Electric field at center of semi-circular bent rod of charge

  • Thread starter caesius
  • Start date
  • #1
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I have the correct answer just the wrong direction

Homework Statement


A uniformly charged insulating rod of length 14.0cm is bent into the shape of a semicircle as shown in figure XXX. The rod has a total charge of 7.5 micro C. Find the magnitude and direction of the electric field at P, the center of the semicircle.


Homework Equations



dE = k.dq.1/r^2


The Attempt at a Solution


I set things up like this
http://img210.imageshack.us/img210/4690/physprobub2.png [Broken]

Since the vertical components would cancel I just looked at the x components.

[tex]
dE_{x} = \frac{k_{e}\cdot dq\cdot\sin\theta}{r^2}
[/tex]

[tex]
= \frac{k_{e}\cdot dx\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dq = dx\cdot\lambda}
[/tex]

[tex]
= \frac{k_{e}\cdot r\cdot d\theta\cdot\lambda\cdot\sin\theta}{r^2} \ \ \mbox{as dx = ds = r\cdot d\theta}
[/tex]

[tex]
= \frac{k_{e}\cdot d\theta\cdot\lambda\cdot\sin\theta}{r} \ \ \mbox{as dx = ds = r\cdot d\theta}
[/tex]

So
[tex]
E = \frac{k_{e}\cdot \lambda}{r} \int d\theta\cdot\sin\theta}
[/tex]
(integral is from zero to pi)

[tex]
E = \frac{k_{e}\cdot \lambda}{r} ( -1 - 1)
[/tex]

[tex]
E = \frac{-2\cdot k_{e}\cdot Q}{L\cdot r} \ \ \mbox{as \lambda = \frac{Q}{L}}
[/tex]
When I substitute the values is I get the correct answer but it's negative. The bent rod is positive so the resultant electric field should be AWAY from the semicircle and therefor positve?!?

What have I done wrong?
 
Last edited by a moderator:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,260
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The integral of sin from 0 to pi is 2. How did you get (-1-1)? BTW the antiderivative of sin(x) is -cos(x).
 
  • #3
Sorry for reactivating this old thread, but I figured better that than starting a whole new thread.

I have gotten to the point of [tex]

E = \frac{2\cdot k_{e}\cdot Q}{L\cdot r} \

[/tex]

but now my question is what is the r value? Of what my reasoning is telling me, it should be [tex]

E = \frac{2\cdot k_{e}\cdot Q}{L^{2}} \
[/tex]

Is this right? Can any one assist me?
 

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