91影视

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

$V_{卤}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}c}\left(\frac{\cos {胃}_{卤}}{r_{卤}}\right)\sin \left[蝇(t鈭�\frac{r_{卤}}{c})\right]$ 鈥︹�︹�︹��. (1)

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

$A_{卤}=鈭�\frac{{渭}_0{p}_0蝇}{4蟺r_{卤}}\sin \left[蝇(t鈭�\frac{r_{卤}}{c})\right]\widehat{z}$

The electric field is given from the section 11.1.2.

$E=鈭�鈭�V鈭�\frac{鈭�A}{鈭�t}$ 鈥︹�︹��.鈥︹�� (3)

The law of cosines is given as follows:

$r_{卤}=\sqrt{r^2+{\left(\frac{d}{2}\right)}^2+2r\left(\frac{d}{2}\right)cos胃}$鈥︹�︹��.鈥︹�� (4)

Here, is the distance.

(a)

Apply the law of cosines in the given figure.

$\begin{array}{c}{r}_{卤}=\sqrt{r^2+{\left(\frac{d}{2}\right)}^2+2r\left(\frac{d}{2}\right)\cos 胃}\\ 鈮�r\sqrt{1鈭�\left(\frac{d}{r}\right)\cos 胃}\\ 鈮�r{(1鈭�\left(\frac{d}{r}\right)\cos 胃)}^{\frac{1}{2}}\\ 鈮�r(1鈭�\frac{d}{2r}\cos 胃)\end{array}$

It follows that

$\begin{array}{c}\frac{1}{r_{卤}}=\frac{1}{r(1鈭�\frac{d}{2r}\cos 胃)}\\ \frac{1}{r_{卤}}=\frac{1}{r}{(1鈭�\frac{d}{2r}\cos 胃)}^{鈭�1}\\ \frac{1}{r_{卤}}=\frac{1}{r}(1卤\frac{d}{2r}\cos 胃)\end{array}$ 鈥︹�︹�︹��. (5)

Now, the angle is

$\cos {胃}_{卤}=\frac{r\cos 鈭�\left(\frac{d}{2}\right)}{r_{卤}}$

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

role="math" localid="1658756836865" $\begin{array}{c}\cos {胃}_{卤}=\frac{1}{r_{卤}}(r\cos 胃鈭�\left(\frac{d}{2}\right))\\ =\frac{1}{r}(1卤\frac{d}{2r}\cos 胃)(r\cos 胃鈭�\frac{d}{2r})\\ =(\cos 胃鈭�\frac{d}{2r})(1卤\frac{d}{2r}\cos 胃)\\ =\cos 胃卤\frac{d}{2r}{\cos }^2胃鈭�\frac{d}{2r}\end{array}$

Solve further as

$\begin{array}{c}\cos {胃}_{卤}=\cos 胃鈭�\frac{d}{2r}(1鈭�{\cos }^2胃)\\ =\cos 胃鈭�\frac{d}{2r}{\sin }^2胃\end{array}$

Let us take

$\sin \left[蝇(t鈭�\frac{r_{卤}}{c})\right]$

Substitute the value of $r_{卤}=r(1鈭�\frac{d}{2r}\cos 胃)$.

$\begin{array}{c}\sin \left[蝇(t鈭�\frac{r(1鈭�\frac{d}{2r}\cos 胃)}{c})\right]=\sin \left[蝇(t鈭�\frac{r}{c}(1鈭�\frac{d}{2r}\cos 胃))\right]\\ =\sin \left\{蝇[t鈭�\frac{r}{c}卤\frac{d}{2c}\cos 胃]\right\}\\ =\sin \left\{蝇[(t鈭�\frac{r}{c})卤\frac{d}{2c}\cos 胃]\right\}\end{array}$ 鈥�.鈥� (6)

