As shown in the diagram, threee equal charges are spaced evenly in a row. the magnitude of each charge is +2e, and the distance between two adjacent charges is 1.50 nm. then the central charge is displaced 0.350 nm to the right while the other two charges are held in place. after the displacement, what is the magnitude and direction of the net force that the outer two charges exert on the central charge?

4.27*10^-10 to the right

4.27*10^-10 to the left

3.20*10^-10 to the left

3.20*10^-10 to the right

The net force will have a magnitude of 4.25 × 10^-10 N and will act to the right of the center charge.

Explanation:

Parameters given:

Charge of leftmost charge, q(1) = +2e

Charge of center charge, q(2) = +2e

Charge of rightmost charge, q(3) = +2e

Distance between leftmost charge and center charge, R = 1.50 + 0.35 nm = 1.85nm

Distance between rightmost charge and center charge = 1.50 - 0.35 nm = 1.15nm

Taking the position of center charge as the origin, the leftmost charge is on the negative x axis while the rightmost charge is on the positive x axis.

Hence, the net force on the centre charge will be

F = -kq(1)q(2)/(R^2) + kq(2)q(3)/(r^2)

Where k = Coulomb's constant

Since q(1) = q(2) = q(3) = q,

F = -(kq^2)/(R^2) + (kq^2)(r^2)

F = -(kq^2)[1/R^2 - 1/r^2]

F = -(9×10^9×4×1.602^2×10^-38)[1/R^2 - 1/r^2]

F = -9.24×10^-28[1/(1.85^2×10^-18) - 1/(1.15^-18)]

F = -9.24×10^-28[(2.9×10^17) - (7.5×10^17)

F = - 9.24×10^-28[-0.46×10^17]

F = 4.25 × 10^-10 N

Hence, the net force will have a magnitude of 4.25 × 10^-10 N and will act to the right of the center charge.