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On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the gravitation
Question
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the gravitational and electric forces on a small dust particle of mass 2.2 ✕ 10−15 g that carries a single electron charge. Fg FE = What is the acceleration (both magnitude and direction) of the dust particle? (Enter the magnitude in m/s2.) magnitude m/s2 direction —Select—
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Physics
3 years
2021-08-29T12:05:46+00:00
2021-08-29T12:05:46+00:00 1 Answers
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Answers ( )
Answer:
a) FE = 0.764FG
b) a = 2.30 m/s^2
Explanation:
a) To compare the gravitational and electric force over the particle you calculate the following ratio:
(1)
FE: electric force
FG: gravitational force
q: charge of the particle = 1.6*10^-19 C
g: gravitational acceleration = 9.8 m/s^2
E: electric field = 103N/C
m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg
You replace the values of all parameters in the equation (1):
Then, the gravitational force is 0.764 times the electric force on the particle
b)
The acceleration of the particle is obtained by using the second Newton law:
you replace the values of all variables:
hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.