Question

Coulumb'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'/d^2    where K= 1/4pie e0 and e0= 8.854*10^-12/ (N*m^2) is the permittivity of free space.

Consider two point charges located on the x axis: one charge,

q1 = -16.0nC , is located atx1 = -1.700m ; the second charge, q2 = 38.5nC ,is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 53.0nC placed between q1 and q2 atx3 = -1.115m ?

Your answer may be positive or negative, depending on the direction of the force.

Express your answer numerically in newtons to three significant figures.

Force on q3?                                N

Answer 1

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else).

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=1/4peo = 8.854*10^-12 nC is the permittivity of free space.

Consider two point charges located on the x axis: one charge, q1 = -19.5nC , is located at x1= -1.680m ; the second charge, q2= 31.5nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 52.5nC placed between q3 and at x3= -1.245m ? Your answer may be positive or negative, depending on the direction of the force.

Answer

The electric force at q3 due to q1 is

F1= (8.9 x 10^9 N.m2/C2)(19.5x10^-9C)(52.5x10^-9C)/(0.435 m)^2

= 4.815x10^-5 N

The electric force at q3 due to q2 is

F2= (8.9 x 10^9 N.m2/C2)(31.5x10^-9C)(52.5x10^-9C)/(1.245 m)^2

= 9.4955x10-6 N

So the net force on q3 is

F = F1+ F2

= -( 4.815x10^-5 N)+(- 9.4955x10-6 N)

= - 5.764557x10^-5 N