- #1

RJLiberator

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## Homework Statement

A semi-circle of radius R has a charge Q uniformly distributed over its length, which provides a line charge density λ. Determine E at the origin.

## Homework Equations

## The Attempt at a Solution

View attachment 105239

I can tell by argument of symmetry that the Electric field will be pointing in +y direction.

If we take a sliver of the charge, call it dq, we will calculate the Electric field.

[tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{dq}{R^2}\right)sin(θ) [/tex]

Also: [tex] \left(\frac{dq}{dθ}\right) = \left(\frac{Q}{π}\right) [/tex]

so: [tex] dq = \left(\frac{Qdθ}{π}\right) [/tex]

Now:

[tex] E = \left(\frac{1}{4πε_0}\right)\left(\frac{Qsin(θ)}{πR^2}\right)dθ [/tex]

After calculating the integral from θ = 0 to θ = π I get the following answer:

[tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{Q}{πR^2}\right)\hat{y} [/tex]

Does this work appear to be correct? I want to make sure I get this easy material down before I go to the more difficult material in electromagnetism. If so, is it fair to say that the linear charge density typically symbolized as λ is equal to Q/π in this problem. Q is the charge in coulombs and pi is the length of the semicircle. So I could represent my answer as:

[tex] E = \left(\frac{1}{2πε_0}\right)\left(\frac{λ}{R^2}\right)\hat{y} [/tex]