The electric field near the Earth's surface has magnitude of about 150 N/C. What is the acceleration experienced by an electron near the surface of the earth? (b) What about a proton?

Given that,

Magnitude of electric field is

E = 150N/C

a. Acceleration of election?

The charge on an electron is

q = - 1.602*10^-19

then, the force in electric field is given as

F=qE

F = 150*1.602*10^-19

F = 2.403*10^-17 N

Using second law of motion

F = ma

Then, a = F/m

Then, mass of electron is

m = 9.11 * 10^-31kg

Therefore,

a = F/m

a = 2.403*10^-17/9.11*10^-31

a = 2.64 * 10^13 m/s²

That's the magnitude of the electron's acceleration.

Since the electron has a negative charge the direction of the force on the electron (and also the acceleration) is opposite the direction of the electric field.

b. Same procedure

Acceleration of proton?

The charge on a proton is

q = + 1.602*10^-19

then, the force in electric field is given as

F=qE

F = 150*1.602*10^-19

F = 2.403*10^-17 N

Using second law of motion

F = ma

Then, a = F/m

Then, mass of proton is

m = 1.67 * 10^-27kg

Therefore,

a = F/m

a = 2.403*10^-17/1.67*10^-27

a = 1.44 * 10^10m/s²

That's the magnitude of the proton's acceleration.

Since the proton has a positive charge the direction of the force on the proton (and also the acceleration) is in the same direction of the electric field.