Coulomb\'s law for the magnitude of the force F between two particles with charg
ID: 1409627 • 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'|/d^2 where K = 1/4 pi epsilon_0 and epsilon_0 = 8.854 Times 10^-12 C^2/(N middot m^2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q_1 = -12.0 nC, is located at x_1 = -1.750 m; the second charge. q_2 = 39.0 nC, is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3 = 49.0 nC placed between q_1 and q_2 at x_3 =-1.185 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 = - 12.0 nC @ (-1.750, 0)
q2 = 39.0 nC @ (0,0)
q3 = 49.0 nC @ (-1.185 , 0)
Force exerted on the thord charge,
F = -( k*q1q3/(d1^2) + k*q2q3/(d2^2) )
F = -( (8.9*10^9 * 12*49*10^-18) / (1.75 - 1.185)^2 + (8.9*10^9 * 39*49*10^-18) / (1.185)^2 )
F = - 2.85 * 10^-5 N
- ve sign means direction which is towards -ve x axis !!