(a) The charge (sign and magnitude) of a particle of mass 1.43 g be for it to remain stationary is 2.12 x 10⁻⁵ C.
(b)The magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight is 10.24 x 10⁻⁸ N/C.
<h3>What is electric field?</h3>
The field developed when a charge is moved. In this field, a charge experiences an electrostatic force of attraction or repulsion depending on the nature of charge.
Given is a particle of mass 1.43 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 N/C.
The electric field and force is related as
F = qE
mg = qE
Substitute the given values in the question, we have
1.43 x 10⁻³ x 9.81 = q x 660
q = 2.12 x 10⁻⁵ C
Thus, the charge of particle is 2.12 x 10⁻⁵ C.
(b)
Given is the electric force on a proton is equal in magnitude to its weight.
Force on proton = mass of proton x acceleration due to gravity
F = 1.67 x 10⁻²⁷ kg x 9.81
F = 16.38 x 10⁻²⁷ N
Charge on proton = 1.6 x 10⁻¹⁹ C
E = F/q
E = 16.38 x 10⁻²⁷ / 1.6 x 10⁻¹⁹
E = 10.24 x 10⁻⁸ N/C
Thus, the magnitude of electric field is 10.24 x 10⁻⁸ N/C.
Learn more about electric field.
brainly.com/question/15800304
#SPJ1
