Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1653201 • 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=8.854×10-12C2/(Nm2) is the permittivity of free space. Consider two point charges located on the x axis: one charge,q1 = -11.0 nC ,is located at x1 = -1.700 m ; the second charge,q2 = 34.0 nC ,is at the origin (x=0.0000). A. What is the net force exerted by these two charges on a third charge q3 = 53.0 nC placed between q1 and q2 at x3 = -1.240 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.
Explanation / Answer
given,. q1 = -11 nC, at x1 = -1.7 m
q2 = 34 nC at origin
q3 = 53 nC, at x = -1.24 m
so net force on q3 due to q1 and q2 from coloumbs law
F = k*q3[-q1/(1.7-1.24)^2 - q2/1.24^2]
F = -k*53*10^-18[11/(1.7-1.24)^2 + 34/1.24^2] = -3.5265*10^-5 N [ towards -ve x axis]