Question

A proton accelerates from rest in a uniform electric field of 610 N/C. At some later time, its speed is 1.4 106 m/s.
(a) Find the magnitude of the acceleration of the proton.
m/s2

(b) How long does it take the proton to reach this speed?
µs

(c) How far has it moved in that interval?
m

(d) What is its kinetic energy at the later time?
J

Answer 1

Concepts and reason

The concept required to solve the given problem is electric field.

For the first part, compute the net force acting on the proton and then rearrange the expression to obtain the acceleration.

For the second part and third part use kinematic equations to determine the time and distance travelled by the proton.

For the last part, calculate the kinetic energy by taking a product of the mass and square of the velocity.

Fundamentals

Electric field: It is the region around a charge where its influence can be felt.

It is given by,

$E = \frac{{{k_e}q}}{{{r^2}}}$

Here, ${k_e}$ is the coulomb’s constant, $q$ is the charge and $r$ is the distance of the point from the charge where the electric field needs to be calculated.

The force acting on a charge of magnitude $q$ in an electric field $E$ is given by,

$F = qE$

Equations of motion: The three equation of motion are:

1.$v = {v_0} + at$

2.$y = {v_0}t + \frac{1}{2}a{t^2}$

3.${v^2} - {u^2} = 2ay$

Here, $v$ is the final velocity, ${v_0}$ is the initial velocity, $a$ is the acceleration due to gravity, $y$ is the distance travelled and $t$ is the time taken.

(a)

The net force acting on the proton released from rest will be given by,

${F_{net}} = qE$

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

The force is also given by,

$F = ma$

Here, $m$ is the mass and $a$ is the acceleration.

Substitute $ma$ for $F$ in equation (1).

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

Substitute $1.6 \times {10^{ - 19}}{\rm{ C}}$ for$q$, $1.67 \times {10^{ - 27}}{\rm{ kg}}$for $m$ and $610{\rm{ N / C}}$ for $E$ in the above equation.

$\begin{array}{c}\\a = \frac{{\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {610{\rm{ N / C}}} \right)}}{{\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right)}}\\\\ = 5.84 \times {10^{10}}{\rm{ m / }}{{\rm{s}}^2}\\\end{array}$

(b)

The time taken by the proton to reach the speed $1.4 \times {10^6}{\rm{ m / s}}$ is given by,

$\begin{array}{l}\\v = u + at\\\\t = \frac{{v - u}}{a}\\\end{array}$

Here, $v$ is the final speed, $u$ is the initial speed and $a$ is the acceleration.

Since the proton starts from rest, the initial speed will be, $u = 0{\rm{ m / s}}$.

The final speed will be, $v = 1.4 \times {10^6}{\rm{ m / s}}$.

The acceleration will be,

$a = 5.84 \times {10^{10}}{\rm{ m / }}{{\rm{s}}^2}$

Substitute $5.84 \times {10^{10}}{\rm{ m / }}{{\rm{s}}^2}$ for $a$, $1.4 \times {10^6}{\rm{ m / s}}$ for $v$and $0{\rm{ m / s}}$ for $u$ in the above equation.

$\begin{array}{c}\\t = \frac{{\left( {1.4 \times {{10}^6}{\rm{ m / s}}} \right) - \left( {0{\rm{ m / s}}} \right)}}{{\left( {5.84 \times {{10}^{10}}{\rm{ m / }}{{\rm{s}}^2}} \right)}}\\\\ = 2.40 \times {10^{ - 5}}{\rm{ s}}\\\end{array}$

(c)

The distance moved by proton in the calculated time interval will be,

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

Here, $u$ is the initial speed, $a$ is the acceleration and $t$ is the time.

Substitute $5.84 \times {10^{10}}{\rm{ m / }}{{\rm{s}}^2}$ for $a$, $2.40 \times {10^{ - 5}}{\rm{ s}}$ for $t$and $0{\rm{ m / s}}$ for $u$ in the above equation.

$\begin{array}{c}\\s = \left( {0{\rm{ m / s}}} \right)\left( {2.40 \times {{10}^{ - 5}}{\rm{ s}}} \right) + \frac{1}{2}\left( {5.84 \times {{10}^{10}}{\rm{ m / }}{{\rm{s}}^2}} \right){\left( {2.40 \times {{10}^{ - 5}}{\rm{ s}}} \right)^2}\\\\ = 16.8{\rm{ m}}\\\end{array}$

(d)

The kinetic energy at the later time will be,

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

Here, $m$ is the mass and $v$ is the velocity.

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

$\begin{array}{c}\\K = \frac{1}{2}\left( {1.67 \times {{10}^{ - 27}}{\rm{ kg}}} \right){\left( {1.4 \times {{10}^6}{\rm{ m / s}}} \right)^2}\\\\ = 1.64 \times {10^{ - 15}}{\rm{ J}}\\\end{array}$

Ans: Part a

The magnitude of acceleration of the proton is$5.84 \times {10^{10}}{\rm{ m / }}{{\rm{s}}^2}$.