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Electric Field at the Origin

  • Thread starter amb0027
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
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Wait? The charge Q is distributed over a semicircle? I'm guessing you're gonna have to use Gauss' law. But could you make the problem statement clearer? Where's the origin?
 
  • #3
11
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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 : [tex]\lambda[/tex] = Q/[tex]\pi[/tex]R
The charge on the slice dq = [tex]\lambda[/tex]Rd[tex]\theta[/tex]

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

Components of dE would be: dEx = dEcos[tex]\theta[/tex], dEy = -dEsin[tex]\theta[/tex]

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

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

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