91Ó°ÊÓ

Given data:

The decay of charge after a certain time interval is $Q\left(t\right)=Q_0{e}^{-t/RC}$.

The capacitance of a capacitor is $C=1pF$.

The resistance of a resistor is $R=1000\mathrm{Î\copyright }$

The position of the charge is $d=0.1\mathrm{mm}$.

(a)

Write the expression for a fraction of energy radiated.

$\frac{{\mathbf{E}}_{\mathbf{r}\mathbf{a}\mathbf{d}}}{{\mathbf{E}}_{\mathbf{0}}}\mathbf{=}\frac{{\mathbf{∫}}_{\mathbf{0}}^{\mathbf{∞}}\mathbf{P}\mathbf{d}\mathbf{t}}{\left({\displaystyle \frac{Q_0^2}{2C}}\right)}$ …… (1)

Here, P is the total radiated power from the parallel plate capacitor, and C is the capacitance of a capacitor.

Write the expression for the total radiated power from the parallel plate capacitor.

$P=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}{\left(\stackrel{¨}{p}\right)}^2$ …… (2)

Here, ${\mathrm{μ}}_0$ is the magnetic permeability, c is the speed of light, and p is the dipole moment.

Write the expression for the dipole moment.

$p=Q\left(t\right)d$

Substitute $Q\left(t\right)=Q_0{e}^{-t/RC}$in the above expression.

$p=Q_0\left(e^{-\frac{t}{RC}}\right)d$

Take the double differentiation of the above equation.

$$\begin{gathered} \stackrel{˙}{p}=-\frac{Q_0}{RC}\left(e^{-\frac{t}{RC}}\right)d \\ \stackrel{¨}{p}=\frac{Q_0}{{\left(RC\right)}^2}\left(e^{\frac{t}{RC}}\right)d \end{gathered}$$

Substitute $\stackrel{¨}{p}=\frac{Q_0}{{\left(RC\right)}^2}\left(e^{-\frac{t}{RC}}\right)d$in equation (2).

$P=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}{\left(\frac{Q_0}{{\left(RC\right)}^2}\left(e^{-\frac{t}{RC}}\right)d\right)}^2$

Now, substitute $P=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}{\left(\frac{Q_0}{{\left(RC\right)}^2}\left(e^{-\frac{t}{RC}}\right)d\right)}^2$to calculate the fraction of the radiated initial energy.

$$\begin{gathered} \frac{E_{rad}}{E_0}=\frac{{∫}_0^{∞}{\displaystyle \frac{{\mathrm{μ}}_0}{6\mathrm{π}c}}{\left({\displaystyle \frac{Q_0}{{\left(RC\right)}^2}}\left(e^{-{\displaystyle \frac{t}{RC}}}\right)d\right)}^2}{\left({\displaystyle \frac{Q_0^2}{2c}}\right)} \\ \frac{E_{rad}}{E_0}=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}\frac{Q_0^2{d}^2}{{\left(RC\right)}^4}{∫}_0^{∞}{\left(e^{-\frac{t}{RC}}\right)}^{2}dt×\left(\frac{2C}{Q_0^2}\right) \\ \frac{E_{rad}}{E_0}=\frac{{\mathrm{μ}}_0}{3\mathrm{π}c}\frac{Q_0^2{d}^2}{{\left(RC\right)}^4}\frac{RC}{2}{\left(e^{-\frac{2t}{RC}}\right)}_0^{∞}×\left(\frac{C}{Q_0^2}\right) \\ \frac{E_{rad}}{E_0}=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}\frac{d^2}{R^3{C}^2}........\left(3\right) \end{gathered}$$

Therefore, the fraction of energy radiated is $\frac{E_{rad}}{E_0}=\frac{{\mathrm{μ}}_0}{6\mathrm{π}c}\frac{d^2}{R^3{C}^2}$.

(b)

Substitute${\mathrm{μ}}_0=4\mathrm{Ï€}×{10}^{-7}\mathrm{H}/\mathrm{m},d\mathit{=}\mathit{0}\mathit{.}\mathit{1}\mathrm{mm},c\mathit{=}\mathit{3}\mathit{×}{\mathit{10}}^{\mathit{8}}\mathit{}\mathrm{m}/\mathrm{s},\mathrm{R}=1000\mathrm{Î\copyright }$ and$C=1pF$ in equation (3).

$$\begin{gathered} \frac{E_{rad}}{E_0}\frac{\left(4\mathrm{Ï€}×{10}^{-7}\mathrm{H}/\mathrm{m}\right)}{6\mathrm{Ï€}\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}\frac{{\left(0.1\mathrm{mm}×{\displaystyle \frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}}\right)}^2}{{\left(1000\mathrm{Î\copyright }\right)}^3{\left(1pF×{\displaystyle \frac{{10}^{-12}F}{1pF}}\right)}^2} \\ \frac{E_{rad}}{E_0}=2.2×{10}^{-9} \end{gathered}$$

As the fraction is very small, so this will be safe to neglect.

Therefore, the fractional energy loss is$2.2×{10}^{-9}$ which is safe to neglect.