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

$\begin{array}{c}\sin \left\{蝇[(t鈭�\frac{r}{c})卤\frac{d}{2c}\cos 胃]\right\}=\sin \left\{蝇[t_0卤\frac{d}{2c}\cos 胃]\right\}\\ =\sin [蝇t_0卤\frac{蝇d}{2c}\cos 胃]\\ =\sin \left(蝇t_0\right)\cos \left(\frac{蝇d}{2c}\cos 胃\right)卤\cos \left(蝇t_0\right)\sin \left(\frac{蝇d}{2c}\cos 胃\right)\\ =\sin \left(蝇t_0\right)卤\frac{蝇d}{2c}\cos 胃\cos \left(蝇t_0\right)\end{array}$鈥� (7)

Substituting all these values in the equation (1)

$\begin{array}{c}{V}_{卤}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}c}\left(\frac{\cos {胃}_{卤}}{r_{卤}}\right)\sin \left[蝇(t鈭�\frac{r_{卤}}{c})\right]\\ V_{卤}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}c}\left[(\cos 胃鈭�\frac{d}{2r}{\sin }^2胃)\frac{1}{r}(1卤\frac{d}{2r}\cos 胃)\right]\left[\sin \left({蝇}_{0}t\right)卤\frac{蝇d}{2c}\cos 胃\cos \left(蝇t_0\right)\right]\\ V_{卤}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}cr}\left[(\cos 胃鈭�\frac{d}{2r}{\sin }^2胃)(1卤\frac{d}{2r}\cos 胃)\right]\left[\sin \left({蝇}_{0}t\right)卤\frac{蝇d}{2c}\cos 胃\cos \left(蝇t_0\right)\right]\\ V_{卤}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}cr}\left[(\cos 胃卤\frac{d}{2r}{\cos }^2胃鈭�\frac{d}{2r}{\sin }^2胃)\right]\left[\sin \left({蝇}_{0}t\right)卤\frac{蝇d}{2c}\cos 胃\cos \left(蝇t_0\right)\right]\end{array}$

Further solved as

$\begin{array}{l}{V}_{卤}鈮�鈭�\frac{p_0蝇}{4蟺{蔚}_{0}cr}\left(\cos 胃\sin \left({蝇}_{0}t\right)卤\frac{蝇d}{2c}{\cos }^2胃\cos \left(蝇t_0\right)卤\frac{d}{2r}{\cos }^2胃\sin \left({蝇}_{0}t\right)鈭�\frac{d}{2r}{\sin }^2胃\sin \left({蝇}_{0}t\right)\right)\\ V_{卤}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}cr}\left(\cos 胃\sin \left({蝇}_{0}t\right)卤\frac{蝇d}{2c}{\cos }^2胃\cos \left(蝇t_0\right)卤\frac{d}{2r}({\cos }^2胃鈭�{\sin }^2胃)\sin \left({蝇}_{0}t\right)\right)\end{array}$

Therefore, the expression for the total voltage is derived as

$\begin{array}{l}{V}_{tot}=V_{+}+V_{鈭�}\\ V_{tot}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}cr}[\frac{蝇d}{c}{\cos }^2胃\cos \left({蝇}_{0}t\right)+\frac{d}{r}({\cos }^2胃鈭�{\sin }^2胃)\sin \left(蝇t_0\right)]\\ V_{tot}=鈭�\frac{p_0蝇}{4蟺{蔚}_{0}cr}脳\frac{蝇d}{c}[{\cos }^2胃\cos \left({蝇}_{0}t\right)+\frac{c}{蝇d}\frac{d}{r}({\cos }^2胃鈭�{\sin }^2胃)\sin \left(蝇t_0\right)]\\ V_{tot}=鈭�\frac{p_0{蝇}^{2}d}{4蟺{蔚}_0{c}^{2}r}[{\cos }^2胃\cos \left({蝇}_{0}t\right)+\frac{c}{蝇r}({\cos }^2胃鈭�{\sin }^2胃)\sin \left(蝇t_0\right)]\end{array}$

