Question

A proton accelerates from rest in a uniform electric field of 660 N/C. At one later moment, its speed is 1.40 Mm/s (nonrelativistic because v is much less than the speed of light).

(a) Find the acceleration of the proton.
m/s²

(b) Over what time interval does the proton reach this speed?
s

(c) How far does it move in this time interval?
m

(d) What is its kinetic energy at the end of this interval?
J

Answer 1

Concepts and reason

The concept required to solve the given problem is electric field, force, and acceleration of charge.

Firstly, find the relation between the acceleration and the electric field of the charge. Then, substitute the values to find the acceleration.

Then, use Newton’s equation of motion to find the time taken by the charge to reach its final speed.

Then, use the Newton’s equation of motion to find the distance moved in the calculated time interval.

Finally, find the kinetic energy of the charge by using the expression of kinetic energy.

Fundamentals

The expression of the electric force is,

$F = qE$

Here, E is the electric field, q is the charge.

The net force exerted on a particle is as follows:

$F = ma$

Here, m is the mass and a is the acceleration of the particle.

The Newton’s equation of motion for constant acceleration are as follows:

$\begin{array}{c}\\s = ut + \frac{1}{2}a{t^2}\\\\{v^2} = {u^2} + 2as\\\\v = u + at\\\end{array}$

Here, s is the distance, v is the final velocity, u is the initial velocity, a is the constant acceleration, and t is the time taken during the motion.

The kinetic energy of the particle of mass m is as follows:

$K = \frac{1}{2}m{v^2}$

Here, v is the velocity of the particle.

a)

The expression of the electric force is,

$F = qE$

Here, E is the electric field, q is the charge.

The net force exerted on a particle is as follows:

$F = ma$

Here, m is the mass and a is the acceleration of the particle.

Equate the electric force with the net force to find the expression of the acceleration.

$\begin{array}{c}\\ma = qE\\\\a = \frac{{qE}}{m}\\\end{array}$

Substitute $1.6 \times {10^{ - 19}}{\rm{ C}}$ for q, 660 N/C for E, and $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for m in the above expression.

$\begin{array}{c}\\a = \frac{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {660{\rm{ N/C}}} \right)}}{{1.67 \times {{10}^{ - 27}}{\rm{ kg}}}}\\\\ = 6.32 \times {10^{10}}{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

b)

The Newton’s equation of motion for constant acceleration is as follows:
‎$v = u + at$

Substitute 0 for u and find the expression of t.

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

Substitute $1.4 \times {10^6}{\rm{ m/s}}$ for v and $6.32 \times {10^{10}}{\rm{ m/}}{{\rm{s}}^2}$for a in the above expression.

$\begin{array}{c}\\t = \frac{{1.4 \times {{10}^6}{\rm{ m/s}}}}{{6.32 \times {{10}^{10}}{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 2.2 \times {10^{ - 5}}{\rm{ s}}\\\end{array}$

c)

The Newton’s equation of motion for constant acceleration is as follows:

$s = ut + \frac{1}{2}a{t^2}$

Substitute 0 for u, $6.32 \times {10^{10}}{\rm{ m/}}{{\rm{s}}^2}$for a, and $2.2 \times {10^{ - 5}}{\rm{ s}}$for t in the above expression.

$\begin{array}{c}\\s = \left( 0 \right)\left( {2.2 \times {{10}^{ - 5}}{\rm{ s}}} \right) + \frac{1}{2}\left( {6.32 \times {{10}^{10}}{\rm{ m/}}{{\rm{s}}^2}} \right){\left( {2.2 \times {{10}^{ - 5}}{\rm{ s}}} \right)^2}\\\\ = 15.3{\rm{ m}}\\\end{array}$

d)

At the end of the time interval t, the velocity of the proton is v. Thus, the kinetic energy of the particle of mass m is as follows:

$K = \frac{1}{2}m{v^2}$

Substitute $1.67 \times {10^{ - 27}}{\rm{ kg}}$ for m $1.4 \times {10^6}{\rm{ m/s}}$ for v in the above expression.

$\begin{array}{c}\\K = \frac{1}{2}\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right){\left( {1.4 \times {{10}^6}{\rm{ m/s}}} \right)^2}\\\\ = 1.63 \times {10^{ - 15}}{\rm{ J}}\\\end{array}$

Ans: Part a

The acceleration is $6.32 \times {10^{10}}{\rm{ m/}}{{\rm{s}}^2}$.

Part b

The time taken by proton to reach its final speed is $2.2 \times {10^{ - 5}}{\rm{ s}}$.

Part c

The distance moved by the proton in time t is 15.3 m.

Part d

The kinetic energy of proton is $1.63 \times {10^{ - 15}}{\rm{ J}}$.