Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1423158 • 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 = -16.5 nC , is located at x1 = -1.650 m ;the second charge, q2 = 37.5 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 53.0 nC placed between q1 and q2 at x3 = -1.110 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
Force on q3 due to q1=kq1q2/r2
r=(1.65-1.11)m=0.54m
F1=Force on q3 due to q1=(9*109 *16.5*10-9 *53*10-9)/0.542 =2.699*10-5N
F2=Force on q3 due to q2=(9*109 *37.5*10-9 *53*10-9)/(1.11)2=1.45*10-5N
Both force acts in same direction =F1+F2=2.699*10-5 +1.45*10-5=4.15*10-5N along negative x axis
Net Force= -4.15*10-5N