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Homework on quiz i took alredy

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Charges +Q and –Q are arranged at the corners of a square as shown. When the electric field and the electric potential V are determined at P, the center of the square, we find that

    2. Relevant equations
    (+Q)---------------(-Q)


    -----------P----------- There is line crossging through P from top left +Q and to bottom right +Q and same for -Q's


    (-Q)---------------(+Q)

    3. The attempt at a solution
    1. E = 0 and V > 0
    2. E ≠ 0 and V < 0
    3. E= 0 and V= 0
    4. E ≠ 0 and V > 0
    5. None of these is correct. 0%
     
    Last edited: Apr 9, 2010
  2. jcsd
  3. Apr 10, 2010 #2

    kuruman

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    Homework Helper
    Gold Member

    If you took this quiz already as you seem to imply in the title, what answer did you give and what was your reasoning?
     
    Last edited: Apr 10, 2010
  4. Apr 10, 2010 #3
    Question seems to be chopped off?
     
  5. Apr 10, 2010 #4
    i choose none of these are correct because none of them seemed correct to me. notic the 0% next to the last option(none of these are correct)

    the question is there in full. it was a multiple choise question.
     
  6. Apr 10, 2010 #5
    That isn't a reason. That is circular reasoning.
     
  7. Apr 10, 2010 #6

    Borek

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    Staff: Mentor

    Let's start with the electric filed - how do you think, what is electric field in P?
     
  8. Apr 10, 2010 #7
    are the two fields canceled by the push and pull of the oppositely charged particles...leading to 0
     
  9. Apr 10, 2010 #8

    Borek

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    Staff: Mentor

    Right. It is even easier if you think in terms of the definition - force acting on the unit charge. If you place any charge in point P is it pulled in two opposite directions - and these pulls cancel out, and it is pushed in two opposite directions - once agin, forces cancels out.

    What about potential?
     
  10. Apr 10, 2010 #9
    Since potential is a scalar, you can just add up the value and account for sign. I can see that for every positive charge at a fixed distance there is a corresponding negative charge. This is why when I add it up, the net potential is zero.

    So E=0 and V=0????
     
  11. Apr 10, 2010 #10
    yes.
     
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