Question

What are the magnitude and direction of a uniform electric field perpendicular to the ground that is able to suspend a particle of mass

m = 1.30 g

carrying a charge of +5.00 µC in midair, assuming gravity and the electrostatic force are the only forces exerted on the particle?

Answer 1

here,

m = 0.0013 kg

charge , Q = 5 * 10^-6 C

let the electric feild be E

for the particle to suspend in air

gravitational force = electric force

m * g = q * E

E = 0.0013 * 9.8 /(5 * 10^-6)

E = 254.8 N/C

the electric field is 254.8 N/C and direction is upwards