Question

three 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 0.350nm 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?

Answer 1

initial position of the charges 1. 5o mm final position of the charges

r₁= distance of the charge in between from the left charge = 1.85 nm = 1.85 x 10^-9 m

r₂ = distance of the charge in between from the right charge = 1.15 nm = 1.15 x 10^-9 m

Q = magnitude of charge on each charge = 2 e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

F₁ = force by charge on left

force by charge on left is given as

F₁ = k Q²/r₁²

F₂ = force by charge on right

force by charge on right is given as

F₂ = k Q²/r₂²

Taking force in left as negative and in right as positive, net force on the charge in between is given as

F = F₁ - F₂

F = (k Q²/r₁² ) - (k Q²/r₂² )

F = (k Q²) ((1/r₁² ) - (1/r₂² ))

inserting the values

F = ((9 x 10⁹) (3.2 x 10^-19 )²) ((1/(1.85 x 10^-9)² ) - (1/(1.15 x 10^-9)² ))

F = - 4.3 x 10^-10 N

the negative sign indicate the direction of net force in left direction

magnitude : 4.3 x 10^-10 N

direction : Left