Question

A particle (mass = 6.7 x 10^-27 kg, charge = 3.2 x 10^-19 C) moves along the positive x axis with a speed of 4.8 x 10^5 m/s. It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field. What is the magnitude of the electric field?

Answer 1

Since the field does work on the charge and the charge will surely gain kinetic energy, the magnitude of the electric field is 1206 N/C

Charged Particle in an Electric Field

A charged particle at rest or moving along the direction of an electric field experiences an electric force.

Given that a  particle (mass = 6.7 x 10^-27 kg, charge = 3.2 x 10^-19 C) moves along the positive x axis with a speed of 4.8 x 10^5 m/s. It enters a region of uniform electric field parallel to its motion and comes to rest after moving 2.0 m into the field.

The given parameters are;

• m = 6.7 x 10^-27 kg

• q = 3.2 x 10^-19 C

• V = 4.8 x 10^5 m/s

• Electric field E = ?

• distance s = 2 m

The field does work on the charge and the charge will surely gain kinetic energy. Therefore,

Work done = force x distance

Force = Eq

Work done = 1/2m$V^{2}$

1/2 m$V^{2}$ = Eq x s

Substitute all the parameters into the formula above

1/2 x 6.7 x $10^{-27}$ x (4.8 x $10^{5}$)^2 = 3.2 x $10^{-19}$ E x 2

7.7184 x $10^{-16}$ = 6.4 x $10^{-19}$ E

Make E the subject of formula

E = (7.7184 x $10^{-16}$) / (6.4 x $10^{-19}$)

E = 1206 N/C

Therefore, the magnitude of the electric field is 1206 N/C.

Learn more about Electric field here: https://brainly.com/question/14372859