As a model for electric quadrupole radiation, consider two oppositely oriented oscillating electric dipoles, separated by a distance $d$, as shown in figure in Fig. 11.19. Use the results of Sect. 11.1.2 for the potentials of each dipole, but note that they are not located at the origin. Keeping only the terms of first order in $d$:

(a) Find the scalars and vector potentials

(b) Find the electric and magnetic fields.

(c) Find the pointing vector and the power radiated

Total scalar potential of two oppositely oriented electric dipoles is given from the section 11.1.2.

$V_{\pm }=\mp \frac{p_0\omega }{4\pi {\epsilon }_{0}c}\left(\frac{\cos {\theta }_{\pm }}{r_{\pm }}\right)\sin \left[\omega (t-\frac{r_{\pm }}{c})\right]$ …………. (1)

Total vector potential of two oppositely oriented electric dipoles is given from the section 11.1.2.

$A_{\pm }=\mp \frac{{\mu }_0{p}_0\omega }{4\pi r_{\pm }}\sin \left[\omega (t-\frac{r_{\pm }}{c})\right]\widehat{z}$

The electric field is given from the section 11.1.2.

$E=-\nabla V-\frac{\partial A}{\partial t}$ ……….…… (3)

The law of cosines is given as follows:

$r_{\pm }=\sqrt{r^2+{\left(\frac{d}{2}\right)}^2+2r\left(\frac{d}{2}\right)cos\theta }$……….…… (4)

Here, is the distance.

(a)

Apply the law of cosines in the given figure.

$\begin{array}{c}{r}_{\pm }=\sqrt{r^2+{\left(\frac{d}{2}\right)}^2+2r\left(\frac{d}{2}\right)\cos \theta }\\ \cong r\sqrt{1\mp \left(\frac{d}{r}\right)\cos \theta }\\ \cong r{(1\mp \left(\frac{d}{r}\right)\cos \theta )}^{\frac{1}{2}}\\ \cong r(1\mp \frac{d}{2r}\cos \theta )\end{array}$

It follows that

$\begin{array}{c}\frac{1}{r_{\pm }}=\frac{1}{r(1\mp \frac{d}{2r}\cos \theta )}\\ \frac{1}{r_{\pm }}=\frac{1}{r}{(1\mp \frac{d}{2r}\cos \theta )}^{-1}\\ \frac{1}{r_{\pm }}=\frac{1}{r}(1\pm \frac{d}{2r}\cos \theta )\end{array}$ …………. (5)

Now, the angle is

$\cos {\theta }_{\pm }=\frac{r\cos \mp \left(\frac{d}{2}\right)}{r_{\pm }}$

Now substitute the value of $\frac{1}{r_{\pm }}=\frac{1}{r}(1\pm \frac{d}{2r}\cos \theta )$.

role="math" localid="1658756836865" $\begin{array}{c}\cos {\theta }_{\pm }=\frac{1}{r_{\pm }}(r\cos \theta \mp \left(\frac{d}{2}\right))\\ =\frac{1}{r}(1\pm \frac{d}{2r}\cos \theta )(r\cos \theta \mp \frac{d}{2r})\\ =(\cos \theta \mp \frac{d}{2r})(1\pm \frac{d}{2r}\cos \theta )\\ =\cos \theta \pm \frac{d}{2r}{\cos }^2\theta \mp \frac{d}{2r}\end{array}$

Solve further as

$\begin{array}{c}\cos {\theta }_{\pm }=\cos \theta \mp \frac{d}{2r}(1-{\cos }^2\theta )\\ =\cos \theta \mp \frac{d}{2r}{\sin }^2\theta \end{array}$

Let us take

$\sin \left[\omega (t-\frac{r_{\pm }}{c})\right]$

Substitute the value of $r_{\pm }=r(1\mp \frac{d}{2r}\cos \theta )$.

$\begin{array}{c}\sin \left[\omega (t-\frac{r(1\mp \frac{d}{2r}\cos \theta )}{c})\right]=\sin \left[\omega (t-\frac{r}{c}(1\mp \frac{d}{2r}\cos \theta ))\right]\\ =\sin \left\{\omega [t-\frac{r}{c}\pm \frac{d}{2c}\cos \theta ]\right\}\\ =\sin \left\{\omega [(t-\frac{r}{c})\pm \frac{d}{2c}\cos \theta ]\right\}\end{array}$ ….… (6)

Put $t_0=t-\frac{r}{c}$in the equation (6).

