Question

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?

Answer 1

(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)