On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 100 n/c. compare the gravitational and electric forces on a small dust particle of mass 3.5×10-19 kg that carries a single electron charge. what is the acceleration (both magnitude and direction) of the dust particle?

55.51 m/s^2 (pointing downward )

Fe/Fg=4.66

Explanation: In order to solve this problem we have to determine the electric and gravitational forces on the dust particle . They are given by:

Fe=e*E=1.6*10^-19*100 N/C=1.6*10^-17N

Fg=m*g=3.5*10^-19*9.8=3.43*10^-18 N

both forces point downward  so the acceleration is given by the second Newrton law:

FT=Fe+Fg=1.6*10^-17N+3.43*10^-18 N=1.94*10^-17 N

a=FT/m=55.51 m/s^2 (pointing downward )

f=ma newton

f=100x2.5=250n