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Homework Help: Finding the electric field at the origin

  1. May 18, 2010 #1
    I'm reviewing for my physics final and going over all of my old exams. I'm trying to figure out how to do this question that I got wrong.

    1. The problem statement, all variables and given/known data
    Six charges that are equal in magnitude, 0.48µC are arranged in a hexagon where each charge is r = 1.4 m from the center. Two of the charges are positive, and four charges are negative.

    (i) What is the electric field at the origin?

    (ii) What is the electric potential at the origin?

    2. Relevant equations

    E = kq/r2
    V(Electric Potential) = kq/r

    3. The attempt at a solution


    In (i) i'm not sure what I'm going wrong, although the answer does seem pretty huge but I'm doing it just like the examples in the book.

    In (ii) I'm pretty sure I forgot the equation on the exam which should be V = kq/r.

    Again, if I do this I get a large answer 4(9.0x10^9)(.48µC)/1.4m = 12,342.
  2. jcsd
  3. May 18, 2010 #2
    Think about what the E field looks like for each of the charges seperately, and which way E points at the origin for each charge seperately, you'll find out that some components cancel so you can't just add the magnitudes of E for every charge. What you have calculated is the E field 1.4m from a spherical charge of 4*.48 micro C. Remember E is a vector field. This should also help you with the second part of the question.
  4. May 18, 2010 #3
    Alright, well what I know is E = kq/r2 for a single point charge, but the left and right most charges cancel each other out.

    So I'm left with the 4 other point charges. Now you're saying I need to find each component of these 4? Well just by looking at the picture I can see that the X components will cancel each other out for these 4 leaving only the y components.

    E = Ey = kq/r2 * sin(60) for one of the point charges.

    = (9x10^9)(.48µC)/1.42 * (sin60) = 1908 N/C. This is only for 1 charge. So the y components for all 4 of the remaining charges is 4(1908) = 7632 N/C.

    Is this correct for part (i)?
    Last edited: May 18, 2010
  5. May 18, 2010 #4
    Yes that's correct.

    I actually misinterpreted part 2. You wrote the correct equation there V = kq/r. Do you know where to go from there?
    Last edited: May 18, 2010
  6. May 18, 2010 #5
    Well, by looking at some examples it looks like nothing cancels for V. Maybe it isn't a vector quantity? I could be wrong but that would make the answer 6kq/r = 1.85 x 10^4
  7. May 18, 2010 #6
    I don't know that your numerical answer is correct but you have the right idea. Precisely because the electric potential is a scalar quantity as you said, you can find the potential due to each charge at the origin and then just add them all up. Of course when you calculate this you have to take the sign of each charge into account. So V=6kq/r isn't quite correct since the charges don't all have the same sign. On the other hand, the electric field is a vector field so unlike the potential you have to calculate the E field at a point by considering the E vectors for each charge, just like you did in part 1.
  8. May 18, 2010 #7
    Ok so V = kq/r + kq/r - kq/r - kq/r - kq/r - kq/r
    V = -2kq/r ?
  9. May 18, 2010 #8
    Yep, looks good.
  10. May 18, 2010 #9
    ok, thanks a lot.
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