$\begin{array}{c}\sin \left\{\omega [(t-\frac{r}{c})\pm \frac{d}{2c}\cos \theta ]\right\}=\sin \left\{\omega [t_0\pm \frac{d}{2c}\cos \theta ]\right\}\\ =\sin [\omega t_0\pm \frac{\omega d}{2c}\cos \theta ]\\ =\sin \left(\omega t_0\right)\cos \left(\frac{\omega d}{2c}\cos \theta \right)\pm \cos \left(\omega t_0\right)\sin \left(\frac{\omega d}{2c}\cos \theta \right)\\ =\sin \left(\omega t_0\right)\pm \frac{\omega d}{2c}\cos \theta \cos \left(\omega t_0\right)\end{array}$… (7)

Substituting all these values in the equation (1)

$\begin{array}{c}{V}_{\pm }=\mp \frac{p_0\omega }{4\pi {\epsilon }_{0}c}\left(\frac{\cos {\theta }_{\pm }}{r_{\pm }}\right)\sin \left[\omega (t-\frac{r_{\pm }}{c})\right]\\ V_{\pm }=\mp \frac{p_0\omega }{4\pi {\epsilon }_{0}c}\left[(\cos \theta \mp \frac{d}{2r}{\sin }^2\theta )\frac{1}{r}(1\pm \frac{d}{2r}\cos \theta )\right]\left[\sin \left({\omega }_{0}t\right)\pm \frac{\omega d}{2c}\cos \theta \cos \left(\omega t_0\right)\right]\\ V_{\pm }=\mp \frac{p_0\omega }{4\pi {\epsilon }_{0}cr}\left[(\cos \theta \mp \frac{d}{2r}{\sin }^2\theta )(1\pm \frac{d}{2r}\cos \theta )\right]\left[\sin \left({\omega }_{0}t\right)\pm \frac{\omega d}{2c}\cos \theta \cos \left(\omega t_0\right)\right]\\ V_{\pm }=\mp \frac{p_0\omega }{4\pi {\epsilon }_{0}cr}\left[(\cos \theta \pm \frac{d}{2r}{\cos }^2\theta \mp \frac{d}{2r}{\sin }^2\theta )\right]\left[\sin \left({\omega }_{0}t\right)\pm \frac{\omega d}{2c}\cos \theta \cos \left(\omega t_0\right)\right]\end{array}$

Further solved as

$\begin{array}{l}{V}_{\pm }\cong \mp \frac{p_0\omega }{4\pi {\epsilon }_{0}cr}\left(\cos \theta \sin \left({\omega }_{0}t\right)\pm \frac{\omega d}{2c}{\cos }^2\theta \cos \left(\omega t_0\right)\pm \frac{d}{2r}{\cos }^2\theta \sin \left({\omega }_{0}t\right)\mp \frac{d}{2r}{\sin }^2\theta \sin \left({\omega }_{0}t\right)\right)\\ V_{\pm }=\mp \frac{p_0\omega }{4\pi {\epsilon }_{0}cr}\left(\cos \theta \sin \left({\omega }_{0}t\right)\pm \frac{\omega d}{2c}{\cos }^2\theta \cos \left(\omega t_0\right)\pm \frac{d}{2r}({\cos }^2\theta -{\sin }^2\theta )\sin \left({\omega }_{0}t\right)\right)\end{array}$

Therefore, the expression for the total voltage is derived as

$\begin{array}{l}{V}_{tot}=V_{+}+V_{-}\\ V_{tot}=-\frac{p_0\omega }{4\pi {\epsilon }_{0}cr}[\frac{\omega d}{c}{\cos }^2\theta \cos \left({\omega }_{0}t\right)+\frac{d}{r}({\cos }^2\theta -{\sin }^2\theta )\sin \left(\omega t_0\right)]\\ V_{tot}=-\frac{p_0\omega }{4\pi {\epsilon }_{0}cr}\times \frac{\omega d}{c}[{\cos }^2\theta \cos \left({\omega }_{0}t\right)+\frac{c}{\omega d}\frac{d}{r}({\cos }^2\theta -{\sin }^2\theta )\sin \left(\omega t_0\right)]\\ V_{tot}=-\frac{p_0{\omega }^{2}d}{4\pi {\epsilon }_0{c}^{2}r}[{\cos }^2\theta \cos \left({\omega }_{0}t\right)+\frac{c}{\omega r}({\cos }^2\theta -{\sin }^2\theta )\sin \left(\omega t_0\right)]\end{array}$

In the radiation zone $r>>\frac{\omega }{c}$then the second term in the above square bracket can be neglected then the above equation becomes

$V_{tot}=-\frac{p_0{\omega }^{2}d}{4\pi {\epsilon }_0{c}^{2}r}{\cos }^2\theta \cos \left(\omega t_0\right)$ ………. (8)

Now, put $t_0=t-\frac{r}{c}$in the equation (8).

