon _0 } ##
## \vec E = \frac {\rho \int_0^\pi d\theta \int_0^R r dr}{2 \pi \epsilon _0 r} ##
## \vec E = \frac {\rho}(2 \pi} \int_0^\pi d \theta \int_0^R dr ##
## \vec E = \frac {\rho R}{2 \epsilon _0} ##
and ## \vec F = \vec E q ##
 
  • #15
You didn't get it. You are integrating the magnitude of ##E##. Vectors are added by adding their components. You have to integrate the components of ##E## separately to find the total and hence the field vector.
 
  • #16
so I just have to add the ##cos \theta ## to them?

## \vec E = \frac {\rho \int_0^\pi cos \theta d\theta \int_0^R r dr}{2 \pi \epsilon _0 r} ##
 
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  • #17
Kosta1234 said:
so I just have to add the ##cos \theta ## to them?

## \vec E = \frac {\rho \int_0^\pi cos \theta d\theta \int_0^R r dr}{2 \pi \epsilon _0 r} ##
As ##\theta## changes you must add together the components parallel to the face of the half-cylinder and separately add the components perpendicular to the face. This means two separate integrals. Your expression in #16 has a vector on the left side and cannot have a scalar on the right side. You need to rethink this and come up with two equations,
##E_{\parallel}=\cdots##
##E_{\perp}=\cdots##
One of these (which one?) should integrate to zero.
 
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Related to Half infinite cylinder uniform charge

1. What is a half infinite cylinder uniform charge?

A half infinite cylinder uniform charge refers to a theoretical scenario in which a cylindrical object has a uniform distribution of electric charge on one half of its surface, while the other half has no charge. This is often used as a simplified model in electrostatics problems.

2. How is the electric field calculated for a half infinite cylinder uniform charge?

The electric field for a half infinite cylinder uniform charge can be calculated using the formula E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space. This formula assumes that the cylinder has an infinite length and a negligible radius.

3. What is the difference between a half infinite cylinder and a finite cylinder in terms of electric fields?

A half infinite cylinder has an infinite length, while a finite cylinder has a finite length. This difference affects the calculation of the electric field, as the electric field for a finite cylinder must take into account the end effects, while the field for a half infinite cylinder does not.

4. How does the electric field vary with distance from a half infinite cylinder uniform charge?

The electric field from a half infinite cylinder uniform charge decreases with distance according to an inverse square law. This means that the electric field is strongest closer to the cylinder and becomes weaker as the distance increases.

5. Can a half infinite cylinder have a non-uniform charge distribution?

Yes, a half infinite cylinder can have a non-uniform charge distribution. In this case, the electric field will also vary along the surface of the cylinder, and the calculation of the field becomes more complex. However, the formula E = σ/2ε0 can still be used as an approximation if the non-uniformity is small.

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