An electron is projected out along the + x axis with initial speed of 3*10^6 m/s. it goes 45cm and stops due to a uniform electric field in the region. the magnitude of the field (in v/m) is: a) 114 b) 57 c) 28.5 d) 14.25

b. 57 V/m

Explanation:

A charged particle that is in a region where there is an electric field will experiment a force equal to:

F = q* E

Where q is the charge of the particle and E is the magnitude of the field.

Force is also equal to the mass of the particle times its acceleration (Newton's second Law). We can find the acceleration that the particle experimented using the equation of velocity:

Vf² = Vo² + 2*a*d

Where Vf is the final velocity, 0 in this case, Vo is the initial velocity, a is the acceleration and d is the distance that the particle has traveled. We replace the data we have and solve for a:

a = (0² - (3*10^6 m/s)²)/(2* 0.45m) = -1 * 10^13 m/s^2

The negative means the particle is slowing down.

The mass of an electron is equal to 9.11 *10^-31 kg

This means the force experimented by the particle will be:

F = m *a = 1*10^13 m/s^2 * 9.11 * 10^-31 kg = 9.11*10^-18 N

The charge 1 of an electron is equal to 1.6 *10^-19.

Then, E will be:

E = F/q = 9.11 * 10^-18/1.6*10^-19 = 56.93 V/m

Or about 57 V/m.

hey , the answer is c