Like

Report

A small object of mass 3.80 $\mathrm{g}$ and charge $-18.0 \mu \mathrm{C}$ is suspended motionless above the ground when immersed in a uniformelectric field perpendicular to the ground. What is the magnitude and direction of the electric field?

$2.07 \times 10^{3} N / C$

You must be signed in to discuss.

Cornell University

Rutgers, The State University of New Jersey

Numerade Educator

Hope College

electric field is given by easing KAL Toh force per unit charge. Therefore, sexting on the the charge is due to the gravitational field, which is equal to its weight, that is. M times. Aggie divided by Q Um, Now this weight should be equal to mass. Times G, which is 9.8 Marcie sir 0.0 380 kilograms times 9.8 Divided by charges 18 times 10 to the power minus six. So this gives us an electric feel off 2.7 constant power three Newton for Coolum.