91影视

Write the expression for the intensity of radiation.

$\mathit{I}\mathbf{=}<S>\mathbf{路}\hat{\mathbf{z}}$ 鈥︹�� (1)

Here, is the Poynting vector which is given as:

Substitute in equation (1).

role="math" localid="1653990441797" $$\begin{gathered} I=\left(\frac{{渭}_0{p}_0^2{蝇}^4}{32{蟺}^{2}c}\right)\frac{{\sin }^2胃}{r^2}\left(\hat{r}路\hat{z}\right) \\ I=\left(\frac{{渭}_0{p}_0^2{蝇}^4}{32{蟺}^{2}c}\right)\frac{{\sin }^2胃}{r^2}\cos 胃 \end{gathered}$$ 鈥︹�� (2)

Here, $p_0=qd$.

(a)

Draw the given situation of the charged particle.

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

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

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

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

For the maximum intensity,

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

Hence, the intensity equation becomes,

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

Therefore, the intensity of the radiation hitting the floor is $\frac{{渭}_0{q}^2{d}^2{蝇}^4{R}^{2}h}{32{蟺}^{2}c{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}$, 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.

$$\begin{gathered} P=鈭�I\left(R\right)da \\ P={鈭�}_0^{鈭�}I\left(R\right)\left(2蟺R\right)dR \end{gathered}$$

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

$$\begin{gathered} P={鈭�}_0^{鈭�}\left(\frac{{渭}_0{q}^2{d}^2{蝇}^4{R}^{2}h}{32{蟺}^{2}c{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)\left(2蟺RdR\right) \\ P=2蟺\left(\frac{{渭}_0{q}^2{d}^2{蝇}^{4}h}{32{蟺}^{2}c}\right){鈭�}_0^{鈭�}\left(\frac{R^3}{{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)dR \end{gathered}$$

Let,

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

Hence, the above equation becomes,

$$\begin{gathered} P=2蟺\left(\frac{{渭}_0{q}^2{d}^2{蝇}^{4}h}{32{蟺}^{2}c}\right){鈭�}_0^{鈭�}\left(\frac{R^3}{{\left(R^2+h^2\right)}^{{\displaystyle \frac{5}{2}}}}\right)dR \\ P=2蟺\left(\frac{{渭}_0{q}^2{d}^2{蝇}^{4}h}{32{蟺}^{2}c}\right)\left(\frac{2}{3h}\right) \\ P=\frac{{渭}_0{q}^2{d}^2{蝇}^4}{24蟺c} \end{gathered}$$

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 $\frac{{渭}_0{q}^2{d}^2{蝇}^4}{24蟺c}$.

(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.

$\frac{dU}{dt}=-2P$ 鈥︹�� (3)

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

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

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

Let$b=\left(\frac{{渭}_0{q}^2{蝇}^4}{6蟺ck}\right)$

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.

$$\begin{gathered} t=\frac{2}{\left({\displaystyle \frac{{渭}_0{q}^2{蝇}^4}{6蟺ck}}\right)} \\ t=\frac{12蟺ck}{{渭}_0{q}^2{蝇}^4} \end{gathered}$$

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

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

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