Half infinite cylinder uniform charge

[Kosta1234]

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?

 

[kuruman]

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?

 

[Kosta1234]

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

 

[kuruman]

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.

 

[Kosta1234]

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..

 

[kuruman]

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?

 

[Kosta1234]

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:

$\vec F = \frac {\rho q R^2 *(sin(\pi/2)-(sin(-pi/2)) * ln(R) }{ 4 \epsilon } \hat x$
$\vec F = \frac {\rho q R^2 ln(R) }{ 2 \epsilon } \hat x$

 

[PeroK]

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.

 

[Kosta1234]

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.

 

[PeroK]

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.

 

[Kosta1234]

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$

 

[PeroK]

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!

 

[kuruman]

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 ...

 

[Kosta1234]

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 $\vec F = \vec E q$

 

[kuruman]

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.

 

[Kosta1234]

so I just have to add the $cos \theta$ to them?

$\vec E = \frac {\rho \int_0^\pi cos \theta d\theta \int_0^R r dr}{2 \pi \epsilon _0 r}$

 

[kuruman]

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.