How Do You Calculate the Electric Field from an Arc?

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Homework Help Overview

The discussion revolves around calculating the electric field generated by an arc, specifically focusing on the integration process involved in deriving the electric field expression. The subject area is electromagnetism, particularly the behavior of electric fields due to charge distributions along curved paths.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the cosine function to account for the symmetry of the arc and question the validity of the original integral limits. There is also a consideration of the relationship between the angle and the sine function in the context of the arc's geometry.

Discussion Status

The discussion is ongoing, with participants providing insights into the integration limits and symmetry considerations. Some guidance has been offered regarding the integration approach, but there is no explicit consensus on the final expression or method.

Contextual Notes

Participants are navigating the complexities of integrating over an arc and are addressing potential assumptions about the geometry and symmetry of the problem. There is a mention of the need to clarify the angle used in the calculations.

VitaX
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Homework Statement



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Homework Equations



E = k*Q/r^2

The Attempt at a Solution



I tried to get the factor through integrating, but it was wrong.I ended up with (k*lambda*2)/R multiplied by the integral of cos(theta) dtheta. Limits were 10pi/21 and 0. I then divided by k*Q/r^2.
 
Last edited:
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0 to 2π is a complete circle.

The arc appears to be symmetric WRT the x-axis, ∴ integrate cosθ from ‒10π/21 to 10π/21 , which should give twice the answer you had.
 
Is theta/sin(theta) the answer to this problem? theta being in radians and half the angle of the arc.
 
Last edited:
VitaX said:
Is theta/sin(theta) the answer to this problem? theta being in radians and half the angle of the arc.

I'm sorry, I only looked at your integral in my earlier post. The 2 in the factor preceding your integral makes up for only integrating over half the arc.

Depending upon what you are using for θ, your answer of θ/sinθ looks good.
 

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