Forces in a Three-Charge System coulomb\'s law for the magnitude of the force F
ID: 1332144 • Letter: F
Question
Forces in a Three-Charge System 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 = 1/4pi 0, and 0 = 8.854Times10-12 C2/(N m2) is the permittivity of free space. consider two point charges located on the x axis: one charge, ql = -11.0 nC , is located at x1 = -1.650 m ; the second charge, q2 = 37.0 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 = 54.0 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. Force on q3 =Explanation / Answer
net force on q2 = F on q1q2 + force on q1q3
q1q2 are separated by 1.650 m to the left of the origin, this is attractive force
q2q3 are seperated by 1.24 m to the right of origin , this is repulsive force
so FNet will ne acting towards the left of q2
Fnet = F12 + F23
Fnet = (9e9 * 11 e-9*37.e-9)/(1.65*1.65) + (9e9 * 37 e-9* 54 e-9)/(1.24 *1.24)
Fnet = 1.3 *10^-5 N