How to calculate the induced charge in an electric field

In summary: However, the expressions given in the document that you posted are very easy to use. You just need to substitute the appropriate values into the expressions given in the document, and you will get the answer that you are looking for.
  • #1
stanlee
5
0
Homework Statement
The aim of project is to measure the electric field with field mill. These two semi-cylindrical conductor rotating in electric field, how can I calculate the induced charge and the current.
Relevant Equations
Below the picture
1598778745720.png

Maybe I should use this?
1598779097035.png
 
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  • #2
1598779285697.png
I don't know whether this is correct.
 
  • #3
You essentially have the problem of finding the induced surface charge density on a long, cylindrical conductor placed in a uniform E-field oriented perpendicular to the axis of the cylinder. This is a fairly standard boundary value problem in electrostatics using separation-of-variables in Laplace's equation. The solution will give you something similar to what is given in equation (3):
1598811533515.png

However, the presence of ##i## in this equation looks odd to me. But, thankfully, this ##i## disappears in equation (5).

##\sigma(\theta)## is supposed to be the "charge per unit length". I'm not sure how to interpret this. If you include an additional factor of the length, ##L##, of the cylinder, then the equation (without the ##i##) would be

##\sigma(\theta) = 2rL\epsilon_0\left(E_x \sin\theta + E_y \cos \theta \right)##

I believe this now represenst the charge per unit angle ##\theta##. Then it would make sense to write ##q_a = \int \sigma(\theta) d \theta## as in equation (4).
 
  • #4
TSny said:
You essentially have the problem of finding the induced surface charge density on a long, cylindrical conductor placed in a uniform E-field oriented perpendicular to the axis of the cylinder. This is a fairly standard boundary value problem in electrostatics using separation-of-variables in Laplace's equation. The solution will give you something similar to what is given in equation (3):However, the presence of ##i## in this equation looks odd to me. But, thankfully, this ##i## disappears in equation (5).

##\sigma(\theta)## is supposed to be the "charge per unit length". I'm not sure how to interpret this. If you include an additional factor of the length, ##L##, of the cylinder, then the equation (without the ##i##) would be

##\sigma(\theta) = 2rL\epsilon_0\left(E_x \sin\theta + E_y \cos \theta \right)##

I believe this now represenst the charge per unit angle ##\theta##. Then it would make sense to write ##q_a = \int \sigma(\theta) d \theta## as in equation (4).
1598819192243.png

If there is only a vertical electric field, can you show me the charge distributed in there surface? And is the current equals to zero in first image? Why?
 
  • #5
The two half-cylinders are connected together by the external circuit such that you can consider the entire cylinder to be a single conductor at one potential. (This assumption seems to lead to the expressions given in the document that you posted.) So, the charge distribution at any instant is just the charge distribution of a full cylinder in the electric field. So, in your first image, ignore for the moment the split in the cylinder. Qualitatively, how would the charge be distributed on the surface? Once you see this, then you should be able to see whether each half-cylinder has any net charge in the first image. Repeat for the other two images. Then we can think about whether or not there is any current for the three images.
 
  • #6
TSny said:
The two half-cylinders are connected together by the external circuit such that you can consider the entire cylinder to be a single conductor at one potential. (This assumption seems to lead to the expressions given in the document that you posted.) So, the charge distribution at any instant is just the charge distribution of a full cylinder in the electric field. So, in your first image, ignore for the moment the split in the cylinder. Qualitatively, how would the charge be distributed on the surface? Once you see this, then you should be able to see whether each half-cylinder has any net charge in the first image. Repeat for the other two images. Then we can think about whether or not there is any current for the three images.
Then every images should have negtive charge in the up semi-cylindrical and positive in the bottom semi-cylindrical.
 
  • #7
Yes.
 
  • #8
TSny said:
Yes.
1598823942561.png

Can you derive this formula in detail or some website I can learn by myself? Thank you very much!
 
  • #9
Have you studied solving Laplace's equation in various coordinate systems using separation of variables? If so, look at the first page here . The general solution in polar coordinates is at the bottom of the page. You need to apply boundary conditions to determine the various constants ##A_n, B_n, C_n,## and ##D_n##.
 

1. What is induced charge in an electric field?

Induced charge refers to the redistribution of electric charges on a conductor when it is placed in an external electric field. This redistribution occurs due to the influence of the external field on the free electrons within the conductor.

2. How do you calculate the induced charge in an electric field?

The induced charge can be calculated by using the formula Q = C x E, where Q is the induced charge, C is the capacitance of the conductor, and E is the strength of the external electric field. This formula is known as the capacitance equation.

3. What factors affect the induced charge in an electric field?

The induced charge is affected by the strength of the external electric field, the size and shape of the conductor, and the dielectric constant of the material the conductor is made of. Additionally, the distance between the conductor and the source of the electric field also plays a role in determining the induced charge.

4. Can the induced charge be negative?

Yes, the induced charge can be negative. This occurs when the external electric field induces a redistribution of charges in the opposite direction of the original charge distribution on the conductor. In this case, the induced charge will have a negative value.

5. How is the induced charge related to the electric potential?

The induced charge is directly proportional to the electric potential. This means that as the electric potential increases, the induced charge also increases. The relationship between the two can be expressed as Q ∝ V, where Q is the induced charge and V is the electric potential.

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