richyw 3 years ago 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.

1. richyw

I don't understand why this is zero. why is it zero?

2. JamesWolf

|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

3. 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?

4. 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

5. TuringTest

slight typo$=\frac q{2\pi\epsilon_0\sqrt{1+\frac14}}=\frac q{\pi\epsilon_0\sqrt5}$

6. Algebraic!

@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)

7. 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.

8. 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.