Electric field due to Half Cylinder

In summary, the conversation discusses the methods to find the electric field at a point on an infinitely long half-cylindrical shell with a uniform charge density. The use of direct integration and Gauss' Law are not applicable in this situation. The method of images is also not a clear solution. The use of Laplace's Equation is suggested, but the boundary conditions are uncertain. An alternative approach is proposed, using the idea of an infinite line charge and Newton's Third Law to find the force per unit length on the half-cylinder, which can then be used to determine the electric field at the cylindrical axis. The use of a finite line charge would not yield the same result.
  • #1
ChowPuppy
8
5

Homework Statement



We have an infinitely long half-cylindrical shell of radius r and charge density σ as shown below
m7vyvAq.png


I am supposed to find the electric field at a point on the cylindrical axis, as seen in the diagram. Consider a coordinate system where the cylinder extends infinitely along the z-axis and is in the upper half of the x-y plane, centered at the origin. Based on the symmetry of the problem, I know that the electric field must point in the -y direction. However I can't find a good method to figure out what the magnitude of the field is.

Direct integration is a no go, because this is an infinite half cylinder.
Gauss' Law doesn't apply here.
I am not sure about the method of images, but I don't see any clear way to match boundary conditions.
Possibly Laplace's Equation would apply? But again the boundary conditions would be odd.

Thanks for any help.
 
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  • #2
ChowPuppy said:
Direct integration is a no go, because this is an infinite half cylinder.

Did you give it a try?

[EDIT: If you want to avoid direct integration, you can try the following. Imagine the z-axis is uniformly charged with a linear charge density. It is easy to find the E-field produced by the line charge (Gauss' law). Find the force that the infinite line charge exerts on a unit length of the half-cylinder. This will require an integration, but it will be simple. Then, think about how you can use this to determine the E-field produced by the half cylinder at the z-axis.]
 
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  • #3
TSny said:
Did you give it a try?

[EDIT: If you want to avoid direct integration, you can try the following. Imagine the z-axis is uniformly charged with a linear charge density. It is easy to find the E-field produced by the line charge (Gauss' law). Find the force that the infinite line charge exerts on a unit length of the half-cylinder. This will require an integration, but it will be simple. Then, think about how you can use this to determine the E-field produced by the half cylinder at the z-axis.]

Hi TSny! :)

I recently came along a similar problem and I am very interested to know how you can determine the E-field without integration.

As per your suggestion, I found the force due to infinite line charge on the unit length of half-cylinder.

It comes out be:
$$\frac{\lambda \sigma}{\pi \epsilon_o}$$
where ##\lambda## is the linear charge density of infinite wire.

I am not sure how I can find the E-field of half-cylinder using the above. I am making a guess, do I have to use Newton's third law?

Many thanks!
 
  • #4
Hi Pranav-Arora.

Pranav-Arora said:
As per your suggestion, I found the force due to infinite line charge on the unit length of half-cylinder.

It comes out be:
$$\frac{\lambda \sigma}{\pi \epsilon_o}$$
where ##\lambda## is the linear charge density of infinite wire.

That looks good.

I am not sure how I can find the E-field of half-cylinder using the above. I am making a guess, do I have to use Newton's third law?

Yes, think about the implication of Newton's 3rd law for the force per unit length on the infinite cylinder compared to the force per unit length on the infinite line of charge.
 
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  • #5
TSny said:
Yes, think about the implication of Newton's 3rd law for the force per unit length on the infinite cylinder compared to the force per unit length on the infinite line of charge.

From Newton's third law, the force acting on the unit length of infinite line charge is also ##\dfrac{\lambda \sigma}{\pi \epsilon_0}##. Hence, the E-field due to cylinder is ##\dfrac{\sigma}{\pi \epsilon_0}##, is this correct?

