Question

A small object of mass 4.10 g and charge −18.4 μc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

Answer 1

force due to electric field is balanced by the force due to its own weight

$F_g = F_e$

$mg = qE$

here

m = 4.10 g

$q = -18.4 \mu C$

now from the above equation we will have

$4.10*10^{-3} * 9.8 = 18.4 * 10^{-6} * E$

$E = 2183.7 N/C$

its direction is downwards along the direction of gravity as its a negative charge given here.