Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1784933 • 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 = -12.5 nC , is located at x1 = -1.695 m ; the second charge, q2 = 33.5 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 = 47.0 nC placed between q1 and q2 at x3 = -1.235 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 q1 on q3:
The distance between q1 and q3 is -1.695m – (-1.235m) = -0.46
F1,3 = K * (12.5 nC * 47.0 nC) / (0.46^2)
Since the charges are in nanocoulombs:
F1,3 = K * (1.25 * 10^-8 * 4.70 * 10^-8) / (0.46^2)
F1,3 = 2.495 * 10^-5 N
Now do the same thing for the effect of charge 2 on charge 3:
The distance between charge 2 and 3 is 1.120m, so:
F2,3 = K * (33.5 nC * 47.0 nC) / (1.235^2)
F2,3 = 9.277 * 10^-6 N
So,
F = -9.277 * 10^-6 N – 2.495 * 10^-5 N
F = -3.42 * 10^-5 N