Two point charges are located on the x axis: one charge, q1 = -11.0 nC, is located at -1.715m ; the second charge, q2 = 34.5 nC , is at the origin (x=0). What is the net force exerted by these two charges on a third charge q3= 52.0 nC placed between q1 and q2 at x3 = -1.145 ? Use 8.85×10−12 for the permittivity of free space.
2007-01-17 18:56:12 · 2 answers · asked by fertig 1 in Science & Mathematics ➔ Physics
By Coulombs law, the force exerted by one charge q1 on another q2 is F=q1*q2 / (4πe0d^2) where d is the distance between the charges.
The force from the charge at -1.715m is
(11.0*10^-9)*(52.0*10^-9) divided by
4πe0(1.715-1.145)^2
The force from the charge at the origin is
(34.5*10^-9)*(52.0*10^-9) divided by
4πe0(1.145)^2
The forces are both in the same direction--the force is attractive to the negative charge on the left, and repulsive from the positive charge on the right. Therefore, the total force is the sum of the two, directed in the negative x direction.
Force from q1 = -1.583*10^-5 newton
Force from q2 = -1.23*10^-5 newton
Total force = -2.813*10^-5 newton
2007-01-17 19:31:23 · answer #1 · answered by gp4rts 7 · 1⤊ 0⤋
What is the net force exerted by these two charges on a third charge q3 = 52.0nC placed between q1 and q2 at x3 = -1.145m ?
Your answer may be positive or negative, depending on the direction of the force.
2013-12-08 15:15:37 · answer #2 · answered by Noor 1 · 0⤊ 0⤋