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Electric field and electric potential problem

  1. Jan 5, 2015 #1
    The question is from AP Physics C Barron's

    Consider the potential-position function shown below(I put it on the attached files)( the potential asymptotically approaches zero as x goes to + infinity or -infinity). The upper limit of the speed that an electron can have as it passes the origin and still remain bound close to the origin is

    A. V=√2q(Vmax)/m
    B. V=0
    C. V=√-2q(Vmax)/m
    D. V=2q(Vmax)/m
    E. none of the above





    and here is the answer-----------------
    Because the charge is negative(electron), the force points away from the origin( rather than toward the origin, as for a positive charge). Therefore. the electron never remains bound close to the origin, but rather always experiences a force away from the origin.



    So my confusion is why it says"the force points away from the origin" here, i think according to the formula E= - dV/dx, the B point and D point might the electron been pushed away, but at point C and E, it would be the opposite, since the slope at those two points is positive, the field would be negative, so for electron, wouldn't the force be points toward the origin at point C and E?
     

    Attached Files:

  2. jcsd
  3. Jan 5, 2015 #2
    and by the way, although i found out that there are a lot surface integrals or line integrals in electromagne part, but i think in AP Physics level, it's still using single variable calculus instead of multi-v calculus. Am i right?
     
  4. Jan 5, 2015 #3
    You're right about the force being away from the origin from points B and D, but it's also away from the origin for the right half of the graph as well. Think about it -- the slope is opposite, yes, but it's also on the other side of the origin. Try working out the sign of the electric field and then the sign of the resulting force on an electron.

    And yes, in AP Physics all the calculus is in one variable. Technically you're doing line and surface integrals, as well as double and triple integrals, but the course uses symmetry to reduce these multivariable problems down to one variable.
     
  5. Jan 7, 2015 #4
    Thank you very much for your help, i really appreciate it:w
     
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