A particle of mass m and charge q is attached to a spring with force constant k, hanging from the ceiling (Fig. 11.18). Its equilibrium position is a distance h above the floor. It is pulled down a distance d below equilibrium and released, at time$\mathit{t}\mathbf{=}\mathbf{0}$.

(a) Under the usual assumptions $(d\ll \lambda \ll h)$, calculate the intensity of the radiation hitting the floor, as a function of the distance R from the point directly below q. [Note: The intensity here is the average power per unit area of floor.]

FIGURE 11.18

At whatR is the radiation most intense? Neglect the radiative damping of the oscillator.

(b) As a check on your formula, assume the floor is of infinite extent, and calculate the average energy per unit time striking the entire floor. Is it what you’d expect?

(c) Because it is losing energy in the form of radiation, the amplitude of the oscillation will gradually decrease. After what time$\mathcal{b}$has the amplitude been reduced to $\mathit{d}\mathbf{/}\mathit{e}$? (Assume the fraction of the total energy lost in one cycle is very small.)

Write the expression for the intensity of radiation.

$\mathit{I}\mathbf{=}<S>\mathbf{·}\hat{\mathbf{z}}$ …… (1)

Here, is the Poynting vector which is given as:

Substitute in equation (1).

role="math" localid="1653990441797" $$\begin{gathered} I=\left(\frac{{\mu }_0{p}_0^2{\omega }^4}{32{\pi }^{2}c}\right)\frac{{\sin }^2\theta }{r^2}\left(\hat{r}·\hat{z}\right) \\ I=\left(\frac{{\mu }_0{p}_0^2{\omega }^4}{32{\pi }^{2}c}\right)\frac{{\sin }^2\theta }{r^2}\cos \theta \end{gathered}$$ …… (2)

Here, $p_0=qd$.

(a)

Draw the given situation of the charged particle.

From the above figure, observe the value of r, $\cos \theta$and $\sin \theta$.

$\begin{array}{rcl}r& =& \sqrt{R^2+h^2}\\ \cos \theta & =& \frac{h}{r}\\ \sin \theta & =& \frac{R}{r}\end{array}$

Substitute $p_0=qd,\cos \theta =\frac{h}{r},\sin \theta =\frac{R}{r}$and$\cos \theta =\frac{h}{r}$ in equation (2).

$$\begin{gathered} I=\left(\frac{{\mu }_0{\left(qd\right)}^2{\omega }^4}{32{\pi }^{2}c}\right)\frac{{\left({\displaystyle \frac{R}{\sqrt{R^2+h^2}}}\right)}^2}{{\left(\sqrt{R^2+h^2}\right)}^2}\left(\frac{h}{\sqrt{R^2+h^2}}\right) \\ I=\left(\frac{{\mu }_0{\left(qd\right)}^2{\omega }^4}{32{\pi }^{2}c}\right){\left(\frac{R}{\sqrt{R^2+h^2}}\right)}^2\left(\frac{1}{{\left(\sqrt{R^2+h^2}\right)}^2}\right)\left(\frac{h}{\sqrt{R^2+h^2}}\right) \\ I=\frac{{\mu }_0{q}^2{d}^2{\omega }^4{R}^{2}h}{32{\pi }^{2}c{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}} \end{gathered}$$

For the maximum intensity,

$\frac{dI}{dR}=0$

Hence, the intensity equation becomes,

$$\begin{gathered} \frac{d}{dR}\left(\frac{{\mu }_0{q}^2{d}^2{\omega }^4{R}^{2}h}{32{\pi }^{2}c{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)=0 \\ \frac{d}{dR}\left(\frac{R^2}{{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)=0 \\ \frac{{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}\left(2R\right)-R^2\left({\displaystyle \frac{5}{2}}\right){\left(R^2+h^2\right)}^{{\displaystyle \frac{3}{2}}}\left(2R\right)}{{\left(R^2+h^2\right)}^5}=0 \\ R=\left(\sqrt{\frac{2}{3}}\right)h \end{gathered}$$

Therefore, the intensity of the radiation hitting the floor is $\frac{{\mu }_0{q}^2{d}^2{\omega }^4{R}^{2}h}{32{\pi }^{2}c{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}$, and at$R=\left(\sqrt{\frac{2}{3}}\right)h$ the radiation will be the most intense.

