Coulombs law for the magnitude of the force F between two particles with charges

Question

Coulombs 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 = 1/4phi epsilon_0 and epsilon_O 8.854 times 10^-12 C^2/(N. m^2) is the permittivity of free space. Consider two point charges located on the x axis: one charge, q_1 = 19.5 nC is located at a1 1.680 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 = 1.105 m Your answer may be positive or negative, depending on the direction of the force.

Explanation / Answer

here,

q1 = - 1.95 * 10^-8 C is at x1 = - 1.68 m

q2= 3.9 * 10^-8 C is at x2 = 0 m

and q3 = 4.9 * 10^-8 C is at x3 = - 1.105 m

the net force on q3 , F = k * q1 * q3 /( 1.68 - 1.105)^2 + k * q2 * q3 /1.105^2

F = 9 * 10^9 * 10^-16 * ( 1.95 * 4.9 /0.575^2 + 3.9 * 4.9 /1.105^2) N

F = 4 * 10^-5 N

the net force exerted on q3 is - 4 * 10^-5 N

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