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Potential Differences in a Uniform Electric Field

  1. Jan 1, 2007 #1
    1. The problem statement, all variables and given/known data

    A 4.00 kg block carrying a charge Q = 50.o micro-C is connected to a spring for which k = 100 N/m. The block lies on a frictionless horizontal track, and the system is immersed in a uniform electric field of magnitude E = 5.00 X 10^5 V/m directed as shown in Figure P25.11. (not shown here, but the block is attached to a spring on it's left and the electric field is pointing to the right). If the block is released from rest when the spring is unstretched (at x = 0), (a) by what maximum amount does the spring expand?

    2. Relevant equations

    [tex]\Delta U = q_0 \Delta V = -q_0 E \cdot x[/tex]
    [tex]U_{spring} = \frac{1}{2}k x^2[/tex]

    3. The attempt at a solution

    [tex]\frac{1}{2}k x^2 = Q E x[/tex]
    [tex] x = \frac {2 Q E}{k}[/tex]

    and that provides the correct solution but why does this work? I realize I'm setting the potentials as equals, but how can I do this?

    We have (a) the change in potential energy from the field and (b) the change in potential energy from the spring. Since the spring is at rest at first, it's total Energy is 0, but when it reaches it's maximum, it has only potential energy. How can we equate this?

    Or, can we say that it does have some energy at first, the potential energy from the electric field?

    [tex]E_i = KE_{spring} + PE_{spring} + PE_{E field} = 0 + 0 + PE_{E field}[/tex]

    and then later we have:

    [tex] E_f = KE_{spring} + PE_{spring} + PE_{E field} = 0 + PE_{spring} + 0[/tex]

    then by the law of conservation of energy, we may equate these? So,

    [tex]E_i = Q E x = \frac{1}{2} k x^2 = E_f[/tex]

    ???

    I made a B last semester; thanks everyone for your help!! :-)
     
  2. jcsd
  3. Jan 1, 2007 #2
    or, should I say that
    [tex]W_{spring} + W_{E field} = 0[/tex]
    [tex]\frac{1}{2} k x^2 - Q E x = 0[/tex]

    when the total work is 0?
     
  4. Jan 1, 2007 #3
    (d) Repeat part (a) if the coefficient of kinetic friction between block and surface is 0.200.

    So I tried [tex]E_f - E_i = -f_k d[/tex]
    [tex]-E Q x - \frac{1}{2} k x^2 = - \mu m g x[/tex]

    but in this case the value for x would be non-real; twiddling with the signs failed to produce the correct answer. What am I not understanding correctly?
     
  5. Jan 1, 2007 #4
    This seemed to do it:
    [tex]E_f + E_i = -f_k x[/tex]
    where [tex]E_f = \frac{1}{2} k x^2[/tex] at the strongest point it's energy is the potential of the spring
    and [tex]E_i = - E Q x[/tex] at x = 0, the energy is the potential of the electric field on the charge

    It seems odd that E_i is negative.
     
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