(b)

Write the expression for the average energy striking per unit time.

$$\begin{gathered} P=\int I\left(R\right)da \\ P={\int }_0^{\infty }I\left(R\right)\left(2\pi R\right)dR \end{gathered}$$

Substitute$I=\frac{{\mu }_0{q}^2{d}^2{\omega }^4{R}^{2}h}{32{\pi }^{2}c{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}$ in the above expression.

$$\begin{gathered} P={\int }_0^{\infty }\left(\frac{{\mu }_0{q}^2{d}^2{\omega }^4{R}^{2}h}{32{\pi }^{2}c{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)\left(2\pi RdR\right) \\ P=2\pi \left(\frac{{\mu }_0{q}^2{d}^2{\omega }^{4}h}{32{\pi }^{2}c}\right){\int }_0^{\infty }\left(\frac{R^3}{{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)dR \end{gathered}$$

Let,

$$\begin{gathered} R^2=x \\ 2RdR=dx \\ RdR=\frac{dx}{2} \end{gathered}$$

Hence, the above equation becomes,

$$\begin{gathered} P=2\pi \left(\frac{{\mu }_0{q}^2{d}^2{\omega }^{4}h}{32{\pi }^{2}c}\right){\int }_0^{\infty }\left(\frac{R^3}{{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)dR \\ P=2\pi \left(\frac{{\mu }_0{q}^2{d}^2{\omega }^{4}h}{32{\pi }^{2}c}\right)\left(\frac{2}{3h}\right) \\ P=\frac{{\mu }_0{q}^2{d}^2{\omega }^4}{24\pi c} \end{gathered}$$

The above equation is half of the total radiated power.

Therefore, considering the floor to be striking in the infinite extent, the average energy per unit time striking the floor is $\frac{{\mu }_0{q}^2{d}^2{\omega }^4}{24\pi c}$.

(c)

Consider the expression for the amplitude of the oscillation as a function of time is as follows:

$x_0\left(t\right)$

Write the expression for the power radiated in terms of potential energy.

$\frac{dU}{dt}=-2P$ …… (3)

Here, $U=\frac{1}{2}k{x}_0^2$.

Substitute$U=\frac{1}{2}k{x}_0^2$and$P=\frac{{\mu }_0{q}^2{d}^2{\omega }^4}{24\pi c}$in equation (3).

$$\begin{gathered} \frac{d}{dt}\left(\frac{1}{2}k{x}_0^2\right)=-2\left(\frac{{\mu }_0{q}^2{x}_0^2{\omega }^4}{24\pi c}\right) \\ \frac{d}{dt}\left(x_0^2\right)=-\left(\frac{{\mu }_0{q}^2{x}_0^2{\omega }^4}{6\pi ck}\right)x_0^2 \end{gathered}$$

Let$b=\left(\frac{{\mu }_0{q}^2{\omega }^4}{6\pi ck}\right)$

Hence, the above equation becomes,

$\frac{d}{dt}\left(x_0^2\right)=-b{x}_0^2$

Write the solution for the above equation.

$$\begin{gathered} x_0^2=d^2{e}^{-bt} \\ {x}_0=d{e}^{\frac{b}{2}t} \end{gathered}$$

For the amplitude$x_0=\frac{d}{e}$write the equation as,

$$\begin{gathered} \frac{d}{e}=d{e}^{-\frac{b}{2}t} \\ {e}^{-1}=e^{-\frac{b}{2}t} \\ t=\frac{2}{b} \end{gathered}$$

Substitute the value of bin the above expression.

$$\begin{gathered} t=\frac{2}{\left({\displaystyle \frac{{\mu }_0{q}^2{\omega }^4}{6\pi ck}}\right)} \\ t=\frac{12\pi ck}{{\mu }_0{q}^2{\omega }^4} \end{gathered}$$

Here,$\sqrt{\frac{k}{m}}=\omega$ rewrite the equation as,

$$\begin{gathered} t=\frac{12\pi ck}{{\mu }_0{q}^2{\left(\sqrt{{\displaystyle \frac{k}{m}}}\right)}^4} \\ t=\frac{12\pi c{m}^2}{{\mu }_0{q}^{2}k} \end{gathered}$$

Therefore, the time after which the amplitude becomes$\frac{d}{e}$ is equals to $\frac{12\pi c{m}^2}{{\mu }_0{q}^{2}k}$.