A 2.4 nC charge is at the origin and a -5.6 nC charge is at x = 1.0 cm .

Question

a)At what x-coordinate could you place a proton so that it would experience no net force?

b)Would the net force be zero for an electron placed at the same position?

Answer 1

F = k Q1 Q2/d^2 for each

we want x so F = 0
distances are x and from 2.4 and 1-x from -5.6

for F = 0
2.4/x^2 - 5.6/(1-x)^2 = 0
(NOTE - by inspection you can see that to get zero you must be left of the origin, negative x)

if you locate so F = 0, then the electric field from the two charges is zero there and the force will be zero on any old charge you put there. Does not matter proton or neutron or Uranium nucleus.

Answer 2

Hey, shouldn't it be (x+1)^2
lets say:
q1=2.4
q2=-5.6
q3=proton

q3--x---q1----d-----q2

d=1

Therefore:
F23=K|q2||q3|/(x+1)^2

Also after solving the quadratic I got
x=1.8956...cm
x=-0.395....cm
The question asks for x-cord, x is the distance from the origin. Which x do I choose and what is my final answer? Also why

Answer 3

Remember. The proton is being pushed by the plus charge and pull by the negative. If you put it between the two charges it would be draw right by the positive and pushed right by the negative. If we put it beyond the second charge it's pulled left by a much stronger negative charge and weakly affected right by the positive. The only place they can balance is left of the first charge. Weaker pushing left, stronger pulling right. ANd yes the distance should be x and x+1.