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

$V_{tot}=鈭�\frac{p_0{蝇}^{2}d}{4蟺{蔚}_0{c}^{2}r}{\cos }^2胃\cos \left(蝇t_0\right)$ 鈥︹�︹��. (8)

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

$V_{tot}=鈭�\frac{p_0{蝇}^{2}d}{4蟺{蔚}_0{c}^{2}r}{\cos }^2胃\cos \left[蝇(t鈭�\frac{r}{c})\right]$

Hence, the scalar potential is $V_{tot}=鈭�\frac{p_0{蝇}^{2}d}{4蟺{蔚}_0{c}^{2}r}{\cos }^2胃\cos \left[蝇(t鈭�\frac{r}{c})\right]$.

Now calculate the vector potential from equation (2).

$A_{卤}=鈭�\frac{{渭}_0{p}_0蝇}{4蟺r_{卤}}\sin \left[蝇(t鈭�\frac{r_{卤}}{c})\right]\widehat{z}$

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

$\begin{array}{l}{A}_{卤}=鈭�\frac{{渭}_0{p}_0蝇}{4蟺}脳\left\{\frac{1}{r}(1卤\frac{d}{2r}\cos 胃)[\sin \left(蝇t_0\right)卤\frac{蝇d}{2c}\cos 胃\cos \left(蝇t_0\right)]\right\}\\ A_{卤}=鈭�\frac{{渭}_0{p}_0蝇}{4蟺r}[\sin \left(蝇t_0\right)卤\frac{蝇d}{2c}\cos 胃\cos \left(蝇t_0\right)卤\frac{d}{2r}\cos 胃\sin \left(蝇t_0\right)]\widehat{z}\end{array}$

Hence, the total scalar potential is

$\begin{array}{l}{A}_{tot}=A^{+}+A^{鈭�}\\ A_{tot}=鈭�\frac{{渭}_0{p}_0蝇}{4蟺r}[\frac{蝇d}{c}\cos 胃\cos \left(蝇t_0\right)卤\frac{d}{r}\cos 胃\sin \left(蝇t_0\right)]\widehat{z}\\ A_{tot}=鈭�\frac{{渭}_0{p}_0{蝇}^2}{4蟺cr}[\cos 胃\cos \left(蝇t_0\right)卤\frac{c}{r蝇}\cos 胃\sin \left(蝇t_0\right)]\widehat{z}\end{array}$

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

$\begin{array}{l}{A}_{tot}=鈭�\frac{{渭}_0{p}_0{蝇}^{2}d}{4蟺cr}\cos 胃\cos \left(蝇t_0\right)\widehat{z}\\ A_{tot}=鈭�\frac{{渭}_0{p}_0{蝇}^{2}d}{4蟺cr}\cos 胃\cos \left[蝇(t鈭�\frac{r}{c})\right]\widehat{z}\end{array}$

Hence, the total vector potential is $A_{tot}=鈭�\frac{{渭}_0{p}_0{蝇}^{2}d}{4蟺cr}\cos 胃\cos \left[蝇(t鈭�\frac{r}{c})\right]\widehat{z}$.

(b)

Consider the expression for the total voltage drop as

$V_{tot}=鈭�\frac{p_0{蝇}^{2}d}{4蟺{蔚}_0{c}^{2}r}{\cos }^2胃\cos \left[蝇(t鈭�\frac{r}{c})\right]$

As we know

$c=\frac{1}{\sqrt{{渭}_0{蔚}_0}}$

where ${蔚}_0$is the vacuum permittivity and ${渭}_0$ is the vacuum permeability.

