Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1530592 • 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 = -1.655 m ; the second charge, q2 = 33.5 nC , is at the origin (x=0.0000).
Part A What is the net force exerted by these two charges on a third charge q3 = 49.0 nC placed between q1 and q2 at x3 = -1.235 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 that
q1 = -17.5 nC
x1 = -1.655 m ;
q2 = 33.5 nC ,
x2 = 0 m
q3 = 49 nC
x3 = -1.235m
Fnet = F31 + F32
F32 = k*q2*q3/(x3-x2)
F32 = - 9*10^9*33.5*10^(-9)*49*10^(-9) / (1.235)^2
F32 = -9.68*10^(-6) N
F31 = k*q1*q3/(x3-x2)^2
F31 = -9*10^9*17.5*10^(-9)*49*10^(-9) / (0.42)^2
F31 = -43.75*10^(-6 ) N
net force on charge 3
Fnet =-9.68*10^(-6) - 43.75*10^(-6 )
Fnet = 5.34*10^(-5) N
the direction along the -ve x axis.