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---

Answer 1

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:

$\frac{F_E}{F_G}=\frac{qE}{mg}$              (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):

$\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764$

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:

$F_E-F_G=ma\\\\a=\frac{qE-mg}{m}$

you replace the values of all variables:

$a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}$

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.