Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 251251 • 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 = -1.680 m ; the second charge, q2 = 32.5 nC , is at the origin (x=0.0000)
What is the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.085 m ?
Explanation / Answer
Since q1 and q3 are of different polarity therefore they will attract each other and force F1 on q3 due to q1 is in (-x) direction and
q2 and q3 are of same polarity therefore they will repel each other and force F2 on q3 due to q2 is in (-x) direction
therefore net force on q3 is addition of both forces F1 + F2 and is in (-x) direction.
now F = [ F1 + F2 ] (direction -x)
F = Kq1*q3 / r12 + Kq2*q3 / r22 = Kq3 [ q1 / r12 + q2 / r22 ]
where r1 = 1.68 - 1.085 = 0.595 meter and r2 = 1.085 meter and we have K = 9*10^9 and
q1 = -17.5 nC and q2 = 32.5 nC and q3 = 55 nC
therefore F = 9*10^9 * 55 *10^(-9) * [ (17.5 *10^(-9)) / (0.595)^2 + (32.5 *10^(-9)) / (1.085)^2 ]
note that negative sign of q1 is omitted because both the forces are in -x direction and we only need to add the magnitudes of the forces
F = 9*55* [ 49.43 + 27.60 ] * 10^(-9) = 3.813 * 10^(-5) Newton and in -x direction