Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1525183 • 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 = -15.5 nC , is located at x1 = -1.730 m ; the second charge, q2 = 36.5 nC , is at the origin (x=0.0000).
A. What is the net force exerted by these two charges on a third charge q3 = 48.5 nC placed between q1 and q2 at x3 = -1.085 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
Given
Charge are
q1 = -15.5 nC
q2 = 36.5 nC
q3 = 48.5 nC
The electric force between the charges q1 and q3 is
F1 = (k*q1*q3)/r^2
= ((9*10^9 Nm^2/C^2)(15.5*10^-9 C)(48.5*10^-9 C))/(1.730 m-1.085 m)^2
= 1.63*10^-5 N
The electric force between the charges q2 and q3 is
F2 = (k*q2*q3)/r^2
= ((9*10^9 Nm^2/C^2)(36.5*10^-9 C)(48.5*10^-9 C))/(1.085 m - 0)^2
= 1.35*10^-5 N
The net electric force is
F net = -(1.63*10^-5 N)-(1.35*10^-5 N)
= - 2.98*10^-5 N