Please read Ch20 in your textbook. (4 pts total)(P20.10) Two small plastic spher

Question

Please read Ch20 in your textbook. (4 pts total)(P20.10) Two small plastic spheres each have a mass of 2.0 g and a charge of 50.0 nC. They are placed 2.0 cm apart. a) What is the magnitude of the electric force between the spheres? (0.5 pts) b) By what factor is the electric force on a sphere larger than its weight? (0.5 pts) c) Now assume a third sphere is placed along the same axis with the first two, as seern 1. in the figure below. If the charge on the 3"d sphere is +50 nC, what is the net electric force on each sphere (i.e., total force on q1, q2 and q3)? (3 pts) 42-50nC 0.5 cm 1.5 cm (3 pts total) Three charges are arranged as shown in Figure below. Find the magnitude and direction of the electrostatic force on the 6-nC charge. 2. 3nC 0.5 m 6nC 0.5m 2nC (3 pts total) A charge of +2q is placed at the origin, and a charge of +4q is placed at x-1.50 m. Locate the point in between the two charges where a charge of +3q should be placed so that the net electric force is zero. 3.

Explanation / Answer

1) a) F = kq1q2 / d^2   = 9 x 10^9 x 50 x 10-9 x50 x 10-9 / 0.02^2
= 0.056 N
b) weight = mg = 2 x 10-3 x 9.81 = 0.0196
force / weight = 0.056 / 0.0196   = 2.85

c) on q1 = kq1q3 / 0.005^2    -     kq1q2 / 0.02^2
    = 0.84 N
on q3
F3 = kq1q3 / 0.005^2   -     kq3q2 / 0.015^2
= 0.8 N
on q2
F2 = kq2q3 /0.015^2 - kq2q1 / 0.02^2
= 0.044 N

2 ) Fnet = [( 9 x 10^9 x6x10-9 x 3 x 10-9 / (2x0.5^2) ) + (9 x10^9 x 6 x 10-9 x 2x10-9 /(2 x 0.5^2))]cos45
= 5.4 x 10-7 N

3) force on +3q due to +2q   = force due +4q
9 x10^9 x 3q x 2q / x^2 = 9 x 10^9 x 3q x 4q / (1.50 -x)^2
(1.50 - x)^2 = 2x^2
x^2 + 3x -2.25 = 0
x = 0.621 m

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