Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1593544 • 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 = -15.0 nC , is located at x1 = -1.685 m ; the second charge, q2 = 33.0 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 52.0 nC placed between q1and q2 at x3 = -1.250 m ?
Explanation / Answer
q1 = -15.0 nC @ x1 = -1.685 m
q2 = 33.0 nC @ x2 = 0
q3 = 52.0 nC @ x3 = - 1.250 m
Net Force Exerted on Charge q3 =
F = k*q1q3/r^2 + k*q2q3/r^2
F = (8.9*10^9 * 10^-9*10^-9) * [(15*52)/(1.685-1.250)^2 + (33.0*52.0)/1.250^2]
F = 4.65 * 10^-5 N
Net force exerted by these two charges on a third charge q3, F = 4.65 * 10^-5 N