$\begin{array}{l}{V}_{tot}=鈭�\frac{p_0{蝇}^{2}d}{4蟺{蔚}_0\left(\frac{1}{{渭}_0{蔚}_0}\right)r}{\cos }^2胃\cos \left[蝇(t鈭�\frac{r}{c})\right]\\ V_{tot}=鈭�\frac{{渭}_0{p}_0{蝇}^{2}d}{4蟺r}{\cos }^2胃\cos \left[蝇(t鈭�\frac{r}{c})\right]\end{array}$

Let us consider $伪=鈭�\frac{{渭}_0{p}_0{蝇}^{2}d}{4蟺}$; then the above equation become

$\begin{array}{l}{V}_{tot}=\frac{伪}{r}{\cos }^2胃\cos \left[蝇(t鈭�\frac{r}{c})\right]\\ V_{tot}=\frac{伪{\cos }^2胃}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right]\end{array}$

Now vector potential in the co-ordinate form

$鈭�V=\frac{鈭�V_{tot}}{鈭�r}\widehat{r}+\frac{1}{r}\frac{鈭�V_{tot}}{鈭�胃}\widehat{胃}$ 鈥︹�︹��. (9)

Put $V_{tot}=\frac{伪{\cos }^2胃}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right]$ in the equation (9).

$\begin{array}{c}鈭�V=\frac{鈭�\left[\frac{伪{\cos }^2胃}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right]\right]}{鈭�r}\widehat{r}+\frac{1}{r}\frac{鈭�\left[\frac{伪{\cos }^2胃}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right]\right]}{鈭�胃}\widehat{胃}\\ 鈭�V=伪{\cos }^2胃\frac{鈭�}{鈭�r}\left[\frac{1}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right]\right]\widehat{r}+\frac{1}{r}\frac{伪}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right]\frac{鈭�}{鈭�胃}\left[{\cos }^2胃\right]\widehat{胃}\\ 鈭�V=\left\{\begin{array}{l}伪{\cos }^2胃\{鈭�\frac{1}{r^2}\cos \left[蝇(t鈭�\frac{r}{c})\right]+\frac{1}{r}[鈭�\sin \left(蝇(t鈭�\frac{r}{c})\right)(鈭�\frac{蝇}{c})]\}\\ +\frac{伪}{r^2}\cos \left[蝇(t鈭�\frac{r}{c})\right](鈭�2\cos 胃\sin 胃)\widehat{胃}\end{array}\right\}\\ 鈭�V=\left\{\begin{array}{l}伪{\cos }^2胃\{鈭�\frac{1}{r^2}\cos \left[蝇(t鈭�\frac{r}{c})\right]+\frac{蝇}{rc}\left[\sin \left(蝇(t鈭�\frac{r}{c})\right)\right]\}\\ +\frac{伪}{r^2}\cos \left[蝇(t鈭�\frac{r}{c})\right](鈭�2\cos 胃\sin 胃)\widehat{胃}\end{array}\right\}\end{array}$

In the radiation zone

$\begin{array}{l}螖V=伪{\cos }^2胃\frac{蝇}{rc}\left[\sin \left(蝇(t鈭�\frac{r}{c})\right)\right]\widehat{r}\\ 螖V=伪\frac{蝇}{c}\frac{{\cos }^2胃}{r}\left[\sin \left(蝇(t鈭�\frac{r}{c})\right)\right]\widehat{r}\end{array}$ 鈥︹�︹�� (10)

Since

$A=\frac{伪}{c}\frac{\cos 胃}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right](\cos 胃\widehat{r}鈭�\sin 胃\widehat{胃})$

Differentiate the equation with respect to time .

$\begin{array}{l}\frac{鈭�A}{鈭�t}=\frac{鈭�}{鈭�t}\left\{\frac{伪}{c}\frac{\cos 胃}{r}\cos \left[蝇(t鈭�\frac{r}{c})\right](\cos 胃\widehat{r}鈭�\sin 胃\widehat{胃})\right\}\\ \frac{鈭�A}{鈭�t}=\frac{伪}{c}\frac{\cos 胃}{r}\{鈭�\sin \left[蝇(t鈭�\frac{r}{c})\left(蝇\right)\right](\cos 胃\widehat{r}鈭�\sin 胃\widehat{胃})\}\\ \frac{鈭�A}{鈭�t}=鈭�\frac{伪蝇}{c}\frac{\cos 胃}{r}\sin \left[蝇(t鈭�\frac{r}{c})\left(蝇\right)\right](\cos 胃\widehat{r}鈭�\sin 胃\widehat{胃})\end{array}$

From equation (3)

$E=鈭�鈭�V鈭�\frac{鈭�A}{鈭�t}$ 鈥︹�︹�� (11)

Put the value of $螖V$ and $\frac{鈭�A}{鈭�t}$in equation (11).

