Question

A proton starts at rest in an electric field of +10 N/C i. If the proton is then free to move, what is its velocity after 15 s?

Answer 1

Given,

the electric field strength, E = +10N/C

the proton's initial velocity, u = 0 m/s

We have to find the proton's final velocity after t = 15s time.

Let the final velocity of the proton be v.

To find its final velocity, we need to first know its acceleration.

To know the proton's acceleration, we should know how much force is exerted by the electric field in which the proton is placed.

So, the force F experienced by a charge q placed in a uniform electric field E, is given by

F = qE

We know that the charge of the proton q = 1.602*10^–19C

Then, the force experienced by the proton due to the field +10N/C is

F = (1.602*10^–19) *(10)

F = 1.602*10^–18N

Now, the acceleration of the proton due to the force 1.602*10^–18N, can be found using Newton's second law of motion, F = ma.

Here, m is the mass of the proton and m = 1.672*10^–27kg.

Now, from F = ma

a = F/m

a = ( 1.602*10^–18) / (1.672*10^–27)

a = 0.958*10⁹ m/s²

So, the acceleration of the proton in the field is 0.958*10⁹ m/s² .

By substituting the values of initial velocity u, time t, and acceleration a, in the equation of motion v = u+at, we can find the proton's final velocity v.

So, v = 0 + (0.958*10⁹)*(15)

v = 14.37*10⁹ m/s

So, the velocity of the proton, which starts moving from rest in an electric field of +10N/C, after 15s time, is 14.37*10⁹ m/s.