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Physics

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Two particles, one with negative charge -Q and the other with positive charge +2Q, are separated by a distance d. Both charges lie on the x-acis with the negative charge at x=0. At what point(s) along the line joining the particles is the potential equal to zero? At what point(s) along the line joining the particles is the electric field zero?

  • Physics - ,

    (a)
    “Zero” point is between Q1 and Q2 and separated by distance ‘x’ from Q1.
    φ=φ1+φ2= - k•Q1/x+k•Q2/(d-x) = 0,
    k•Q1/x= k•Q2/(d-x)
    Q/x=2Q/(d-x),
    x=d/3.
    (b) Assume that “zero” point is to the left from Q1 and separated by distance ‘x’ from Q1.
    E1 is directed to the right, E2 is directed to the left
    kQ1/ x²= kQ2/(d+x)²
    Q/x² = 2Q/(d+x)²
    x²-2dx- d²=0
    x= d±√(d²+d²) =
    = d±d√2.
    x1= d(1+1.41) = 2.41d (this is the sought value)
    x2 = d(1-1.41)= - 0.41d (extraneous root)

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