$V_{tot}=-\frac{p_0{\omega }^{2}d}{4\pi {\epsilon }_0{c}^{2}r}{\cos }^2\theta \cos \left[\omega (t-\frac{r}{c})\right]$

Hence, the scalar potential is $V_{tot}=-\frac{p_0{\omega }^{2}d}{4\pi {\epsilon }_0{c}^{2}r}{\cos }^2\theta \cos \left[\omega (t-\frac{r}{c})\right]$.

Now calculate the vector potential from equation (2).

$A_{\pm }=\mp \frac{{\mu }_0{p}_0\omega }{4\pi r_{\pm }}\sin \left[\omega (t-\frac{r_{\pm }}{c})\right]\widehat{z}$

Substitute equation (5) and (7) in equation (2).

$\begin{array}{l}{A}_{\pm }=\mp \frac{{\mu }_0{p}_0\omega }{4\pi }\times \left\{\frac{1}{r}(1\pm \frac{d}{2r}\cos \theta )[\sin \left(\omega t_0\right)\pm \frac{\omega d}{2c}\cos \theta \cos \left(\omega t_0\right)]\right\}\\ A_{\pm }=\mp \frac{{\mu }_0{p}_0\omega }{4\pi r}[\sin \left(\omega t_0\right)\pm \frac{\omega d}{2c}\cos \theta \cos \left(\omega t_0\right)\pm \frac{d}{2r}\cos \theta \sin \left(\omega t_0\right)]\widehat{z}\end{array}$

Hence, the total scalar potential is

$\begin{array}{l}{A}_{tot}=A^{+}+A^{-}\\ A_{tot}=\mp \frac{{\mu }_0{p}_0\omega }{4\pi r}[\frac{\omega d}{c}\cos \theta \cos \left(\omega t_0\right)\pm \frac{d}{r}\cos \theta \sin \left(\omega t_0\right)]\widehat{z}\\ A_{tot}=\mp \frac{{\mu }_0{p}_0{\omega }^2}{4\pi cr}[\cos \theta \cos \left(\omega t_0\right)\pm \frac{c}{r\omega }\cos \theta \sin \left(\omega t_0\right)]\widehat{z}\end{array}$

In the radiation zone $r>>\frac{\omega }{c}$then the second term in the above square bracket can be neglected then the above equation becomes

$\begin{array}{l}{A}_{tot}=-\frac{{\mu }_0{p}_0{\omega }^{2}d}{4\pi cr}\cos \theta \cos \left(\omega t_0\right)\widehat{z}\\ A_{tot}=-\frac{{\mu }_0{p}_0{\omega }^{2}d}{4\pi cr}\cos \theta \cos \left[\omega (t-\frac{r}{c})\right]\widehat{z}\end{array}$

Hence, the total vector potential is $A_{tot}=-\frac{{\mu }_0{p}_0{\omega }^{2}d}{4\pi cr}\cos \theta \cos \left[\omega (t-\frac{r}{c})\right]\widehat{z}$.

(b)

Consider the expression for the total voltage drop as

$V_{tot}=-\frac{p_0{\omega }^{2}d}{4\pi {\epsilon }_0{c}^{2}r}{\cos }^2\theta \cos \left[\omega (t-\frac{r}{c})\right]$

As we know

$c=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$

where ${\epsilon }_0$is the vacuum permittivity and ${\mu }_0$ is the vacuum permeability.

