A positive charge q is fired head-on at a distant positive charge Q (which is held stationary), with an initial velocity${\mathit{v}}_{\mathbf{0}}$ . It comes in, decelerates to $\mathit{v}\mathbf{=}\mathbf{0}$, and returns out to infinity. What fraction of its initial energy$\left(\frac{1}{2}m{v}_0^2\right)$ is radiated away? Assume ${\mathit{v}}_{\mathbf{0}}\mathbf{\ll }\mathit{c}$, and that you can safely ignore the effect of radiative losses on the motion of the particle. [ Answer $\left(\frac{16}{45}\right)\left(\frac{q}{Q}\right){\left(\frac{v_0}{c}\right)}^{\mathbf{3}}$. ]

Write the expression for the relation of conservation of energy for the point charge.

$\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{q}\mathbf{Q}}{\mathbf{4}\mathbf{\Pi }{\mathbf{\epsilon }}_{\mathbf{0}}\mathbf{x}}$ …… (1)

Here, v is the final velocity, x is the distance of closest approach towards charge Q, m is the mass of a charge,$v_0$ is the initial velocity of the charge q,${\epsilon }_0$ is the permittivity of free space, q is the fired charge, and Q is the stationary charge.

Rearrange the equation (1),

$\begin{array}{rcl}\frac{1}{2}m{v}^2& =& \frac{1}{2}m{v}_0^2-\frac{qQ}{4\pi {\epsilon }_{0}x}\\ v^2& =& v_0^2-\frac{2qQ}{4m\pi {\epsilon }_{0}x}......\left(2\right)\end{array}$

Here,$x_0=\frac{2qQ}{4m\pi {\epsilon }_0{v}_0^2}$

Let,

$k=\frac{qQ}{4m\pi {\epsilon }_0}$

Hence, the above equation becomes,

And$x_0=\frac{2k}{v_0^2}$

Substitute $k=\frac{qQ}{4m\pi {\epsilon }_0}$in equation (2).

$v=\sqrt{v_0^2-\frac{2k}{x}}$

Using Larmor’s formula, write the expression for the total power radiated.

$dW=\frac{{\mu }_0{q}^2{a}^2}{6\pi c}dt$ …… (3)

Here, a is the acceleration of the charge due to Coulomb’s repulsive force, which is given as:

$a=\frac{k}{x^2}$

Substitute $a=\frac{k}{x^2}$in equation (2).

$$\begin{gathered} dW=\frac{{\mu }_0{q}^2{\left({\displaystyle \frac{k}{x^2}}\right)}^2}{6\pi x}dt \\ W=\frac{{\mu }_0{q}^2}{6\pi c}\int \frac{k^2}{x^4}dt \\ W=\frac{{\mu }_0{q}^2{k}^2}{6\pi c}\int \frac{1}{x^4}\frac{dx}{v} \\ W=\frac{{\mu }_0{q}^2{k}^2}{6\pi c}\int \frac{1}{x^4}\frac{dx}{\sqrt{v_0^2-{\displaystyle \frac{2k}{x}}}} \end{gathered}$$

On further solving, the above equation becomes,

$$\begin{gathered} W=\frac{{\mu }_0{q}^2{k}^2}{6\pi c}\int \frac{1}{x^4}\frac{dx}{\sqrt{{\displaystyle \frac{2k}{x_0}}-{\displaystyle \frac{2k}{x}}}} \\ W=\frac{{\mu }_0{q}^2{k}^2}{6\pi c}{\int }_{x_0}^{\infty }\frac{1}{x^4}\frac{dx}{\sqrt{{\displaystyle \frac{2k}{x_0}}-{\displaystyle \frac{2k}{x}}}} \\ W=\frac{2{\mu }_0{q}^2{k}^2}{6\pi c\sqrt{2k}}{\int }_{x_0}^{\infty }\frac{dx}{x^4\sqrt{{\displaystyle \frac{1}{x_0}}-{\displaystyle \frac{1}{x}}}} \\ W=\frac{2{\mu }_0{q}^2{k}^2}{6\pi c\sqrt{2k}}\frac{16}{{\left(15x_0\right)}^{5/2}} \end{gathered}$$

Again on further solving,

$$\begin{gathered} W=\frac{{\mu }_0{q}^2\sqrt{2}{k}^{3/2}}{6\pi c}\frac{16}{{\left(15x_0\right)}^{5/2}} \\ W=\frac{2^{2}k}{2^{2}k}\left(\frac{{\mu }_0{q}^2\sqrt{2}{k}^{3/2}}{6\pi c}\frac{16}{{\left(15x_0\right)}^{5/2}}\right) \\ W=\frac{{\mu }_0{q}^2}{24\pi ck}\frac{16}{15}{\left(\frac{2k}{x_0}\right)}^{5/2} \\ W=\frac{{\mu }_0{q}^2}{24\pi ck}\frac{16}{15}{v}_0^5 \end{gathered}$$

Substitute the value of k in the above expression.

$$\begin{gathered} W=\frac{{\mu }_0{q}^2}{24\pi c\left({\displaystyle \frac{qQ}{4m\pi {\epsilon }_0}}\right)}\frac{16}{15}{v}_0^5 \\ W=\frac{8{\mu }_0{\epsilon }_{0}q}{45cQ}{v}_0^5 \\ W=\frac{8qm{v}_0^5}{45c^{3}Q} \end{gathered}$$

Calculate the fractional of the initial energy of charge q.

$$\begin{gathered} f=\frac{W}{W_i} \\ f=\frac{\left({\displaystyle \frac{8qm{v}_0^5}{45c^{3}Q}}\right)}{\left({\displaystyle \frac{1}{2}}m{v}_0^2\right)} \\ f=\left(\frac{8qm{v}_0^5}{45c^{3}Q}\right)\times \left(\frac{2}{m{v}_0^2}\right) \\ f=\left(\frac{16}{45}\right)\left(\frac{q}{Q}\right){\left(\frac{v_0}{c}\right)}^3 \end{gathered}$$

Therefore, the fraction of the radiated initial energy is $\left(\frac{16}{45}\right)\left(\frac{q}{Q}\right){\left(\frac{v_0}{c}\right)}^3$.