Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1524405 • Letter: C
Question
Coulomb's law for the magnitude of the force F between two particles with charges Q and Q separated 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 = -10.0 nC , is located at x1 = -1.650 m ; the second charge, q2 = 40.0 nC , is at the origin (x=0.0000).
Part A
What is the net force exerted by these two charges on a third charge q3 = 53.5 nC placed between q1and q2 at 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
for the charge placed in the middle
F3 = F13 + F23
F3 = 9 *10^9 * 53.3 *10^-9 * 10^-9 * (-10/(1.650 - 1.060)^2 - 40/1.060^2)
F3 = -3.086 *10^-5 N
the net force acting on the third charge is -3.086 *10^-5 N