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---

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.

this means that the temperature of both substances will be equal

a. are