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Basic electric force/field question I can't get

  1. May 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle has a charge of +1.7 μC and moves from point A to point B, a distance of 0.26 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA - EPEB = +8.1 x 10-4 J. (a) Find the magnitude of the electric force that acts on the particle. (b) Find the magnitude of the electric field that the particle experiences.

    2. Relevant equations
    F= qV/d

    3. The attempt at a solution
    What's wrong with this reasoning:
    F = V/d and V=Ed so E=F/q=V/d
    Therefore F=qV/d
    substitute all those values into that formula to get the electrostatic force and it's wrong?

    I also tried b) using E=kq/d^2 and that didn't work either! (I got 226331.3609 as the answer)

    By the way, there are no rounding requirements or anything, only a 2% margin of error so it's not that.

    Thanks a lot for your help in advance!
     
  2. jcsd
  3. May 14, 2016 #2

    blue_leaf77

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    What you are given is ##\Delta E_p## which is equal to ##q\Delta V##. By the way, what's the correct answer?
     
  4. May 14, 2016 #3
    Right so I tried going:
    deltaE/q = deltaV
    Then using F=(q*deltaV)/d to get the electrostatic force and I got the wrong answer again
     
  5. May 14, 2016 #4

    blue_leaf77

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    What's the correct answer according to your source?
     
  6. May 14, 2016 #5

    I don't know sorry
     
  7. May 14, 2016 #6

    blue_leaf77

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    Then how did you know that your answer was wrong? Is it a homework given to you?
     
  8. May 14, 2016 #7

    It's this homework website and you get 5 attempts at getting the correct answer, it tells you when it is wrong but doesn't give you hints or the correct answer.
     
  9. May 14, 2016 #8
    Could someone please help with this question? Really struggling
     
  10. May 14, 2016 #9

    blue_leaf77

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    Can you post your own answer?
     
  11. May 14, 2016 #10

    I did it's in the OP. Do you know how to solve the question?
     
  12. May 14, 2016 #11

    Charles Link

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    They give you the change in potential energy ## \Delta W=F \cdot d=q V ##. The change in voltage is often also referred to as the change in potential, but here they were quite specific that the particle experienced a change in its "potential energy", so that ## \Delta W=q V ##. (using ## W ## here for energy to save the letter ## E ## for the electric field. Notice also that ## V ## stands for voltage in electrical problems, and only becomes energy when multiplied by the charge ## q ##.)
     
    Last edited: May 14, 2016
  13. May 14, 2016 #12

    NascentOxygen

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    You start with F=V/d and show that F=qV/d? Something is wrong there. The units on each side of F=V/d don't match, so it can't be right.

    What value did you use for V?
     
  14. May 14, 2016 #13

    blue_leaf77

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    You mean this: 226331.3609 ? Isn't that your answer for (b). I asked for your answer for part (a).
     
  15. May 14, 2016 #14


    Thanks so much for this, I got it now.
     
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