Electric potential at center of semicircle

  • #1
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Homework Statement


A total charge Q is uniformly distributed on a thread of length L. This thread forms a semicircle. what is the potential at the center? (assume V=0 at large distances)

Homework Equations



V = -∫Edl
E = kQ/r^2

The Attempt at a Solution


V = - ∫Edr = -E∫dr = -Er = - kQr/r^2 = -kQ/r
is this correct? I am not sure if the sign should be negative
 

Answers and Replies

  • #2
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E is a vector and it is not constant. Finding the electric field is more complicated, it is sufficient to calculate the potential here but that has a different formula.
Also, using r both for radius and for the length along the circle is confusing and leads to wrong results.
 
  • #3
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hmmm so would it be correct to say
V = ∫dV = k∫dQ/r = kQ/r because Q is constant?
 
  • #4
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The result is right but you'll have to justify the steps you used.
 
  • #5
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The result is right but you'll have to justify the steps you used.
Does this mean i have to plug something in for dQ?
 
  • #6
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11,510
The length should appear somewhere.

Oh, and I guess the final result should be expressed in terms of L instead of r as well I guess.
 
  • #7
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I think i can use arc length formula L = theta r to relate r and L. I am not sure how I can plug that in to the equation for V. I think this doesn't seem like it should be a tough problem but I must be missing something obvious =[
 
  • #8
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This is not a tough problem. Yes you can use the arc length formula.

If a charge Q is uniformly distributed over a length L, what is the charge per length?
 
  • #9
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I think it would simply be Q/L so dQ in terms of the charge density would be dQ = λdL = (Q/L)dL

V = k∫dQ/r = k∫(Q/L)dL/r = kQL/Lr = kQ/r ??
 
  • #10
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Looks fine. Now you just have to express r in terms of L.
The semicircle has a length of L.
 
  • #11
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so since L = theta r then r = L/theta so the final step would be
kQ/r = kQtheta/L
but since they dont give me theta can I get away with doing this?
 
  • #13
516
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assuming a semicircle is a half circle (forgive me if this is incorrect I am not 100 percent sure) then L would equal the perimeter or L = (pi/2)r^2
so r = sqrt(2L/pi) so it would be kQ/(sqrt(2L/pi))
 
  • #14
Nathanael
Homework Helper
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assuming a semicircle is a half circle (forgive me if this is incorrect I am not 100 percent sure)
Yes, semicircle just means half-circle

L = (pi/2)r^2
I think you're mixing up the area and circumference formulas. On the left side you have units of length but on the right side you have units of length squared.
 
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  • #15
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oops sorry L = 2(pi/2)r = pi r
so r = L/pi
so kQ/r = kQpi/L
 
  • #18
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i feel like i did something wrong. if my integral was k∫(Q/L)dL/r then wouldnt it be
k∫(Q/L)dL/r = kQ/r∫dL/L(because Q and r are constant) = kQ/r (ln(L)) ??
 
  • #19
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or If i put everything in terms of L instead of r from the begining:
k∫((Q/L)dL)/(L/pi) = k∫piQdL/L^2 = kpiQ∫dL/L^2 = - kpiQ/L
so it would be negative?
 
  • #20
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If you use L like this, you have to integrate over a different variable, like L' (or x, or whatever). The L in Q/L does not change in the integral.
 
  • #21
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the L in r = L/pi changes? How would I have to set that up for it to be correct?
 
  • #22
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No it does not change!
That's why it is better to use a different integration variable along the length of the semicircle, like x.
 

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