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Find the speed of the charge.

  1. Sep 11, 2015 #1
    1. The problem statement, all variables and given/known data

    Positive point charge (Q = XXX μC, mass m = XXX g) is fixed at point (Y2 cm,0). A second identical charge q is constrained to slide on a friction-less wire along the y-axis. Assume: the only force on q is the electrostatic force. If q starts at (0, YYY) and is released from rest, find its speed when it reaches (0,YYY cm), in m/s.

    2. Relevant equations

    W = ΔV * q

    ΔV = -kQ (1/r2 - 1/r1) ----( this equation was already derived from integrating with limits A to B - ∫ E * dr )

    W = ΔK

    3. The attempt at a solution


    I understand, sort of, that I need to find energy needed to move the charge for that distance (y2 - y1). In order to get that, I need to find ΔV so that I could multiply the answer to charge q and make it equal to [(1/2) (m*Vf^2)] to get the Vf.

    I used ΔV = -kQ (1/r2 - 1/r1)
    and I got ΔV = 104500. 5716 V. Used [W = ΔV * q] to get W. Then set W = 1/2 mvf^2. Found vf = 2.647 m/s

    Is this an OK logic to solve this problem? :/
    I need an advice regarding the formulas I am using... not the correct answer...

    Thank you.
     
  2. jcsd
  3. Sep 11, 2015 #2

    TSny

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    You have the correct approach to the problem.

    However, you might have a couple of sign errors. If W represents the work done on the charge q by the electric force, then check your notes to see if W = +ΔVq or W = -ΔVq.

    Also, check if ΔV = +kQ(1/r2 - 1/r1) or ΔV = -kQ(1/r2 - 1/r1).
     
  4. Sep 12, 2015 #3
    I believe it's with the "-" sign in front:
    $$ (\frac {qQ} {4 π ε_o}) \int_{a}^{b} \frac {1} {r^2} dx = (\frac {qQ} { 4πε_o}) ((\frac {-1} { b}) - (\frac {-1} {a})) = - (\frac {qQ} { 4πε_o}) [(\frac {1} { b}) - (\frac {1} {a})] $$

    The work done is by the charge that has a fixed position on the x - axis, I am saying?

    Also, the integration that I just did gives me ## \ W ## ... but why do I have both ## q ## and ## Q ## involved in the equation?

    My professor gave me an equation similar to the one that I have derived, but it only includes Q as a reference charge (Electric field created by point charge Q) and gives off almost exact result. To be precise it follows like this:

    $$ \Delta V = V_b - V_a = - \int_{a}^{b} E \cdot dr = - (\frac {Q} { 4πε_o}) [(\frac {1} { b}) - (\frac {1} {a})] $$

    So I am predicting, to find the ## W ## done by ## Q ## on ## q ## I have to use the one I derived (from Coulomb's Law) and include both ## Q ## and ## q ##. Whereas to find the ## \Delta V ## I need to use the second equation that my professor provided? Well, in fact, I can use the second equation and still find the ## W ## by multiplying the result to the value of charge ## q ## (the charge that Work is being done upon)

    So, to conclude, one finds ## W ## done by Q on q and another finds ## \Delta V ## in the case of charge Q creating an electric field which is not constant.

    Hoping that my thoughts do not have flaws...

    P.S. Since ## a ## is less than ## b ## , the fraction will work out to be ## negative ## there fore the ## W ## done will be ## positive ## if that's what you meant to emphasize on. Since the charges would naturally move away from each other, I believe ## W ## must be positive in this case.
     
    Last edited: Sep 12, 2015
  5. Sep 12, 2015 #4

    TSny

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    Yes, that all looks good! So, for ##b>a## the electric force does a positive amount of work on q as q moves from ##r = a## to ##r = b##.

    Because the electric force is proportional to ##Qq## according to Coulomb's law.

    The integration of ##\frac{1}{r^2}## here will produce a minus sign that cancels the minus sign in front of the integral. So, the result should not have the minus sign out front. Thus ##\Delta V = V_b - V_a = - \int_{a}^{b} E \cdot dr = +(\frac {Q} { 4πε_o}) [(\frac {1} { b}) - (\frac {1} {a})]##. If you think about it, this makes sense. Since ##b > a##, this gives ##\Delta V < 0##. But that's what you expect as you move farther away from Q.

    Yes, but to get the signs to work out you need to use the correct relation ##W = -q \Delta V## rather than ##W = +q \Delta V##. The general relation between the work done by a conservative force and the change in potential energy is ##W = - \Delta U##, with a minus sign. For electrostatics, ##U = qV##, so ##W = - q\Delta V##

    Yes, ##W## should be positive in this problem. This comes about because ##W = -q \Delta V## while ##\Delta V < 0##.
     
  6. Sep 12, 2015 #5
    Yep. It's all clear. Thank you :)
     
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