What is the electric field at the center of curvature of an arc?

In summary, the problem is to determine the electric field at the center of curvature of an arc with an arbitrary angle \alpha, given the equations E=\frac{kQ}{R^2}\widehat{r}, S=R\alpha, and \lambda=\frac{Q}{S}. The solution involves dividing the arc into small pieces and solving the integral for the y-component of the electric field. This may be due to the arc being oriented with the y-axis, rather than the x-axis as indicated in the given picture. By symmetry, the electric field must be along the line from the center of curvature to the midpoint of the arc, and whether it has x and y components depends on its orientation to the axis.
  • #1
Thomas_
21
0

Homework Statement


Determine the field at the center of curvature of an arc of arbitary angle [tex]\alpha[/tex]

([tex]\alpha[/tex] is with the x-axis)


Homework Equations


[tex]E=\frac{kQ}{R^2}\widehat{r}[/tex]

[tex]S=R\alpha[/tex]

[tex]\lambda=\frac{Q}{S}[/tex]

The Attempt at a Solution


I divide the arc into small pieces ds. [tex]ds=rd\alpha[/tex]

[tex]\lambda = \frac{dQ}{ds}[/tex]

This gives me: [tex]dE = \frac{k\lambda d \alpha}{R}\widehat{r}[/tex]
I would have to integrate this / its components (X and Y)

the solution I'm given is: [tex]E_y=\frac{2k \lambda sin(\frac{\alpha}{2})}{R}[/tex]

However, I don't understand the following: Why is there only a solution for Y? As far as I can tell the X components don't cancel since its an arc of arbitrary length, not a half or full circle. Also, shouldn't the Y component be in terms of cos? The angle alpha is with the X axis, I don't understand why E_y would be in terms of sin (since it become cos after integration).

Thank you!
 
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  • #2
Looks to me like they oriented the arc so that it's centered on the y-axis. (The statement that "[itex]\alpha[/itex] is with the x-axis" does seem a bit odd.) Of course the easy way to solve the problem regardless of the orientation of the arc, is to align it with some axis, solve the integral as show, then transform it back to the given axis.
 
  • #3
Well, they give me a picture with the arc starting from the positive x axis, going through an angle alpha which is counterclockwise from the positive x-axis. That's what is confusing me...

Also, shouldn't there be two components to the elctric field? X and y? Regardless of where the arc is oriented?
 
  • #4
Thomas_ said:
Well, they give me a picture with the arc starting from the positive x axis, going through an angle alpha which is counterclockwise from the positive x-axis. That's what is confusing me...
That does seem to contradict the answer.

Also, shouldn't there be two components to the elctric field? X and y? Regardless of where the arc is oriented?
By symmetry, you know the electric field must be along the line from the center of curvature to the midpoint of the arc. To find the magnitude, just choose an axis along that direction to do your integration.

Whether the field has x and y components depends on its orientation to the axis. If the arc were centered on the x-axis, it would only have an x-component. For whatever reason, in contradiction to the picture you were given, they chose to align the arc with the y-axis. But according to how you described the picture, it would have both an x and y component.
 

Related to What is the electric field at the center of curvature of an arc?

What is the "E-Field at center of arc"?

The "E-Field at center of arc" refers to the electric field that is present at the center of an arc, which is a curved shape often seen in nature and in man-made structures. This electric field is caused by the presence of charged particles, such as electrons, within the arc.

How is the "E-Field at center of arc" calculated?

The "E-Field at center of arc" can be calculated using the formula E = kQ/r², where E is the electric field strength, k is the Coulomb's constant, Q is the charge of the particles in the arc, and r is the distance from the center of the arc to the point where the electric field is being measured.

What factors affect the strength of the "E-Field at center of arc"?

The strength of the "E-Field at center of arc" is affected by the charge of the particles in the arc, the distance from the center of the arc, and the shape and size of the arc itself. Additionally, the presence of other charged particles or objects near the arc can also impact the electric field strength.

What are some real-world applications of the "E-Field at center of arc"?

The "E-Field at center of arc" has various applications in different fields. For example, in physics and engineering, understanding the electric field at the center of an arc is crucial for designing and optimizing electrical circuits and systems. In atmospheric science, the electric field at the center of an arc is studied to better understand and predict lightning strikes.

How does the "E-Field at center of arc" relate to other concepts in electromagnetism?

The "E-Field at center of arc" is closely related to other concepts in electromagnetism, such as electric charge, electric field, and Coulomb's law. It can also be connected to the phenomenon of electric potential, as the electric field is the gradient of electric potential. Additionally, the "E-Field at center of arc" can interact with magnetic fields, resulting in complex electromagnetic effects.

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