Question

Question 76)

Part 1)

A charged cloud system produces an electric field in the air near Earth's surface. A particle of charge -2.1 × 10^-9 C is acted on by a downward electrostatic force of 4.0 × 10^-6 N when placed in this field. (a) What is the magnitude of the electric field? (b) What is the magnitude of the electrostatic force on a proton placed in this field? (c) What is the gravitational force on the proton? (d) What is the ratio of the electrostatic force to the gravitational force in this case?

Part 2)

In the figure a nonconducting rod of length L = 8.34 cm has charge -q = -4.42 fC uniformly distributed along its length. (a) What is the linear charge density of the rod? What are the (b) magnitude and (c) direction (positive angle relative to the positive direction of the x axis) of the electric field produced at point P, at distance a = 13.7 cm from the rod? What is the electric field magnitude produced at distance a = 63 m by (d) the rod and (e) a particle of charge -9 = -4.42 fC that replaces the rod? X

Answer 1

(76) charge on a particle q = - 2.1 x 10^-9 C

downward electrostatic force, Fe = 4 x 10^-6 N

(a) magnitude of electric field;

Fe = qE;

E = Fe/q;

E = 4 x 10^-6 / -2.1 x 10^-9

E = - 1904.762 N/C (negative sign shows thatelectic field is in downward direction)

(b) electrostatic force on proton;

F = qE

= 1.6 x 10^-19 x 1904.762

= 3.05 x 10^-16 N

(c) Gravitational force on proton

Fg = mg

= 1.67 x 10^-27 x 9.8

Fg = 1.64 x 10^-26 N

(d) ratio of the electrostatic force to the gravitational force

r = electrostatic force/gravitational force

r = 3.05 x 10^-16/ 1.64 x 10^-26

r = 1.859 x 10^10