What is the Electric Field at the Origin in a Semicircle?

AI Thread Summary
The discussion focuses on calculating the electric field at the origin due to a semicircular charge distribution. Initially, there was confusion regarding the application of formulas, but it was clarified that the charge Q is distributed over a semicircle. The solution involves dividing the charge into infinitesimal segments and calculating the electric field contributions from each segment. The charge per unit length is defined as λ = Q/πR, and the resulting electric field components are derived through integration. Ultimately, the vertical component of the electric field is found to be -2kλ/R, while the horizontal component cancels out to zero.
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Homework Statement


You have a semicircle of radius R and charge Q. We also have available a test charge +qo.

Find the Magnitude and direction of the electric field at the origin.

Homework Equations


Electric Field:
F = k (Q * qo)/(R^2)
E = F/qo = kQ/R^2


The Attempt at a Solution


I'm not sure how to solve this, I assumed it would be kQ/R^2 but I am mistaken. Can someone please explain? I'm looking for an explanation on line but not having any luck
 
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Wait? The charge Q is distributed over a semicircle? I'm guessing you're going to have to use Gauss' law. But could you make the problem statement clearer? Where's the origin?
 
Sorry, I've figured it out. I just thought it was a more simple problem than this... For future reference, this is how you solve it.

It's a continuous Distribution of charge so
1) Divide the charge into segments dQ for which you already know the field
2) Find the field of each dQ
3) Find E by summing all dQ

So, the charge per unit length would be : \lambda = Q/\piR
The charge on the slice dq = \lambdaRd\theta

The field generated by the slice would be dE = k dq/R^2 = k \lambda/R d\theta

Components of dE would be: dEx = dEcos\theta, dEy = -dEsin\theta

Add them all up you get:
Ex = k\lambda/R \int from 0 to pi of cos\theta d\theta = k\lambda/R sin\theta from 0 to pi which = 0

Ey = -k\lambda/R \int from 0 to pi of sin\theta d\theta = k\lambda/R cos\theta from 0 to pi which = -2k\lambda/R
 
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