$\begin{array}{l}{V}_{tot}=-\frac{p_0{\omega }^{2}d}{4\pi {\epsilon }_0\left(\frac{1}{{\mu }_0{\epsilon }_0}\right)r}{\cos }^2\theta \cos \left[\omega (t-\frac{r}{c})\right]\\ V_{tot}=-\frac{{\mu }_0{p}_0{\omega }^{2}d}{4\pi r}{\cos }^2\theta \cos \left[\omega (t-\frac{r}{c})\right]\end{array}$

Let us consider $\alpha =-\frac{{\mu }_0{p}_0{\omega }^{2}d}{4\pi }$; then the above equation become

$\begin{array}{l}{V}_{tot}=\frac{\alpha }{r}{\cos }^2\theta \cos \left[\omega (t-\frac{r}{c})\right]\\ V_{tot}=\frac{\alpha {\cos }^2\theta }{r}\cos \left[\omega (t-\frac{r}{c})\right]\end{array}$

Now vector potential in the co-ordinate form

$\nabla V=\frac{\partial V_{tot}}{\partial r}\widehat{r}+\frac{1}{r}\frac{\partial V_{tot}}{\partial \theta }\widehat{\theta }$ ………. (9)

Put $V_{tot}=\frac{\alpha {\cos }^2\theta }{r}\cos \left[\omega (t-\frac{r}{c})\right]$ in the equation (9).

$\begin{array}{c}\nabla V=\frac{\partial \left[\frac{\alpha {\cos }^2\theta }{r}\cos \left[\omega (t-\frac{r}{c})\right]\right]}{\partial r}\widehat{r}+\frac{1}{r}\frac{\partial \left[\frac{\alpha {\cos }^2\theta }{r}\cos \left[\omega (t-\frac{r}{c})\right]\right]}{\partial \theta }\widehat{\theta }\\ \nabla V=\alpha {\cos }^2\theta \frac{\partial }{\partial r}\left[\frac{1}{r}\cos \left[\omega (t-\frac{r}{c})\right]\right]\widehat{r}+\frac{1}{r}\frac{\alpha }{r}\cos \left[\omega (t-\frac{r}{c})\right]\frac{\partial }{\partial \theta }\left[{\cos }^2\theta \right]\widehat{\theta }\\ \nabla V=\left\{\begin{array}{l}\alpha {\cos }^2\theta \{-\frac{1}{r^2}\cos \left[\omega (t-\frac{r}{c})\right]+\frac{1}{r}[-\sin \left(\omega (t-\frac{r}{c})\right)(-\frac{\omega }{c})]\}\\ +\frac{\alpha }{r^2}\cos \left[\omega (t-\frac{r}{c})\right](-2\cos \theta \sin \theta )\widehat{\theta }\end{array}\right\}\\ \nabla V=\left\{\begin{array}{l}\alpha {\cos }^2\theta \{-\frac{1}{r^2}\cos \left[\omega (t-\frac{r}{c})\right]+\frac{\omega }{rc}\left[\sin \left(\omega (t-\frac{r}{c})\right)\right]\}\\ +\frac{\alpha }{r^2}\cos \left[\omega (t-\frac{r}{c})\right](-2\cos \theta \sin \theta )\widehat{\theta }\end{array}\right\}\end{array}$

In the radiation zone

$\begin{array}{l}\Delta V=\alpha {\cos }^2\theta \frac{\omega }{rc}\left[\sin \left(\omega (t-\frac{r}{c})\right)\right]\widehat{r}\\ \Delta V=\alpha \frac{\omega }{c}\frac{{\cos }^2\theta }{r}\left[\sin \left(\omega (t-\frac{r}{c})\right)\right]\widehat{r}\end{array}$ ……… (10)

Since

$A=\frac{\alpha }{c}\frac{\cos \theta }{r}\cos \left[\omega (t-\frac{r}{c})\right](\cos \theta \widehat{r}-\sin \theta \widehat{\theta })$

Differentiate the equation with respect to time .

$\begin{array}{l}\frac{\partial A}{\partial t}=\frac{\partial }{\partial t}\left\{\frac{\alpha }{c}\frac{\cos \theta }{r}\cos \left[\omega (t-\frac{r}{c})\right](\cos \theta \widehat{r}-\sin \theta \widehat{\theta })\right\}\\ \frac{\partial A}{\partial t}=\frac{\alpha }{c}\frac{\cos \theta }{r}\{-\sin \left[\omega (t-\frac{r}{c})\left(\omega \right)\right](\cos \theta \widehat{r}-\sin \theta \widehat{\theta })\}\\ \frac{\partial A}{\partial t}=-\frac{\alpha \omega }{c}\frac{\cos \theta }{r}\sin \left[\omega (t-\frac{r}{c})\left(\omega \right)\right](\cos \theta \widehat{r}-\sin \theta \widehat{\theta })\end{array}$

From equation (3)

$E=-\nabla V-\frac{\partial A}{\partial t}$ ……… (11)

Put the value of $\Delta V$ and $\frac{\partial A}{\partial t}$in equation (11).

