Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1626108 • Letter: C
Question
Coulomb's law for the magnitude of the force Fbetween two particles with charges Q and Qseparated by a distance d is
|F|=K|QQ|d2, where K=140, and 0=8.854×1012C2/(Nm2) is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q1 = -11.5 nC , is located at x1 = -1.720 m ; the second charge, q2 = 35.0 nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2at x3 = -1.060 m ?
Your answer may be positive or negative, depending on the direction of the force.
Express your answer numerically in newtons to three significant figures.
Explanation / Answer
q1 = -11.5 * 10-9 C at x1 = -1.720m
q2 = 35.0* 10-9 C at x2 = 0 m
k = 1/ 4??0 = 8.99 * 109 N.m2 / C2
q3 = 55.0* 10-9 C at x3 = -1.060m
The net force exerted by these two charges on a third charge q3 = 55. 0nC placed between q1and q2 at x3 = -1.060m is,
F = - k q3 q1 / r132 - k q3 q2 / r232
F = - 8.99 * 109 * 55.0* 10-9 [11.5 * 10-9 / (-1.060+ 1.720)2 + 35.0 * 10-9 / (0 + 1.060)2 ]
F = - 2.845571* 10-5 N
This force is directed along the negative x axis and hence the negative sign.