Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1651256 • 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 = -16.0 nC , is located at x1 = -1.665 m ; the second charge, q2 = 34.5 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 45.5 nC placed between q1and q2 at x3 = -1.170 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 will attract q3 towards it.
F31 = - k q1 q3 / d13^2
= - (9 * 10^9) (16 * 10^-9) (45.5*10^-9) / (1.665- 1.170)^2
= - 2.67 * 10^-5 N
q2 will repel q3.
F23 = - (9 * 10^9) (34.5 * 10^-9)(45.5
*10^-9) / (1.170^2)
=- 1.03*10^-5 N
F = F31 + F23= 3.7*10^-5 N ............Ans