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Electric Field and minimum velocity

  1. Jul 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Electric field given by the vector E = E0/l (xi + yj) N/C is present in the xy plane. A small ring of mass M carrying charge +Q, which can slide freely on a smooth non-conducting rod, is projected along the rod from the point (0,l) such that it can reach the other end of the rod at (l,0) . Assuming there is no gravity in the region, what minimum velocity should be given to the ring, if QE0l/M = 8


    2. Relevant equations


    3. The attempt at a solution
    1/2 mv^2 = QΔV
    dV = E.ds (ds is the displacement)
    dV ={ E0/l (xi + yj) } {dx i + dy j}
    = E0/l (xdx + ydy)
    V = E0/l ∫xdx + ∫ydy ( ∫xdx from 0 to l, ∫ydy from l to 0)
    = 0

    Hence velocity required is zero.
    Which is terribly wrong.
     
  2. jcsd
  3. Jul 10, 2015 #2

    Nathanael

    User Avatar
    Homework Helper

    Look (visually) at the field xi+yj maybe you will get some insight as to why the answer came out to be zero.

    Edit:
    What is the minimum velocity needed to get over a hill?
    By analogy, your method is to integrate the force of gravity (dot ds) over the entire hill.
    But of course, this yields an answer of zero (if the hill is the same height on both sides).
     
    Last edited: Jul 10, 2015
  4. Jul 10, 2015 #3
    Got this one, just had to integrate till the point of highest potential, i.e. (l/2, l/2).
    Thanks :)
     
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