Answer:
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.