$\begin{array}{l}E=-\alpha \frac{\omega }{c}\frac{{\cos }^2\theta }{r}\left[\sin \left(\omega (t-\frac{r}{c})\right)\right]\widehat{r}+\frac{\alpha \omega }{c}\frac{\cos \theta }{r}\sin \left[\omega (t-\frac{r}{c})\left(\omega \right)\right](\cos \theta \widehat{r}-\sin \theta \widehat{\theta })\\ E=\frac{\alpha \omega }{c}\frac{\cos \theta }{r}\left[\sin \omega (t-\frac{r}{c})\right](\cos \theta \widehat{r}-\sin \theta \widehat{\theta })-\frac{\alpha \omega }{c}\frac{{\cos }^2\theta }{r}\sin \left[\omega (t-\frac{r}{c})\right]\widehat{r}\\ E=\frac{\alpha \omega }{c}\frac{\cos \theta }{r}\left[\sin \omega (t-\frac{r}{c})\right][\cos \theta \widehat{r}-\sin \theta \widehat{\theta }-\cos \theta \widehat{r}]\\ E=\frac{\alpha \omega }{c}\frac{\cos \theta }{r}\left[\sin \omega (t-\frac{r}{c})\right][-\sin \theta \widehat{\theta }]\end{array}$

Further solve as

$\begin{array}{l}E=\frac{\alpha \omega }{c}\frac{\cos \theta \sin \theta }{r}\left[\sin \omega (t-\frac{r}{c})\right]\widehat{\theta }\\ E=-\frac{\alpha \omega }{cr}\cos \theta \sin \theta \left[\sin \omega (t-\frac{r}{c})\right]\widehat{\theta }\end{array}$ ……… (12)

Thus, the electric field is .$E=-\frac{\alpha \omega }{cr}\cos \theta \sin \theta \left[\sin \omega (t-\frac{r}{c})\right]\widehat{\theta }$

The magnetic field is given by

$\begin{array}{l}B=\nabla \times A\\ B=\frac{1}{r}[\frac{\partial }{\partial r}\left(r{A}_0\right)-\frac{\partial A_r}{\partial \theta }]\widehat{\varphi }\\ B=\frac{\alpha }{cr}\{\frac{\partial }{\partial r}\left[\cos \theta \cos \left(\omega (t-\frac{r}{c})\right)(-\sin \theta )\right]-\frac{\partial }{\partial \theta }\frac{{\cos }^2\theta }{r}\cos \left(\omega (t-\frac{r}{c})\right)\}\widehat{\varphi }\\ B=\frac{\alpha }{cr}\left\{(-\sin \theta \cos \theta )[-\sin \left(\omega (t-\frac{r}{c})\right)]\left(\frac{-\omega }{c}\right)-\frac{\partial }{\partial \theta }\frac{{\cos }^2\theta }{r}\cos \left(\omega (t-\frac{r}{c})\right)\right\}\widehat{\varphi }\end{array}$

Further solved as

$B=\frac{\alpha }{cr}\left\{(-\sin \theta \cos \theta )\frac{\omega }{c}\sin \left(\omega (t-\frac{r}{c})\right)-\frac{\partial }{\partial \theta }\frac{{\cos }^2\theta }{r}\cos \left(\omega (t-\frac{r}{c})\right)\right\}\widehat{\varphi }$

Consider the radiation zone.

$\begin{array}{l}B=\frac{\alpha }{cr}(-\sin \theta \cos \theta )\frac{\omega }{c}\sin \left[\omega (t-\frac{r}{c})\right]\widehat{\varphi }\\ B=-\frac{\alpha \omega }{c^{2}r}\sin \theta \cos \theta \sin \left[\omega (t-\frac{r}{c})\right]\widehat{\varphi }\end{array}$

Thus, the magnetic field is .$B=-\frac{\alpha \omega }{c^{2}r}\sin \theta \cos \theta \sin \left[\omega (t-\frac{r}{c})\right]\widehat{\varphi }$

(c)

Consider the formula for the magnetic field as

$B=\frac{1}{c}(\widehat{r}\times E)$

Here, $E\cdot \widehat{r}=0$.

