A parallel-plate capacitor C, with plate separation d, is given an initial charge $(\pm ){\mathit{Q}}_{\mathbf{0}}$. It is the0n connected to a resistor R, and discharges, $\mathit{Q}\left(t\right)\mathbf{=}{\mathit{Q}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{R}\mathbf{C}}$.

(a) What fraction of its initial energy$\left(Q_0^2/2C\right)$ does it radiate away?

(b) If $\mathit{C}\mathbf{=}\mathbf{1}\mathit{p}\mathit{F}\mathbf{}\mathbf{,}\mathbf{}\mathit{R}\mathbf{=}\mathbf{1000}\mathbf{}\mathit{\Omega }\mathbf{,}$and role="math" localid="1653972749344" $\mathit{d}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{}\mathit{m}\mathit{m}$, what is the actual number? In electronics we don’t ordinarily worry about radiative losses; does that seem reasonable, in this case?

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 $R=1000\Omega$

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

(a)

Write the expression for a fraction of energy radiated.

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

$P=\frac{{\mu }_0}{6\pi c}{\left(\ddot{p}\right)}^2$ …… (2)

Here, ${\mu }_0$ 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.

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

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

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

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

$$\begin{gathered} \frac{E_{rad}}{E_0}=\frac{{\int }_0^{\infty }{\displaystyle \frac{{\mu }_0}{6\pi c}}{\left({\displaystyle \frac{Q_0}{{\left(RC\right)}^2}}\left(e^{-{\displaystyle \frac{t}{RC}}}\right)d\right)}^2}{\left({\displaystyle \frac{Q_0^2}{2c}}\right)} \\ \frac{E_{rad}}{E_0}=\frac{{\mu }_0}{6\pi c}\frac{Q_0^2{d}^2}{{\left(RC\right)}^4}{\int }_0^{\infty }{\left(e^{-\frac{t}{RC}}\right)}^{2}dt\times \left(\frac{2C}{Q_0^2}\right) \\ \frac{E_{rad}}{E_0}=\frac{{\mu }_0}{3\pi c}\frac{Q_0^2{d}^2}{{\left(RC\right)}^4}\frac{RC}{2}{\left(e^{-\frac{2t}{RC}}\right)}_0^{\infty }\times \left(\frac{C}{Q_0^2}\right) \\ \frac{E_{rad}}{E_0}=\frac{{\mu }_0}{6\pi c}\frac{d^2}{R^3{C}^2}........\left(3\right) \end{gathered}$$

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

(b)

Substitute${\mu }_0=4\pi \times {10}^{-7}\mathrm{H}/\mathrm{m},d\mathit{=}\mathit{0}\mathit{.}\mathit{1}\mathrm{mm},c\mathit{=}\mathit{3}\mathit{\times }{\mathit{10}}^{\mathit{8}}\mathit{}\mathrm{m}/\mathrm{s},\mathrm{R}=1000\mathrm{\Omega }$ and$C=1pF$ in equation (3).

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

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

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