Chapter 39, Problem 39.88CP

A particle with electric charge q moves along a straight line in a uniform electric field $\stackrel{\to }{E}$ with speed u. The electric force exerted on the charge is $q\stackrel{\to }{E}$. The velocity of the particle and the electric field are both in the x direction. (a) Show that the acceleration of the particle in the x direction is given by

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

(b) Discuss the significance of the dependence of the acceleration on the speed. (c) What If? If the particle starts from rest it x = 0 at t = 0, how would you proceed to find the speed of the particle and its position at time t?

To determine

To show: The acceleration of the particle in the $x$ direction is $a=\frac{du}{dt}=\frac{qE}{m}{\left(1-\frac{u^2}{c^2}\right)}^{3/2}$

Answer to Problem 39.88CP

The acceleration of the particle in the $x$ direction is $a=\frac{du}{dt}=\frac{qE}{m}{\left(1-\frac{u^2}{c^2}\right)}^{3/2}$.

Explanation of Solution

The formula to calculate the relative momentum is,

$p=\frac{mu}{\sqrt{1-{\left(u/c\right)}^2}}$

Here,

$m$ is the mass of the electric charge.

$c$ is the speed of the light.

$u$ is the speed of the electric charge.

The formula to calculate the force on the electric charge is,

$F=qE$

Here,

$q$ is the charge of the electric charge.

$E$ is the magnitude of the electric field.

The formula to calculate the Force due to motion is,

$F=\frac{dp}{dt}$

The force on the electric charge due to motion must be equal to that of the force due to electric field.

Substitute $qE$ for $F$ in above equation to find $\frac{dp}{dt}$.

$qE=\frac{dp}{dt}$

Substitute $\frac{mu}{\sqrt{1-{\left(u/c\right)}^2}}$ for $p$ in above equation.

$\begin{array}{l}qE=\frac{d}{dt}\left(\frac{mu}{\sqrt{1-{\left(u/c\right)}^2}}\right)\\ qE=\frac{d}{dt}\left[mu{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{1}{2}}\right]\\ qE=m{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{1}{2}}\frac{du}{dt}+\frac{1}{2}mu{\left(1-\frac{u^2}{c^2}\right)}^{-\frac{3}{2}}\left(\frac{2u}{c^2}\right)\frac{du}{dt}\\ \frac{qE}{m}=\frac{du}{dt}\left[\frac{1}{{\left(1-{\left(u/c\right)}^2\right)}^{\frac{3}{2}}}\right]\end{array}$

Further solve the above equation.

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

The formula to calculate the acceleration is,

$a=\frac{du}{dt}$

Substitute $a$ for $\frac{du}{dt}$ in equation (1).

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

Conclusion

Therefore, the acceleration of the particle in the $x$ direction is $a=\frac{du}{dt}=\frac{qE}{m}{\left(1-\frac{u^2}{c^2}\right)}^{3/2}$.

To determine

The significance of the dependence of the acceleration on the speed.

Answer to Problem 39.88CP

The significance of the dependence of the acceleration on the speed is that when the speed of the charge is very small as compared to that of the speed of light the relative expression is transformed to the classical expression.

Explanation of Solution

The formula to calculate the acceleration of the charge is,

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

As the speed of charge approaches to the speed of light, the acceleration approaches to zero.

When the speed of the charge is very small as compared to that of the speed of the light the above equation can be transformed.

$a=\frac{du}{dt}=\frac{qE}{m}$

So the relative expression is transformed to the classical expression when the speed of the charge is very small as compared to that of the speed of the light.

Conclusion

Therefore, the significance of the dependence of the acceleration on the speed is that when the speed of the charge is very small as compared to that of the speed of light the relative expression is transformed to the classical expression.

To determine

The speed and the position of the charge particle at time $t$ .

Answer to Problem 39.88CP

The speed of the charge particle at time $t$ is $\frac{qEct}{\sqrt{m^2{c}^2+q^2{E}^2{t}^2}}$ and the position of the charge particle at time $t$ is $\frac{c}{qE}\left(\sqrt{m^2{c}^2+q^2{E}^2{t}^2}-mc\right)$ .

Explanation of Solution

The formula to calculate the acceleration of the charge is,

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

Integrate the above equation from velocity $0$ to $v$ and time $0$ to $t$ to find the total velocity.

$\begin{array}{c}{\displaystyle \underset{0}{\overset{u}{\int }}\frac{du}{{\left(1-\frac{u^2}{c^2}\right)}^{3/2}}}={\displaystyle \underset{0}{\overset{t}{\int }}\frac{qE}{m}dt}\\ \frac{u}{{\left(1-\frac{u^2}{c^2}\right)}^{1/2}}=\frac{qEt}{m}\\ u^2={\left(\frac{qEt}{m}\right)}^2\left(1-\frac{u^2}{c^2}\right)\\ u=\frac{qEct}{\sqrt{m^2{c}^2+q^2{E}^2{t}^2}}\end{array}$

Thus the speed of the particle at time $t$ is $\frac{qEct}{\sqrt{m^2{c}^2+q^2{E}^2{t}^2}}$.

The formula to calculate the position of the particle is,

$\frac{dx}{dt}=u$

Substitute $\frac{qEct}{\sqrt{m^2{c}^2+q^2{E}^2{t}^2}}$ for $u$ in above equation to find the value of $x$.

$\frac{dx}{dt}=\frac{qEct}{\sqrt{m^2{c}^2+q^2{E}^2{t}^2}}$

Integrate the above equation from position $0$ to $x$ and time $0$ to $t$ to find the final position.

$\begin{array}{c}{\displaystyle \underset{0}{\overset{x}{\int }}dx}={\displaystyle \underset{0}{\overset{t}{\int }}\frac{qEct}{\sqrt{m^2{c}^2+q^2{E}^2{t}^2}}dt}\\ x=\frac{c}{qE}{\left[\sqrt{m^2{c}^2+q^2{E}^2{t}^2}\right]}_0^t\\ =\frac{c}{qE}\left(\sqrt{m^2{c}^2+q^2{E}^2{t}^2}-mc\right)\end{array}$

Conclusion

Therefore, the speed of the charge particle at time $t$ is $\frac{qEct}{\sqrt{m^2{c}^2+q^2{E}^2{t}^2}}$ and the position of the charge particle at time $t$ is $\frac{c}{qE}\left(\sqrt{m^2{c}^2+q^2{E}^2{t}^2}-mc\right)$.