But can you please explain why this method works? Why only an infinite line charge, why not some finite line charge or something else? :confused:
 
  • #6
If you had an infinite half-cylinder but a finite line charge, then the 3rd law would tell you that the magnitude of the net force on the cylinder would equal the net force on the line charge. But the 3rd law would not be able to tell you very much about the force per unit length (i.e., the force density) on the cylinder. The force density would not be uniform on the cylinder.

However, when both the half-cylinder and the line charge are infinite, then you can use symmetry along with the 3rd law to reason that the force densities must be uniform and of equal magnitude for each object.
 
  • #7
Pranav-Arora said:
From Newton's third law, the force acting on the unit length of infinite line charge is also ##\dfrac{\lambda \sigma}{\pi \epsilon_0}##. Hence, the E-field due to cylinder is ##\dfrac{\sigma}{\pi \epsilon_0}##, is this correct?

Yes.
 
  • #8
TSny said:
If you had an infinite half-cylinder but a finite line charge, then the 3rd law would tell you that the magnitude of the net force on the cylinder would equal the net force on the line charge. But the 3rd law would not be able to tell you very much about the force per unit length (i.e., the force density) on the cylinder. The force density would not be uniform on the cylinder.

Umm...I have trouble understanding why are we concerned with the force density and not the net force itself. I am sorry if this is a dumb question.
 
  • #9
Pranav-Arora said:
Umm...I have trouble understanding why are we concerned with the force density and not the net force itself.

When both the half-cylinder and line charge are infinitely long, then the net force on each is infinite. But the force densities are finite.

If you use a finite line charge with an infinite half-cylinder, then the net force on each will be finite. But then you cannot use the simple expression for the electric field of an infinite line charge in your analysis.
 
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  • #10
TSny said:
If you use a finite line charge with an infinite half-cylinder, then the net force on each will be finite. But then you cannot use the simple expression for the electric field of an infinite line charge in your analysis.

I am not sure if I understand it correctly but if I use a finite line charge, would that still give me the right answer?
 
  • #11
Yes, it would give you the right answer if you use a finite line charge. But it seems to me that you would then have to do essentially the same integration as you would if you just integrate over the half-cylinder directly to find the E field on the axis produced by the half-cylinder.

For example, suppose you shrunk the finite line charge down to a point charge on the axis of the cylinder. Would there be any advantage to using the point charge at all?
 
  • #12
TSny said:
For example, suppose you shrunk the finite line charge down to a point charge on the axis of the cylinder. Would there be any advantage to using the point charge at all?

I don't think there would be any advantage, that would be similar to finding E-field of cylinder through direct integration.

I was wondering that instead, if we had a solid half-cylinder, can the same method be used?
 
  • #13
Yes, you could use the infinite line charge for the infinitely long, solid half-cylinder.
 
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  • #14
TSny said:
Yes, you could use the infinite line charge for the infinitely long, solid half-cylinder.

Thanks a lot TSny! :smile:
 

What is an electric field?

An electric field is a physical field that surrounds an electrically charged particle or object. It is a force field that describes the influence that the charged particle or object has on other particles or objects in its vicinity.

What is a half cylinder?

A half cylinder is a three-dimensional shape that is formed by cutting a cylinder in half lengthwise. It has a curved surface on one side and a flat surface on the other side.

How is the electric field due to a half cylinder calculated?

The electric field due to a half cylinder can be calculated using the formula E = (σ/2ε0) * (1 + cosθ), where σ is the surface charge density, ε0 is the permittivity of free space, and θ is the angle between the point where the electric field is being measured and the axis of the half cylinder.

What factors affect the electric field due to a half cylinder?

The electric field due to a half cylinder is affected by the surface charge density, the permittivity of free space, and the distance from the half cylinder. The angle between the point where the electric field is being measured and the axis of the half cylinder also plays a role in determining the strength of the electric field.

What are some real-world applications of the electric field due to a half cylinder?

The electric field due to a half cylinder is used in various engineering and scientific fields, such as in the design of capacitors and in the study of electromagnetic fields. It is also used in the development of technologies such as wireless charging and medical imaging.

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