Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1366953 • 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=1/40, and 0=8.854×1012C2/(Nm2) is the permittivity of free space.
Consider two point charges located on the x axis: one charge, q1 = -10.0 nC , is located at x1 = -1.750 m ;the second charge, q2 = 35.5 nC , is at the origin (x=0.0000).
What is the net force exerted by these two charges on a third charge q3 = 47.5 nC placed between q1 and q2 at x3 = -1.080 m ?
Your answer may be positive or negative, depending on the direction of the force.
Explanation / Answer
F13 = k*q1*q3/r13^2
= 9*10^9*10*10^-9*47.5*10^-9/(1.75 - 1.08)^2
= 9.52*10^-6 N (towards -x axis)
F23 = k*q2*q3/r23^2
= 9*10^9*35.5*10^-9*47.5*10^-9/(1.08 - 0)^2
= 1.3*10^-5 N (towards -x axis)
so, Fnet = - (F13 + F23)
= - (9.52*10^-6 + 1.3*10^-5 )
= -2.252*10^-5 N