Question

What must the charge (sign and magnitude) of a particle of mass 1.48 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/C

Answer 1

`q=-2.26* 10^(-5)\ C`

Step-by-step explanation:

Given that,

The mass of a particle, m = 1.48 g = 0.00148 kg

The electric field, E = 640 N/C

We need to find the charge of the particle when placed in a downward-directed electric field.

The force of gravity is balanced by the electric force such that,

mg = qE

Where

q is the charge of the particle

`q=(mg)/(E)\\\\q=(0.00148* 9.8)/(640)\\\\q=2.26* 10^(-5)\ C`

q must be negative, the force must be upward (opposite direction of the electric field).