Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1525215 • 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 = -16.0 nC , is located at x1 = -1.715 m ; the second charge, q2 = 34.0 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 = 55.0 nC placed between q1 and q2 at x3 = -1.095 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 x 10^9) (16 x 10^-9) ( 55 x 10^-9) / (1.715 - 1.095)^2
= - 2.06 x 10^-5 N
q2 will repel q3.
F23 = - (9 x 10^9) (34 x 10^-9)(55 x 10^-9) / (1.095^2)
=- 1.44 x 10^-5 N
F = F31 + F32 = 3.50 x 10^-5 N ............Ans