posted by Michael on .
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?
“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,
(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
x= d±√(d²+d²) =
x1= d(1+1.41) = 2.41d (this is the sought value)
x2 = d(1-1.41)= - 0.41d (extraneous root)