Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1525199 • 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 = -14.5 nC , is located at x1 = -1.710 m ; the second charge, q2 = 33.5 nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 54.0 nC placed between q1 and q2 at x3 = -1.105 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
q1 = - 14.5 nC
q1 = -14.5 * 10^-9 C
x1 = - 1.710 m
q2 = 33.5 nC
q2 = 33.5 * 10^-9 C
x2 = 0
q3 = 54 nC
q3 = 54 * 10^-9 C
x3 = - 1.105 m
the net force exerted by these two charges on a third charge, F = K * q1 * q3/(x1-x3)2 + K * q2 * q3/(x3)2
F = 9 * 10^9 *14.5 * 10^-9* 54 * 10^-9/(0.605)2 + 9 * 10^9 *33.5 * 10^-9* 54 * 10^-9/(1.105)^2
F = 3.29 * 10^-5 N
the net force exerted by these two charges on a third charge q3 = 54 nC is 3.29 * 10^-5 N towards the charge q1