Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1528347 • 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 = -20.0 nC , is located at x1 = -1.725 m ; the second charge, q2 = 36.5 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 53.5 nC placed between q1and q2 at x3 = -1.130 m ?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
step;1
Given that
charge q1=-20 nC
charge q2=36.5 nC
charge q3=53.5 nC
step;2
Now we find net force exerted by thoes charges
the force b/w charge q1 and q3=>F1=9*10^9*-20*53.5*10^-18/(-1.130+1.725)^2
=-27.2*10^-6 N
the force b/w charge q3 and q2 =>F2=9*10^9*36.5*53.5*10^-18/(-1.13)^2=13.8*10^-6 N
the net force Fnet=(-27.2+13.8)*10^-6=-13.4*10^-6 N