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Homework Help: Electric Potential and Kinetic Energy

  1. Jul 25, 2015 #1
    1. The problem statement, all variables and given/known data
    A charged particle (q = -8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is released from rest at point A. At point B the kinetic energy of the particle is equal to 4.8 J. What is the electric potential difference VB-VA?

    2. Relevant equations
    EPE = Delta V (volts) = Delta U (potential energy) / q0

    3. The attempt at a solution
    Using this equation, I came up with the solution of -.6 kV, however, the answer is positive. Why is this?
  2. jcsd
  3. Jul 25, 2015 #2
    Which point do you think is at a higher potential ?
  4. Jul 25, 2015 #3
    Ahhh. Point VB has higher potential energy, therefore, in the equation VB - VA, it is positive. Thank you for your response.

    I also have another question.

    A particle (charge = +2.0 mC) moving in a region where only electric forces act on it has a kinetic energy of 5.0 J at point A. The particle subsequently passes through point B which has an electric potential of +1.5 kV relative to point A. Determine the kinetic energy of the particle as it moves through point B.

    I'll be honest with you, I do not know where to start with this problem. I know my equations however, I cannot picture the concept of potential energy caused by an electric field in relative to points A and B. If someone could clarify I would greatly appreciate it. Thank you.
  5. Jul 25, 2015 #4
    Simply , q*ΔV = ΔKE .
  6. Jul 25, 2015 #5
    Thanks again Qwerty. I must have missed out on that equation when I was copying my notes.
  7. Jul 25, 2015 #6
    1. The problem statement, all variables and given/known data
    Identical point charges (+50 μC) are placed at the corners of a square with sides of 2.0-m length. How much external energy is required to bring a fifth identical charge from infinity to the geometric center of the square?

    2. Relevant equations
    Electric Potential for Multiple Charges

    ΔU = K∑(Qi / Ri)
    K(50x10^-6 C) / 2cos(45°) yields 3.17 x 10^5 V.

    3. The attempt at a solution
    Each point charge by itself has an electric potential of 3.17 x 10^5 V (V is the correct unit, right?).

    However, I'm sure I'm doing this wrong because my answer choices are in a completely different magnitude (10^1).

    Thanks for the help.
  8. Jul 25, 2015 #7
    ΔU = K∑(Qi*Q/Ri) .

    Also , I think you haven't inputted Ri correctly either .
  9. Jul 25, 2015 #8
    What should the equation be then for all 4 charges?

    K(Q1 + Q2 + Q3 + Q4) / 2cos(45)?

    Also, what would Ri be then?
  10. Jul 25, 2015 #9
    Ri equals distance between any charge and the new charge ( Use diagram ) .

    KQ(Q1 + Q2 + .. ) .
  11. Jul 25, 2015 #10


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    Are you sure? sqrt(2) looks right to me.
    TAMUwbEE, I'm not sure whether you have understood Qwertywerty's other point. You calculated 3.17 x 10^5 V (as the potential due to one existing charge). You are asked for an energy, not a potential. That explains the orders of magnitude discrepancy.
  12. Jul 25, 2015 #11
    Then that's my mistake .
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