Question

What must the charge (sign and magnitude) of a particle of mass 1.50 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 670 N/CN/C ?

Answer 1

q = -2.19 x 10⁻⁵ C

Step-by-step explanation:

Given;

mass of the particle, m = 1.5 g = 0.0015 kg

magnitude of electric field, E = 670 N/C

Electric field is given by;

`E = (F)/(q)`

where;

q is the magnitude of the

f is the force of the charge

f = mg

`E = (F)/(q)\\\\E = (mg)/(q)\\\\q = (mg)/(E)\\\\q = (0.0015*9.8)/(670) \\\\q = 2.19*10^(-5) \ C`

Since the electric field is acting downward, the force on the charge must be acting upward. Therefore, the charge must be negative

q = -2.19 x 10⁻⁵ C