1. Coulomb\'s law for the magnitude of the force F between two particles with ch

Question

1. 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 = -20.0 nC , is located at x1 = -1.660 m ; the second charge, q2 = 39.0 nC , is at the origin (x=0.0000). What is the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1 and q2 at x3 = -1.050 m ?

2. Two small charged balls have a repulsive force of 0.18 N when they are separated by a distance of 0.80 m . The balls are moved closer together, until the repulsive force is 0.68 N . How far apart are they now? Express your answer to two significant figures and include the appropriate units.

3. A point charge +6.1 C is placed at the origin, and a second charge –4.0 C is placed in the x-y plane at (0.40 m , 0.45 m ). Where should a third charge be placed in the plane so that the net force acting on it is zero? Express your answers using two significant figures. Enter the x and y coordinates of the charge separated by a comma.

Explanation / Answer

1. force on q3 due to q1 :

F1 = k q1 q3 / d^2

k = 1 / 4 pi e0 = 9 x 10^9

F1 = (9 x 10^9 x 20 x 10^-9 x 47 x 10^-9) / (1.660 - 1.050)^2

F1 = 2.274 x 10^-5 N toward negative x axis

force on q3 due to q2.

F2 = (9 x 10^9 x 47 x 10^-9 x 39 x 10^-9 ) / (1.050)^2

F2 = 1.50 x 10^-5 N toward negative x axis.

Fnet = F1 + F2 = 3.77 x 10^-5 N towards negative x axis.

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