Question

1. (a) What must the charge (sign and magnitude) of 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? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? 1. ( a ) What must the charge ( sign and magnitude ) of 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 ? ( b ) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight ?​

Answer 1

(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.

What is electric field?

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