Coulomb\'s law for the magnitude of the force F between two particles with charg

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 elementof_0, and elementof_0 = 8.854 times 10^-12 C^2/(N middot m^2) is the permittivity of free What is the net force exerted by these two charges on a third charge q_3 = 48.5 nC placed between q_1 and q_2 at x_3 = -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.

Explanation / Answer

Net charge on charge 3 will be

F3 = F13 + F23

F3 = kq1q3/r13^2 + kq2q3/r23^2

q1 = -8 nC

q2 = 38.5 nC

q3 = 48.5 nC

r13 = 1.730 - 1.240 = 0.49 m

r23 = 0 - (-1.240) = 1.24 m

So,

F3 = 9*10^9*18*48.5*10^-18/0.49^2 + 9*10^9*38.5*10^-9*48.5*10^-9/1.24^2

F3 = 4.36*10^-5 N

Since the net force is in -ve x-axis, So

Fnet = -4.36*10^-5 N

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