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Potential Difference from Electric field

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    A uniform electric field with a magnitude of 1100 N/C points in the positive x direction as shown above. A positive charge of +4.4 μC starts at point A, then moves to point B, then to point C and then back to point A.
    What is the difference in electrical potential between points B and C.
    [PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/Phys1202/fall/homework/Ch-20-Potential/V_from_E_field/E-field.jpg [Broken]


    2. Relevant equations
    E=-Δ/ΔS
    https://www.physicsforums.com/newthread.php?do=newthread&f=153 [Broken]

    3. The attempt at a solution
    Well, I used the Pythagorean theorem to find the 3rd side of triangle which was 10. Then I divided that by 100 to get it into meters. Then I multiplied it by the electric field magnitude (1100) to get -110 V. But I know that I have to multiply by cos of some angle. From this picture it looks like that the angle is 135 degrees. But when I multiply this by 110 i get 77.78 V. This answer is wrong. Please show me what I am doing wrong. I know this is a super ez problem.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 13, 2010 #2
    This is a super ez problem if you remember one thing: voltage is path independent! Remember that

    V=E∆Scos(Ø)
     
  4. Sep 13, 2010 #3

    rl.bhat

    User Avatar
    Homework Helper

    Hi hrs90, welcome to PF.

    Any line perpendicular to the uniform electric field is called equi potential line. There fore the potentials at A and C are the same.

    Hence electric potential difference VB - VA = E*AB.
     
  5. Sep 13, 2010 #4
    Thanks for the quick response, but I am still confused. I already got the potential difference for VB-VA which is -88V btw. Now in order to get the difference from C to B, don't I just multiply E by .10 and then multiply that by cos of whatever angle that is being made with the horizontal electric field line.
     
  6. Sep 13, 2010 #5
    What we are saying is that going from VB to VA is the same thing since as going from VB to VC. Not only is voltage path independent, so you could do VB to VA and then VA to VC. Additionally, you also know that from VA to VC is zero because they are at the same potential. Mathematically going from VA to VC gives a 90 degree angle so the cosine term will go to zero, which says that there is no potential contribution to go from VA to VC, VA to VC = 0.

    However, with all that being said, since the potential is path independent you will get the same answer by going along the diagonal.

    VBC=-10*cosØ where Ø=arctan(6/8)
     
    Last edited: Sep 13, 2010
  7. Sep 14, 2010 #6
    Hey thank you so much! Now I understand it.
     
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