Coulomb\'s law for the magnitude of the force F between two particles with charg

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=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.

**Please solve with a numerical answer. I've tried this several times and this is my last attempt on mastering physics.

Explanation / Answer

q1 = -19.5*10^-9 C at x1 = -1.68 m q2 = 31.5*10^-9 C at x2 = 0 q3 = 52.5*10^-9 C at x3 = -1.245 m Seperation between q1 and q3, r13 = 1.68 - 1.245 = 0.435 m Seperation between q2 and q3, r23 = 1.245 m Force on q3 due to q1, F1 = Kq1q3/(r13)^2 (- i)along the -ve X axis                                          = (9*10^9*19.5*10^-9*52.5*10^-9)/(0.435)^2(- i)                                          = - 48692.03*10^-9 i N Force on q3 due to q2, F2 = Kq2q3/)r23)^2 ( - i)                                           = (9*10^9*31.5*10^-9*52.5*10^-9)/(1.245)^2 (- i)                                           = - 9602.26*10^-9 N i along the -ve X axis Net force on q3, F = F1 + F2                              = - 48692.03*10^-9 i N - 9602.26*10^-9 i N                              = - 58294.29*10^-9 i N Magnitude F = - 58294.29*10^-9 N Magnitude F = - 58294.29*10^-9 N   

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