A charged cloud system produces an electric field in the air near the Earth\'s s

Question

A charged cloud system produces an electric field in the air near the Earth's surface. A particle of charge -1.7 ? 10-9 C is acted on by a downward electrostatic force of 3.0 ? 10-6 N when placed in this field.

1-What is the Magnitude of the Electric Field (N/C)

2- What are the magnitude and direction of the electrostatic force exerted on a proton placed in this field?Down, up, right, left

3- What is the gravitational force on the proton? (N)

4-What is the ratio of the magnitude of the electrostatic force to the magnitude of the gravitational force over here?

Explanation / Answer

1)

F = qE

( 3.0 X 10-6) = (1.7 X 10-9)E

E = 1765 N/C

2)

The proton would be forced Downward. The proton is a positive charge and the Electric Field points in the direction of force on a positive charge

F = qE

F = (1.6 X 10-19)(1765)

F = 2.82 X 10-16 N

3)

F = mg = (1.67 X 10-27)(9.8)

F = 1.64 X 10-26 N

4)

2.82 X 10-16 /1.64 X 10-26

Ratio = 1.72 X 1010

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