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 |Q Q'|/d^2. Where K = 7/4 pi e_0, and sum_0 = 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 = -12.5 nC, is located at x_1 = -1.725 m; the second charge, q_2 = 40.0 nC, is at the origin (x = 0.0000). What is the net force exerted by these two charges on a third charge q_3 = 48.0 nC placed between q_1 and q_2 at x_3 = -1.230 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

here,

q1 = - 12.5 * 10^-9 C at x1 = - 1.725 m

and

q2 = 40 * 10^-9 C at x2 = 0 m

q3 = 48 * 10^-9 C at x3 = -1.23 m

as the charge q3 is placed between them , both charges exert force in -x direction

net force on q3 , F = k * q1 * q3 /( x1 - x3)^2 + k * q2 * q3 /( x3^2)

F = 9 * 10^9 * 48 * 10^-9 * 10^-9 * ( 12.5 /0.495^2 + 40/1.23^2)

F = 3.35 * 10^-5 N

the force on charge q3 is 3.35 * 10^-5 N towrds -ve x axis

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