Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 2002232 • 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 = -17.5 nC , is located at x1 = -1.705 m ; the second charge, q2 = 37.0 nC , is at the origin (x=0.0000)
What is the net force exerted by these two charges on a third charge q3 = 48.5 nC placed between q1 and q2 at x3 = -1.055 m ?
Express answer to 3 significant figures.
Explanation / Answer
As F=kqq'/r^2
here q1 = -17.5 nC , x1 = -1.705 m, q2 = 37.0 nC, x2 = 0 m, q3 = 48.5 nC, x3 = -1.055 m
Net force on q3 = F = kq3[(q1/0.65^2) + (q2/1.055^2)]
= (9*10^9*48.5*10^-9)*10^-9 [(17.5/0.4225) + (37/1.113025)]
F = 3.26*10^-5 N