Electric field at the semi-circle's center

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Homework Help Overview

The problem involves calculating the electric field at the center of a semicircle with a uniformly distributed charge around its perimeter. The subject area pertains to electrostatics and vector fields.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to account for the vector nature of the electric field, considering how different elements of charge contribute to the field at the center. There are attempts to set up integrals to calculate the electric field, with questions about the appropriate limits and components to include.

Discussion Status

Some participants have provided guidance on considering the vector components of the electric field, and there is an acknowledgment of the need to adjust the integral to account for the orientation of the semicircle. Multiple interpretations of the setup and calculations are being explored.

Contextual Notes

There is uncertainty regarding the limits of integration and the treatment of the semicircle's orientation, which are under discussion. Participants are also reflecting on the cancellation of components due to symmetry.

Colts
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Homework Statement



Charge Q is uniformly distributed around the perimeter of a semicircle of radius R. What is the electric field at the circle's center?

Homework Equations


E=kQ/R^2


The Attempt at a Solution


λ=Q/(∏R)

∫(kλ)/R^2 dθ from 0 to pi

and get the answer to be (kQ)/R^3

or is it

∫(kλ)θR/R^2 dθ from 0 to pi

and get (kλ∏^2)/2R

The second one I was trying to put the arc length in, but I'm not sure if you need to or not.
 
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Remember that electric field is a vector quantity. Different elements of the semicircle produce electric field at the center in different directions. You need to take that into account. You have probably seen other examples where this is true, such as the field from a straight line of charge.
 
I forgot about that. Thanks. So Since it is a semi-circle. All the charges of the i or j direction (depending on how the circle is orientated) will cancel out. So would I multiply the first integral by cos(θ) to get the "vertical" component (once again depending how it is orientated, but I am thinking of it as a bowl right now). But what would the range of the integral? -pi/2 to pi/2?
 
Colts said:
I forgot about that. Thanks. So Since it is a semi-circle. All the charges of the i or j direction (depending on how the circle is orientated) will cancel out. So would I multiply the first integral by cos(θ) to get the "vertical" component (once again depending how it is orientated, but I am thinking of it as a bowl right now). But what would the range of the integral? -pi/2 to pi/2?

Yes, that sounds good.
 

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