Semi-circular cylinder and line charge

[Angela G]

Homework Statement

Determine the amount of the force per unit length acting on the line charge from a semi-circular cylinder with a radius R. The axis of the cylinder is perpendicular to
the plan of the figure. The cylinder is very ("infinitely") long and has a surface charge density σ. Theris a line charge with longitudinal charge density, λ_0, at the center.

Relevant Equations

$E = \frac{1}{4\pi \epsilon_0} \int \frac{dq}{r^2}$
$F = qE$

Hi!
I think I solved the problem but I would like you to take a look at it please, is it right? is there a better way to solve it?

 

[Delta2]

I am not sure where you go wrong but I think your final answer shouldn't depend on z, because the problem has z-symmetry.

There is a much easier (I believe) way of solving this. Calculate the force that the line charge (of the center) applies to the infinite half cylinder and then use Newton's 3rd law to equate this force with the force requested by the problem's statement.

 

[Delta2]

BTW, when you integrate, you set the limits of integration from 0 to z, I think they should be $-\infty$ to $+\infty$.

 

[Angela G]

I followed your advice and I got that the electric field due to a charge line at a distance R is $$ E = \frac{\lambda}{2 \pi \epsilon_0 R} $$ ( I'm searching the magnitude of the force per unit charge not the dircetion)

And that the Force per unit length is $$ \frac{F}{L} = \frac{\sigma\lambda R}{\pi\epsilon_0 L}$$

is that right? I maybe can write the force per unit length as
$$ \frac{F}{L} = \frac{\sigma Q R}{\pi\epsilon_0} $$ where Q is the charge of the rod

I realized that I have to use $\lambda$ so forget the expression with Q

 

[PeroK]

I followed your advice and I got that the electric field due to a charge line at a distance R is $$ E = \frac{\lambda}{2 \pi \epsilon_0 R} $$ ( I'm searching the magnitude of the force per unit charge not the dircetion)

And that the Force per unit length is $$ \frac{F}{L} = \frac{\sigma\lambda R}{\pi\epsilon_0 L}$$

is that right? I maybe can write the force per unit length as
$$ \frac{F}{L} = \frac{\sigma Q R}{\pi\epsilon_0} $$ where Q is the charge of the rod

I realized that I have to use $\lambda$ so forget the expression with Q

This seems to have got a bit confused. In any case, there shouldn't be an $R$ in the answer. I think your first answer was closer, as you had the correct integrand.

https://www.physicsforums.com/threads/electric-field-due-to-half-cylinder.718319/

 

[Delta2]

Ah damn, sorry I see now I messed up myself, it's not that simple because the force changes in direction (it is in the radial direction). I think to find the correct answer you should find the x-component of the E-field (produced by the linear charge) and integrate it through the surface of the cylinder. (the y components seem to cancel out must do a figure to show you what i mean).

 

[Angela G]

I'm sorry I think I don't understand wasnt it what I did? to determine the E- fied and then integrate along the cylinder area? Could you please tell me where did I wrong?

 

[PeroK]

Ah damn, sorry I see now I messed up myself, it's not that simple because the force changes in direction (it is in the radial direction). I think to find the correct answer you should find the x-component of the E-field (produced by the linear charge) and integrate it through the surface of the cylinder. (the y components seem to cancel out must do a figure to show you what i mean).

Yours was a good idea - especially if you already know the electric field of an infinite wire. That also makes it a good way to confirm the answer. Overall, you have the same integration either way.

 

[PeroK]

I'm sorry I think I don't understand wasnt it what I did? to determine the E- fied and then integrate along the cylinder area? Could you please tell me where did I wrong?

The factors of $R$ should have canceled in the integration: you have $\frac 1 R$ in the electric field and $\sigma Rd\theta$ in the surface charge of the cylinder.

 

[Angela G]

ahh, I see I was unsure how to write #dq# due to the cylinder, $dq = R \sigma d\theta$?
which method do yo think I should use?

 

[PeroK]

ahh, I see I was unsure how to write #dq# due to the cylinder, $dq = R \sigma d\theta$?

Yes, that's the infinitesimal charge on a strip of the cylinder of unit length.

 

[Angela G]

Yes, that's the infinitesimal charge on a strip of the cylinder of unit length.

so, then I will not need to divide the force by L?

 

[PeroK]

so, then I will not need to divide the force by L?

You can introduce $L$ if you want. But, to get the force per unit length, that's going to cancel as well.

@Delta2 's idea amounts to: the force per unit length on the cylinder must be the same magnitude as the force per unit length on the wire.

 

[Angela G]

so I changed dq in the integral, is my calclation correct? I mean that F is the force per unit length right?

 

[PeroK]

so I changed dq in the integral, is my calclation correct? I mean that F is the force per unit length right?

If that says $\frac{\lambda \sigma}{\pi \epsilon_0}$ then yes.

 

[Angela G]

It does, Thank you everyone