$\begin{array}{l}E=鈭�伪\frac{蝇}{c}\frac{{\cos }^2胃}{r}\left[\sin \left(蝇(t鈭�\frac{r}{c})\right)\right]\widehat{r}+\frac{伪蝇}{c}\frac{\cos 胃}{r}\sin \left[蝇(t鈭�\frac{r}{c})\left(蝇\right)\right](\cos 胃\widehat{r}鈭�\sin 胃\widehat{胃})\\ E=\frac{伪蝇}{c}\frac{\cos 胃}{r}\left[\sin 蝇(t鈭�\frac{r}{c})\right](\cos 胃\widehat{r}鈭�\sin 胃\widehat{胃})鈭�\frac{伪蝇}{c}\frac{{\cos }^2胃}{r}\sin \left[蝇(t鈭�\frac{r}{c})\right]\widehat{r}\\ E=\frac{伪蝇}{c}\frac{\cos 胃}{r}\left[\sin 蝇(t鈭�\frac{r}{c})\right][\cos 胃\widehat{r}鈭�\sin 胃\widehat{胃}鈭�\cos 胃\widehat{r}]\\ E=\frac{伪蝇}{c}\frac{\cos 胃}{r}\left[\sin 蝇(t鈭�\frac{r}{c})\right][鈭�\sin 胃\widehat{胃}]\end{array}$

Further solve as

$\begin{array}{l}E=\frac{伪蝇}{c}\frac{\cos 胃\sin 胃}{r}\left[\sin 蝇(t鈭�\frac{r}{c})\right]\widehat{胃}\\ E=鈭�\frac{伪蝇}{cr}\cos 胃\sin 胃\left[\sin 蝇(t鈭�\frac{r}{c})\right]\widehat{胃}\end{array}$ 鈥︹�︹�� (12)

Thus, the electric field is .$E=鈭�\frac{伪蝇}{cr}\cos 胃\sin 胃\left[\sin 蝇(t鈭�\frac{r}{c})\right]\widehat{胃}$

The magnetic field is given by

$\begin{array}{l}B=鈭�脳A\\ B=\frac{1}{r}[\frac{鈭�}{鈭�r}\left(r{A}_0\right)鈭�\frac{鈭�A_r}{鈭�胃}]\widehat{蠒}\\ B=\frac{伪}{cr}\{\frac{鈭�}{鈭�r}\left[\cos 胃\cos \left(蝇(t鈭�\frac{r}{c})\right)(鈭�\sin 胃)\right]鈭�\frac{鈭�}{鈭�胃}\frac{{\cos }^2胃}{r}\cos \left(蝇(t鈭�\frac{r}{c})\right)\}\widehat{蠒}\\ B=\frac{伪}{cr}\left\{(鈭�\sin 胃\cos 胃)[鈭�\sin \left(蝇(t鈭�\frac{r}{c})\right)]\left(\frac{鈭�蝇}{c}\right)鈭�\frac{鈭�}{鈭�胃}\frac{{\cos }^2胃}{r}\cos \left(蝇(t鈭�\frac{r}{c})\right)\right\}\widehat{蠒}\end{array}$

Further solved as

$B=\frac{伪}{cr}\left\{(鈭�\sin 胃\cos 胃)\frac{蝇}{c}\sin \left(蝇(t鈭�\frac{r}{c})\right)鈭�\frac{鈭�}{鈭�胃}\frac{{\cos }^2胃}{r}\cos \left(蝇(t鈭�\frac{r}{c})\right)\right\}\widehat{蠒}$

Consider the radiation zone.

