Question

1. As shown in the diagram, three equal charges are spaced evenly in a row. The magnitude of (1 point) 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? 1.50 1.50 n +26 04.27x 10-10N to the right 04.27 x 1010 N to the left O3.20 x 10-10 N to the left 03.20 x 10-10 N to the right 2. Two small spheres are spaced 20.0 cm apart and have equal negative charge. The force of point) repulsion between them is 4.57 x 10-21 N. How many excess electrons are present on each ○ 562 09.14 x 103 O1.01 x 106 0891

Answer 1

1) The distance between the left charge and the center charge = 1.50+0.35 nm = 1.85 nm
The distance between the right charge and the center charge = 1.50-0.35 nm = 1.15 nm

The force exerted on the center charge by the left charge = k(2e)(2e)/(1.85nm*1.85nm) = 9*10^9*4*1.6*1.6*10^-38/1.85*1.85*10^-18 = 2.69*10^-10 N to the right

The force exerted on the center charge by the right charge = k(2e)(2e)/(1.85nm*1.85nm) = 9*10^9*4*1.6*1.6*10^-38/1.15*1.15*10^-18 = 6.97*10^-10 N to the left

Net force = (6.97-2.69)*10^-10 N to the left = 4.27*10^-10 N to the left

2) The repulsive force exerted = k(q)(q)/(r*r) = 9*10^9*q*q/0.2*0.2 = 4.57*10^-21 N
q = 1.4*10^-16 C
excess electrons = 1.4*10^-16 C/1.6*10^-19 C = 891 electrons