Two point charges are placed on the x-axis as follows: one positive charge, q_1 = 4.01 nC, is located to the right of the origin at x= 0.202 m, and a second positive charge, q_2 = 4.98 nC, is located to the left of the origin at x= -0.296 m.
What is the total force (magnitude and direction) exerted by these two charges on a negative point charge, q_3 = -5.98 nC, that is placed at the origin?
I think you are supposed to use couloumbs law to calculate the answer. I got 8.35 x 10^-6 C but its wrong for some reason. Please help
2007-02-03 20:40:08 · 2 answers · asked by wtfitsnguyen 2 in Science & Mathematics ➔ Physics
make equation
k*4..98*5.98/(.296)^2 + k * 4.02* 5.98/(.202)^2
2007-02-03 21:07:11 · answer #1 · answered by n nitant 3 · 0⤊ 0⤋
Subtract the magnitude of the two Fe's since the point charges pull the middle charge in opposite directions. And then get the direction of the force with higher value.
2007-02-03 22:20:56 · answer #2 · answered by crabmeat 2 · 0⤊ 0⤋