Question

A proton accelerates from rest in a uniform electric field of 680 N/C. At one later moment, its speed is 1.30 Mm/s (nonrelativistic because v is much less than the speed of light). (a) Find the acceleration of the proton.

Answer 1

Answer:

Acceleration, $a=6.51\times 10^{10}\ m/s^2$                                                

Explanation:

Given that,

Electric field, E = 680 N/C

Speed of the proton, v = 1.3 Mm/s

We need to find the acceleration of the proton. We know that the force due to motion is balanced by the electric force as :

$qE=ma$

a and m are the acceleration and mass of the proton.

$a=\dfrac{qE}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 680}{1.67\times 10^{-27}}$

$a=6.51\times 10^{10}\ m/s^2$

So, the acceleration of the proton is $a=6.51\times 10^{10}\ m/s^2$. Hence, this is the required solution.