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Half infinite cylinder uniform charge

  • Thread starter Kosta1234
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  • #1
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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 devide 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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Answers and Replies

  • #2
kuruman
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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
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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
kuruman
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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
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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
kuruman
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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
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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
PeroK
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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
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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
PeroK
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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
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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
PeroK
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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
kuruman
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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
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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
kuruman
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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
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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
kuruman
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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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