- #1
thursdaytbs
- 53
- 0
So what I did was:+6.8microC -1.7microC
(q1) ------ 1m ------ (q2)
"Two charges are separated by 1.0 m as shown. q1 has a positive charge of 6.8micro-C, and q2 has a negative charge of 1.7micro-C. At what position is the net electric field that is produced by both q1 and q2 equal to zero?
If a +5.0 micro-C charge is placed at this position, what force would act on it as a result of q1 and q2?
Kq1/(1+r) + Kq2/r = 0, making r =the distance away from q2 outside of the system to be the equilibrium point. so q1 is 1m + r distance away from the point, and q2 = r distance away from the point.
K's cancel out, you get:
q1/(1+r) + q2/r = 0
Some algebra, turns out r = -0.2m.
Does that mean the point is 0.2m inwards from q2->q1 instead of outside of the system?
aka: (q1) -----0.8m----(qe)--0.2--(q2), where qe = equilibrium point?
Also, arent there two points which are in equliibrium of the system? We did a problem in class similar to this and had 2 pointsat equlibrium. Any help is greatly appreciated.
Last edited: