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Negative in a Electric Potential Energy/Kinetic Energy Question

  1. Jul 22, 2014 #1
    1. The problem statement, all variables and given/known data

    My questions reads:

    An electron is pulled away from a fixed charge of 1.3μC. The electron is moved from the positive charge to 4.0 cm away from the charge. If the electron is released from the 4.0 c mark, what is the max velocity of the electron?

    2. Relevant equations

    Ep = k q1q2/r - electric potential energy

    k = 9x10^9
    q1 = -1.6 x 10^-19 C
    q2 = 1.3 x 10^-6 C
    r = 4.0 cm = 0.04 m

    Ek = 1/2 mv² - kinetic energy

    m = 9.11 x 10^-31 kg
    v = ?

    3. The attempt at a solution

    Ep = Ek because all the electric potential energy is converted to kinetic energy when velocity is at its maximum.

    kq1q2/r = 1/2 mv²

    Since q1 is the charge of the electron, it is negative, so the left hand side of the equation is negative. That makes taking the square root impossible. But if I ignore the negative, I get the right answer in the text.

    Does that mean the sign of the Ep is not important, or that the magnitude (the absolute value) is all that matters?

    Or is it because the first equation should be the net energy =0 and the Ep gets moved algebraically and the negative disappears?

    The answer I get when I ignore the negative, and the answer given in the text is 3.2 x 10^8 m/s

    This is faster than the speed of light, which also confuses me a bit. Maybe just not a well written question?
  2. jcsd
  3. Jul 22, 2014 #2
    This is a case where the distinction between the electric potential function and electric potential energy is important. Many different functions can be used as the electric potential function for the same system, since it is merely a type of antiderivative of the electric field. It is because of this status that it can be used to evaluate the work integral that tells us the total change in energy of the electron when moved between two points in the electric field (the fundamental theorem of calculus). This change in electric potential energy is the quantity you are equating to change in kinetic energy of the electron.
    The change in potential energy is positive if we pull the electron away from the proton, and therefore negative if the electron moves in the opposite direction. When the electron is released, since there are no other forces acting on it, the Work-Energy theorem allows us to equate the change in electric potential energy with its change in kinetic energy. But remember it is change that is being equated, not particular values of potential energy or kinetic energy. Therefore, even if you got a negative change in potential energy, this would correspond to a decrease in kinetic energy, not an equation between a particular kinetic energy and a negative value. In symbols, then, your negative change in electric potential is equated to [tex]\frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2 = 0 - \frac{1}{2}mv_f^2[/tex] since the electron's initial velocity was 0. This should alleviate your worries about signage.
    As for the problem with the speed of light, it is indeed not a prudent type of question, unless the author is going to use it as an example later when they mention special relativity.
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