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Electric Potential Difference

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

    What is the difference in electric potential between a point midway between the charges and a point 0.05 meters away from the positive charge but along the line between the charges?

    2. Relevant equations

    UE = K * (Q1*Q2)/r

    3. The attempt at a solution

    K = 9.0x109Nm2/C2

    UE = (9.0x109Nm2/C2)(-2.0x10-6C)(Qcenter)/.1 + (9.0x109Nm2/C2)(2.0x10-6C)(Qmidpoint)/.1

    Am I doing this right? If not, how could I go about solving this?
     
  2. jcsd
  3. Jun 9, 2010 #2
    You used potential energy formula, you have to calculate potential...
    potential due to a charge is K*Q/r
     
    Last edited by a moderator: Jun 9, 2010
  4. Jun 9, 2010 #3
    We have two charges, and then one in the middle. Which one goes into Q?
     
  5. Jun 9, 2010 #4

    vela

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    You don't have a charge in the middle. You have a point in the middle, and at that point, there's an electric potential due to the two charges. Use the principle of superposition to find the potential at that point.
     
  6. Jun 9, 2010 #5
    u have two charges +2micro and -2micro charges,
    potential due to a charge is K*Q/r
    so potential due to +2micro is = K*(+2micro)/r (r is the distance from the charge here it is +2micro)
    similarly find potential for -2micro...
    but we have calculated potential of individual charges for any distance r,but here we have both the charge what we do..just add individual potential of the charges at that point where u want to find the potential for the 2 charge system...
     
  7. Jun 9, 2010 #6
    So, for the first part of the question where the point is in the middle, it would be 0, right? Since you have the same charges, opposite signs; if you add them up, you'll get 0.
     
  8. Jun 9, 2010 #7
    yup you r right..do the same for point at 0.05 meters away from +2micro but be carefull that while taking individual potential for -2micro in this case the distance is 0.1+0.05 from the -2micro charge
     
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