Radiant flux density $S=\frac{1}{\propto_0}(\mathbf{E} \times \mathbf{B})=C^2 \varepsilon_0(\mathbf{E} \times \mathbf{B})$

$$\left[\because c=\frac{1}{\sqrt{\propto_0 \varepsilon_0}}\right]$$

Suppose electromagnetic waves be propagating along $x$-axis. The electric field vector of electromagnetic wave be along $y$-axis and magnetic field vector be along $z$-axis. Therefore,

$$\begin{aligned} \mathbf{E}_0 & =\mathbf{E}_0 \cos (k x-\omega t) \\ \text{and}\quad\mathbf{B} & =\mathbf{B}_0 \cos (k x-\omega t) \\ \mathbf{E} \times \mathbf{B} & =\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \\ \mathbf{S} & =\mathbf{c}^2 \varepsilon_0(\mathbf{E} \times \mathbf{B}) \\ & =c^2 \varepsilon_0\left(\mathbf{E}_0 \times \mathbf{B}_0\right) \cos ^2(k x-\omega t) \end{aligned}$$

Average value of the magnitude of radiant flux density over complete cycle is

$$\begin{array}{rlrl} S_{\mathrm{av}} & =c^2 \varepsilon_{\mathrm{o}}\left|\mathbf{E}_0 \times \mathbf{B}_0\right| \frac{1}{T} \int_0^T \cos ^2(k x-\omega t) d t \\ & =c^2 \varepsilon_0 E_0 B_0 \times \frac{1}{T} \times \frac{T}{2} \quad\left[\because \int_0^T \cos ^2(k x-\omega t) d t=\frac{T}{2}\right] \\ \Rightarrow \quad S_{\mathrm{av}} & =\frac{c^2}{2} \varepsilon_0 E_0\left(\frac{E_0}{c}\right) \quad\left[\text { As, } C=\frac{E_0}{B_0}\right]\\ & =\frac{c}{2} \varepsilon_0 E_0^2=\frac{c}{2} \times \frac{1}{c^2 \alpha_0} E_0^2 \quad {\left[C=\frac{1}{\sqrt{\alpha_0 \varepsilon_0}} \text { or } \varepsilon_0=\frac{1}{c^2 \alpha_0}\right]}\\ \Rightarrow \quad S_{\mathrm{av}} & =\frac{E_0^2}{2 \alpha_0 c} \end{array}$$

Hence proved.

Given, capacitance of capacitor $C=2 \propto F$,

Displacement current $I_d=1 \mathrm{~mA}$

Charge $$q=C V$$

$$\begin{array}{rlr} I_d d t & =C d V & {[\because q=i t]} \\ \text{or} \quad I_d & =C \frac{d V}{d t} \\ 1 \times 10^{-3} & =2 \times 10^{-6} \times \frac{d V}{d t} \\ \text{or}\quad\frac{d V}{d t} & =\frac{1}{2} \times 10^{+3}=500 \mathrm{~V} / \mathrm{s} & \end{array}$$

So, by applying a varying potential difference of 500 V/s, we would produce a displacement current of desired value.

$$\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}$$

Force is the rate of change of momentum i.e.,

$$\begin{aligned} \text{i.e.,}\quad F & =\frac{d p}{d t} \\ \text{Energy in time dt,}\quad U & =p \cdot C \text { or } p=\frac{U}{C} \\ \therefore\quad\text { Pressure } & =\frac{1}{A} \cdot \frac{U}{C \cdot d t} \\ \text { Pressure } & =\frac{I}{C} \quad\left[\because I=\text { Intensity }=\frac{U}{A \cdot d t}\right] \end{aligned}$$

As the distance is doubled, the area of spherical region $\left(4 \pi r^2\right)$ will become four times, so the intensity becomes one fourth the initial value $\left(\because I \propto \frac{1}{r^2}\right)$ but in case of laser it does not spread, so its intensity remain same.Geometrical characteristic of LASER beam which is responsible for the constant intensity are as following

(i) Unidirection

(ii) Monochromatic

(iii) Coherent light

(iv) Highly collimated

These characteristic are missing in the case of light from the bulb.

Since, electric field of an EM wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle. Hence, electric field is not responsible for radiation pressure.