Then, the pointing vector is given as

$\begin{array}{l}S=\frac{1}{{\mu }_0}(E\times B)\\ S=\frac{1}{{\mu }_0}[E\times \frac{1}{c}(\widehat{r}\times E)]\\ S=\frac{1}{{\mu }_{0}c}[E\times (\widehat{r}\times E)]\\ S=\frac{1}{{\mu }_{0}c}(E^2\widehat{r}-(E.\widehat{r})E)\end{array}$

Substitute $E.\widehat{r}=0$ in the above expression.

$\begin{array}{l}S=\frac{1}{{\mu }_{0}c}[E^2\widehat{r}-\left(0\right)E]\\ S=\frac{E^2}{{\mu }_{0}c}\widehat{r}\end{array}$

Substitute equation (12) in the expression.

$S=\frac{{\{-\frac{\alpha \omega }{cr}\cos \theta \sin \theta \left[\sin \omega (t-\frac{r}{c})\right]\}}^2}{{\mu }_{0}c}\widehat{r}$

Thus, the pointing vector is

$S=\frac{{\{-\frac{\alpha \omega }{cr}\cos \theta \sin \theta \left[\sin \omega (t-\frac{r}{c})\right]\}}^2}{{\mu }_{0}c}\widehat{r}$

The power radiated is given as

$\begin{array}{c}P=\int ⟨S⟩.da\\ =\frac{1}{{\mu }_{0}c}{\left(\frac{\alpha \omega }{c}\right)}^2\int {\sin }^2\theta {\cos }^2\theta \sin \theta d\theta d\varphi \\ =\frac{1}{2{\mu }_{0}c}{\left(\frac{\alpha \omega }{c}\right)}^2\left(2\pi \right)\underset{0}{\overset{\pi }{\int }}(1-{\cos }^2\theta ){\cos }^2\theta \sin \theta d\theta \end{array}$

Let$u=\cos \theta$; then $du=-\sin \theta d\theta$.

Upper limit is $\theta =\pi$; then

$\begin{array}{l}u=\cos \pi \\ u=-1\end{array}$

Lower limit is $\theta =0$; then

$\begin{array}{l}u=\cos 0\\ u=1\end{array}$

Therefore

$\begin{array}{c}\underset{1}{\overset{-1}{\int }}(1-u^2)u^2(-du)=\underset{-1}{\overset{1}{\int }}(1-u^2)u^2\left(du\right)\\ =\underset{-1}{\overset{1}{\int }}(u^2-u^4)\left(du\right)\\ =\underset{-1}{\overset{1}{\int }}{u}^2\left(du\right)-\underset{-1}{\overset{1}{\int }}{u}^4\left(du\right)\end{array}$

Now integrate the expression

$\begin{array}{c}\underset{-1}{\overset{1}{\int }}{u}^2\left(du\right)-\underset{-1}{\overset{1}{\int }}{u}^4\left(du\right)={{\left[\frac{u^3}{3}\right]}^{+1}}_{-1}-{{\left[\frac{u^5}{5}\right]}_{-1}}^{+1}\\ =[\frac{1}{3}-(-\frac{1}{3})]-[\frac{1}{5}-(-\frac{1}{5})]\\ =\frac{2}{3}-\frac{2}{5}\\ =\frac{4}{15}\end{array}$

Now, power radiated is

$\begin{array}{l}P=\frac{1}{2{\mu }_{0}c}\frac{{\omega }^2}{c}\left(2\pi \right)\left(\frac{4}{15}\right)\\ P=\frac{1}{2{\mu }_{0}c}\frac{{\omega }^2}{c}\left(2\pi \right)\left(\frac{4}{15}\right){\alpha }^2\end{array}$

Since $\alpha =-\frac{{\mu }_0{p}_0{\omega }^{2}d}{4\pi }$, then the above equation becomes

$\begin{array}{c}P=\frac{1}{2{\mu }_{0}c}\frac{{\omega }^2}{c}\left(2\pi \right)\left(\frac{4}{15}\right){[-\frac{{\mu }_0{p}_0{\omega }^{2}d}{4\pi }]}^2\\ =\frac{1}{2{\mu }_{0}c}\frac{{\omega }^2}{c}\left(2\pi \right)\left(\frac{4}{15}\right)\frac{{{\mu }_0}^2{p_0}^2{\omega }^4{d}^2}{16{\pi }^2}\\ =\frac{{\mu }_0}{60\pi c^3}{\left(p_{0}d\right)}^2{\omega }^6\end{array}$

Thus, the power radiated is $P=\frac{{\mu }_0}{60\pi c^3}{\left(p_{0}d\right)}^2{\omega }^6$.