Chapter 2, Problem 4P

A charged particle moves along a straight line in a uniform electric field E with a speed v. If the motion and the electric field are both in the x direction, (a) show that the magnitude of the acceleration of the charge q is given by

$a=\frac{dv}{dt}=\frac{qE}{m}{\left(1-\frac{v^2}{c^2}\right)}^{3/2}$

(b) Discuss the significance of the dependence of the acceleration on the speed. (c) If the particle starts from rest at x = 0 at t = 0, find the speed of the particle and its position after a time t has elapsed. Comment on the limiting values of v and x as t →∞.

To determine

The magnitude of acceleration of the charge.

Answer to Problem 4P

It is proved that the acceleration of the charged particle is $a=\left(qE/m\right){\left(1-\left(v^2/c^2\right)\right)}^{3/2}$.

Explanation of Solution

Write the equation for the relativistic momentum.

    $p=\gamma mv=\frac{mv}{{\left[1-\left(v^2/c^2\right)\right]}^{1/2}}$        (I)

Here, $p$ is the relativistic momentum of the particle, $m$ is the mass of the particle, $v$ is the velocity of the particle and $c$ is the speed of light.

Write the equation for relativistic force.

    $F=\frac{dp}{dt}$        (II)

Here, $F$ is the relativistic force, $p$ is the relativistic momentum that changes with time $t$.

Substitute equation (I) in (II).

    $\begin{array}{c}F=\frac{d}{dt}\left\{\frac{mv}{{\left[1-\left(v^2/c^2\right)\right]}^{1/2}}\right\}\\ =\frac{m}{{\left[1-\left(v^2/c^2\right)\right]}^{3/2}}\left(\frac{dv}{dt}\right)\end{array}$        (III)

Write the equation for the force in terms of electric field.

    $F=qE$        (IV)

Here, $F$ is the force on the charged particle, $q$ is the charge of the particle and $E$ is the electric field.

Conclusion:

Substitute equation (IV) in (III) and rearrange.

    $\begin{array}{c}qE=\frac{m}{{\left[1-\left(v^2/c^2\right)\right]}^{3/2}}\left(\frac{dv}{dt}\right)\\ a=\frac{dv}{dt}=\left(\frac{qE}{m}\right){\left(1-\frac{v^2}{c^2}\right)}^{3/2}\end{array}$        (V)

Hence, the given equation for the acceleration of the charged particle is proved.                   

To determine

The significance of dependence of acceleration on speed.

Answer to Problem 4P

It signifies that no particle can move with a speed greater than the speed of light.

Explanation of Solution

Equation (V) gives the expression for the acceleration of the charged particle.

Conclusion:

From equation (V), as $v\to c$, $a\to 0$. Hence, it supports the fact that no speed can exceed the speed of light. 

To determine

The speed and position of the particle.

Answer to Problem 4P

The speed of the particle is $v=qEct/{\left[{\left(mc\right)}^2+{\left(qEt\right)}^2\right]}^{1/2}$ and the position of the particle is $x=\left(c/qE\right)\left\{{\left[{\left(mc\right)}^2+{\left(qEt\right)}^2\right]}^{1/2}-mc\right\}$.

Explanation of Solution

Rearrange equation (V) to separate the variables.

    $\frac{dv}{{\left(1-v^2/c^2\right)}^{3/2}}=\left(\frac{qE}{m}\right)dt$

Conclusion:

Integrate the above equation by giving proper limits.

    $\begin{array}{c}{\displaystyle \underset{0}{\overset{v}{\int }}\frac{dv}{{\left(1-v^2/c^2\right)}^{3/2}}}={\displaystyle \underset{0}{\overset{t}{\int }}\frac{qE}{m}dt}\\ {\frac{v}{{\left(1-v^2/c^2\right)}^{1/2}}|}_0^v=\frac{qEt}{m}\\ \frac{v^2}{{\left(1-v^2/c^2\right)}^{1/2}}={\left(\frac{qEt}{m}\right)}^2=\frac{qEt}{m}\\ v^2={\left(\frac{qEt}{m}\right)}^2-\left(\frac{v^2}{c^2}\right){\left(\frac{qEt}{m}\right)}^2\end{array}$

Simplify further.

    $\begin{array}{c}{v}^2\left[1+{\left(\frac{qEt}{m}\right)}^2\right]={\left(\frac{qEt}{m}\right)}^2\\ v^2=\frac{{\left(qEt/mc\right)}^2}{1+{\left(qEt/mc\right)}^2}\\ v^2=\frac{{\left(qEct\right)}^2}{{\left(mc\right)}^2+{\left(qEt\right)}^2}\end{array}$

The limiting behavior of v as $t\to 0$ and $t\to \infty$ is reasonable.

    $\begin{array}{c}v=\frac{dx}{dt}=\frac{qEct}{{\left[{\left(mc\right)}^2+{\left(qEt\right)}^2\right]}^{1/2}}\\ x={qEc{\left[{\left(mc\right)}^2+{\left(qEt\right)}^2\right]}^{1/2}\left[\frac{1}{{\left(qE\right)}^2}\right]|}_0^t\\ =\frac{c}{qE}\left\{{\left[{\left(mc\right)}^2+{\left(qEt\right)}^2\right]}^{1/2}-mc\right\}\end{array}$

Here, as $t\to 0$, $x\to 0$. Also, as $t\to \infty$, $x\to ct$.