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Determine the magnitude of the electric field

  1. Aug 26, 2007 #1
    1. The problem statement, all variables and given/known data

    P and Q are points within a uniform electric field that are separated by a distance of 0.1 meters as shown. The potential difference between P and Q is 50 V. Determine the magnitude fo this electric field.


    2. Relevant equations

    E=F/q
    E=(k)(q)/r^2


    3. The attempt at a solution

    This problem doesn't fit the pattern of the other "electric field" problems that I have worked. The formulas above are for electric field, and electric field of a point charge; neither one fits this application because I'm not dealing with point charges so there is no value for "q" and I'm not given enough info to find "F". I'm told that the 50 V is the potential difference between P and Q but how does that info fit into my problem?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Aug 26, 2007 #2
    What's the relationship between force and work?
     
  4. Aug 26, 2007 #3
    Thanks for the quick response!

    1. Work = (F)(d); but how does this help? Should I consider the consider the 50 V between points P and Q as the force?

    2. When I do so, I come up with:
    E = (8.99 x 10^9)(50 x 10^-6)/0.1^2

    When I plug that into my calculator I get 44950000 or 4.495 x 10^7. My soloution sheet says the answer should be 500. Neither of the numbers I got should be rounded up that far. Am I missing something?
     
  5. Aug 26, 2007 #4
    yep, you will find that out.

    So, now you know what's F equal to, and d is given.

    and assuming that initially the energy between two charges was 0, what would the work equal to?
     
  6. Aug 26, 2007 #5
    I'm sorry,...but I'm not following. Isn't the work simply equal to 50 V x .1 meters?
     
  7. Aug 26, 2007 #6
    no, rather think about potential energy, and the way it is related to the work,
    and then find a relationship between potential energy and voltage
    and you just found the relationship between Electric field and force,and
    between force and the work.
     
  8. Aug 26, 2007 #7
    Okay, I got it!...my mistake was that I was trying to use E=kq/r^2. When I plugged the numbers into E=F/q it came out correct. Big thanks to rootX for the help!
     
  9. Aug 26, 2007 #8
    so, here's the derivation in proper way:
    U=qV
    F=qE
    Fd=U

    hence V=Ed
     
  10. Aug 26, 2007 #9
    I spoke to soon...I didn't come out correctly my way! Hope you can bear with my so that I can see if I follow your last post
     
  11. Aug 26, 2007 #10
    yep, i knew
     
  12. Aug 26, 2007 #11
    What does "U" stand for in your equations?
     
  13. Aug 26, 2007 #12
    potential energy

    as work = potential energy-0
     
  14. Aug 26, 2007 #13
    I would have never been able to come up with the derivations that you listed! I didn't know that work is equal to potential energy -0; or that Fd = U. F = qE is in my textbook so I was aware of that one. I'm worried because I don't understand how you came up with those derivations.
     
  15. Aug 26, 2007 #14
    I assumed that intially the points are at infinite distance from each other. (a common assumption)
    and, then their potential energy changed to U (some value as distance between them is finite).

    and work = change in potential energy

    and as work = Fd

    so U = Fd
    U = q* voltage
    and qV=qEd
     
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