Question

"A proton is placed in a uniform electric field of 2750 N/C. You may want to review (Page) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Electron in a uniform field. Calculate the magnitude of the electric force felt by the proton. Express your answer in newtons.(F = ? )
Calculate the proton's acceleration.
( a= ? m/s2 )

Calculate the proton's speed after 1.40 {\rm \mu s} in the field, assuming it starts from rest.
( V= ? m/s )"

Answer 1

To solve this problem we will start from the definition of Force, as the product between the electric field and the proton charge. Once the force is found, it will be possible to apply Newton's second law, and find the proton acceleration, knowing its mass. Finally, through the linear motion kinematic equation we will find the speed of the proton.

PART A ) For the electrostatic force we have that is equal to

$F=qE$

Here

q= Charge

E = Electric Force

$F=(1.6*10^{-19}C)(2750N/C)$

$F = 4.4*10^{-16}N$

PART B) Rearrange the expression F=ma for the acceleration

$a = \frac{F}{m}$

Here,

a = Acceleration

F = Force

m = Mass

Replacing,

$a = \frac{4.4*10^{-16}N}{1.67*10^{-27}kg}$

$a = 2.635*10^{11}m/s^2$

PART C) Acceleration can be described as the speed change in an instant of time,

$a = \frac{v_f-v_i}{t}$

There is not $v_i$ then

$a = \frac{v_f}{t}$

Rearranging to find the velocity,

$v_f = at$

$v_f = (2.635*10^{11})(1.4*10^{-6})$

$v_f = 3.689*10^{5}m/s$