richyw
  • richyw
charged particles each of magnitude \(2.00 \mu C\) are located on the x axis, one is at x=1.00m, the other is at x=-1.00m. Determine the electric potential on the y axis at y=0.500m.
Physics
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SOLVED
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katieb
  • katieb
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richyw
  • richyw
I don't understand why this is zero. why is it zero?
anonymous
  • anonymous
|dw:1349753624257:dw| Im not sure either. I would expect it to be zero if this were the case|dw:1349753700882:dw| hope you find out
richyw
  • richyw
I'm pretty sure it's not zero and my book has the wrong answer. All I did was say \[V=\frac{1}{4\pi\epsilon_0}\sum^i_{i=1}\frac{q_i}{r_i}\]\[V=\frac{2q_1}{4\pi\epsilon_0r_1}\]Since \(r_1=r_2\) and \(q_1=q_2\) Is this correct?

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TuringTest
  • TuringTest
|dw:1349763953576:dw|horizontal components cancel leaving vertical\[\vec E=\frac{2q}{4\pi\epsilon_0r^2}\sin\theta\hat j=\frac{2qy}{4\pi\epsilon_0r^3}\hat j\]bringing a point charge along y from +infty\[V=\int_{\infty}^{1/2}Edy=\frac q{2\pi\epsilon_0}\int_\infty^{1/2}\frac y{(1+y^2)^{3/2}}dy\]\[=\frac q{2\pi\epsilon_0\sqrt{1+\frac14}}=\frac {q_0}{\pi\epsilon\sqrt5}\]a nice answer, but not zero I agree your book is wrong
TuringTest
  • TuringTest
slight typo\[=\frac q{2\pi\epsilon_0\sqrt{1+\frac14}}=\frac q{\pi\epsilon_0\sqrt5}\]
anonymous
  • anonymous
@turingtest That's the right answer. You don't need to go through all that however, it's simply 2kq/r where r= sqrt(5/4)
TuringTest
  • TuringTest
lol I know, I just wanted to be rigorous. I always feel better about disagreeing with the book when I can prove something from scratch.
richyw
  • richyw
thanks, you think paying $250 for a textbook and two solution manuals they could put some effort into explaining solutions (or at least give correct answers!) haha.

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