Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 584671 • 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 = -17.0 nC , is located at x1 = -1.660 m ; the second charge, q2 = 34.0 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 46.0 nC placed between q1 and q2 at x3 = -1.210 m ?
Explanation / Answer
q1 = -17.0 nC, @ x1 = -1.660 m
q2 = 34.0 nC , @ x2 = 0.0
q3 = 46.0nC @ x3 = -1.210 m
Net Force, F = q*E
Electric Field at that point due to two charges =
E = k*q1/r1^2 + k*q2/r2^2
E = 8.9*17/(1.66 - 1.21)^2 + (8.9*34.0)/1.21^2
E = 953.84 N/C
F = 46.0 * 10^-9 * 953.84 N
F = 4.39 * 10^-5 N
Net Force, F = 4.39 * 10^-5 N