Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1527666 • 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 = -12.0 nC , is located atx1 = -1.725 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 = 45.0 nC placed between q1and q2 at x3 = -1.205 m ?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
Given that
charge q1=-12nC
charge q2=34 nc
distance x1=-1.725 m
distance x3=-1.205 m
charge q3=45 nC
step;2
now we find the net force ecerted by the two charges
F1=9*10^9*q1*q3/r^2=9*10^9*45*-12*10^-18/(-1.725+1.205)^2=17.97*10^-6 N
the force F2=9*10^9*34*45*10^-12/(-1.205)^2=9.5*10^-6 N
the net force Fnet=F1+F2=[9.5*10^-6-17.97*10^-6]=-8.5*10^-6 N