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Homework Help: Work done by an external force to move a charged particle

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data

    A particle of (positive) charge Q is assumed to have a fixed position at P. A second particle of mass m and (negative) charge -q moves at constant speed in a circle of radius r1, centered at P. Derive an expression for the work W that must be done by an external agent on the second particle in order to increase the radius of the circle of motion, centered at P, to r2.

    2. Relevant equations

    W=∆U+∆K where U is the potential energy and K the kinetic energy.
    [itex]F_{coulomb}=\frac{1}{4πε_{0}} \frac{Q(-q)}{r^{2}}[/itex]

    3. The attempt at a solution
    From the fact that [itex]F_{coulomb}=F_{centrifugal}[/itex] I know that v changes as r changes. So I get the two values of v at r1 and r2.

    Then I simply wrote W=∆K+∆U explicitly but my result is 3 times the book one. What may be wrong?
  2. jcsd
  3. Dec 3, 2011 #2


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    Check your signs. It is the centripetal force which is equal to the Coulomb force, both pointing inward, towards the centre.

  4. Dec 3, 2011 #3
    Thanks for the answer, but there's something I can't get from your explanation.
    If both forces pointed towards the centre, there would be a resultant inward force on the particle that make it accelerate along the radial direction, while it is actually still with respect to that direction.
  5. Dec 3, 2011 #4


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    The centripetal force does not "act". Performing circular motion, the particle accelerates towards the centre, this is the centripetal acceleration. The force needed to it is the centripetal force. The attractive force between the charges, the Coulomb force supplies the force needed for the circular motion. So you need to write that the centripetal force needed=Coulomb force (acting)


    The centripetal force does not "act". It is not one force among other forces. It is the result of the real forces, which is needed to move a particle of mass m and speed v along a circle of radius r.

    Last edited: Dec 3, 2011
  6. Dec 3, 2011 #5
    Sure! That's the point. Thanks a lot!
  7. Dec 3, 2011 #6


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