$\begin{array}{l}B=\frac{伪}{cr}(鈭�\sin 胃\cos 胃)\frac{蝇}{c}\sin \left[蝇(t鈭�\frac{r}{c})\right]\widehat{蠒}\\ B=鈭�\frac{伪蝇}{c^{2}r}\sin 胃\cos 胃\sin \left[蝇(t鈭�\frac{r}{c})\right]\widehat{蠒}\end{array}$

Thus, the magnetic field is .$B=鈭�\frac{伪蝇}{c^{2}r}\sin 胃\cos 胃\sin \left[蝇(t鈭�\frac{r}{c})\right]\widehat{蠒}$

(c)

Consider the formula for the magnetic field as

$B=\frac{1}{c}(\widehat{r}脳E)$

Here, $E鈰�\widehat{r}=0$.

Then, the pointing vector is given as

$\begin{array}{l}S=\frac{1}{{渭}_0}(E脳B)\\ S=\frac{1}{{渭}_0}[E脳\frac{1}{c}(\widehat{r}脳E)]\\ S=\frac{1}{{渭}_{0}c}[E脳(\widehat{r}脳E)]\\ S=\frac{1}{{渭}_{0}c}(E^2\widehat{r}鈭�(E.\widehat{r})E)\end{array}$

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

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

Substitute equation (12) in the expression.

$S=\frac{{\{鈭�\frac{伪蝇}{cr}\cos 胃\sin 胃\left[\sin 蝇(t鈭�\frac{r}{c})\right]\}}^2}{{渭}_{0}c}\widehat{r}$

Thus, the pointing vector is

$S=\frac{{\{鈭�\frac{伪蝇}{cr}\cos 胃\sin 胃\left[\sin 蝇(t鈭�\frac{r}{c})\right]\}}^2}{{渭}_{0}c}\widehat{r}$

The power radiated is given as

$\begin{array}{c}P=鈭�鉄�S鉄�.da\\ =\frac{1}{{渭}_{0}c}{\left(\frac{伪蝇}{c}\right)}^2鈭�{\sin }^2胃{\cos }^2胃\sin 胃d胃d蠒\\ =\frac{1}{2{渭}_{0}c}{\left(\frac{伪蝇}{c}\right)}^2\left(2蟺\right)\underset{0}{\overset{蟺}{鈭�}}(1鈭�{\cos }^2胃){\cos }^2胃\sin 胃d胃\end{array}$

Let$u=\cos 胃$; then $du=鈭�\sin 胃d胃$.

Upper limit is $胃=蟺$; then

$\begin{array}{l}u=\cos 蟺\\ u=鈭�1\end{array}$

Lower limit is $胃=0$; then

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

Therefore

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

Now integrate the expression

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

Now, power radiated is

$\begin{array}{l}P=\frac{1}{2{渭}_{0}c}\frac{{蝇}^2}{c}\left(2蟺\right)\left(\frac{4}{15}\right)\\ P=\frac{1}{2{渭}_{0}c}\frac{{蝇}^2}{c}\left(2蟺\right)\left(\frac{4}{15}\right){伪}^2\end{array}$

Since $伪=鈭�\frac{{渭}_0{p}_0{蝇}^{2}d}{4蟺}$, then the above equation becomes

$\begin{array}{c}P=\frac{1}{2{渭}_{0}c}\frac{{蝇}^2}{c}\left(2蟺\right)\left(\frac{4}{15}\right){[鈭�\frac{{渭}_0{p}_0{蝇}^{2}d}{4蟺}]}^2\\ =\frac{1}{2{渭}_{0}c}\frac{{蝇}^2}{c}\left(2蟺\right)\left(\frac{4}{15}\right)\frac{{{渭}_0}^2{p_0}^2{蝇}^4{d}^2}{16{蟺}^2}\\ =\frac{{渭}_0}{60蟺c^3}{\left(p_{0}d\right)}^2{蝇}^6\end{array}$

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