Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1398666 • 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 = -10.5 nC , is located atx1 = -1.750 m ; the second charge,q2 = 36.5 nC , is at the origin (x=0.0000).
A)
What is the net force exerted by these two charges on a third charge q3 = 51.0 nC placed between q1and q2 at x3 = -1.065 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
let,
q1=-10.5 nC at x1=-1.75 m
q2=36.5 nc at x=0 (origin)
and
q3=51nc at x3=-1.065 m
force between q1 and q3 is,
F13=kq1*q3/r13^2 (attractive force)
=9*10^9*10.5*51*10^-18/(1.75-1.065)^2
=1.027*10^-5 N
and
force between q2 and q3 is,
F23=kq2*q3/r23^2 (repulsive force)
=9*10^9*36.5*51*10^-18/(1.065)^2
=1.477*10^-5 N
now,
net force, Fnet=F13+F23
=(1.027+1.477)*10^-5
=2.504*10^-5 N