Show that the radiation pressure exerted by an EM wave of intensity $I$ on a surface kept in vacuum is $\frac{I}{C}$.
$$\text { Pressure }=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}$$
Force is the rate of change of momentum i.e.,
$$\begin{aligned} \text{i.e.,}\quad F & =\frac{d p}{d t} \\ \text{Energy in time dt,}\quad U & =p \cdot C \text { or } p=\frac{U}{C} \\ \therefore\quad\text { Pressure } & =\frac{1}{A} \cdot \frac{U}{C \cdot d t} \\ \text { Pressure } & =\frac{I}{C} \quad\left[\because I=\text { Intensity }=\frac{U}{A \cdot d t}\right] \end{aligned}$$
What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of room, its intensity essentially remain constant. What geometrical characteristic of LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?
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.
Even though an electric field E exer ts a force $q \mathrm{E}$ on a charged particle yet electric field of an EM wave does not contribute to the radiation pressure (but transfers energy). Explain.
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.
An infinitely long thin wire carrying a uniform linear static charge density $\lambda$ is placed along the $z$-axis (figure). The wire is set into motion along its length with a uniform velocity $v=v \hat{\mathbf{k}}_z$. Calculate the pointing vector $\mathbf{S}=\frac{1}{\alpha_0}(\mathbf{E} \times \mathbf{B})$
$$\begin{aligned} \text{Given,}\quad\mathbf{E} & =\frac{\lambda \hat{\mathbf{e}}_S}{2 \pi \varepsilon_0 a} \hat{\mathbf{j}} \\ \mathbf{B} & =\frac{\propto_0 \mathrm{i}}{2 \pi a} \hat{\mathbf{i}}=\frac{\propto_0 \lambda V}{2 \pi a} \hat{\mathbf{i}} \quad\left[\because I=\lambda_V\right]\\ \therefore\quad\mathbf{S} & =\frac{1}{\propto_0}[\mathbf{E} \times \mathbf{B}]=\frac{1}{\propto_0}\left[\frac{\lambda_{\hat{\mathbf{j}}}}{2 \pi \varepsilon_0 a} \times \frac{\propto_0}{2 \pi a} \lambda \sqrt{\hat{\mathbf{i}}}\right] \\ & =\frac{\lambda^2 V}{4 \pi^2 \varepsilon_0 a^2}(\hat{\mathbf{j}} \times \hat{\mathbf{i}})=-\frac{\lambda^2 V}{4 \pi^2 \varepsilon_0 \mathrm{a}^2} \hat{\mathbf{k}} \end{aligned}$$
Sea water at frequency $v=4 \times 10^8 \mathrm{~Hz}$ has permittivity $\varepsilon \approx 80 \varepsilon_0$, permeability $\alpha \approx \alpha_0$ and resistivity $\rho=0.25 \mathrm{~m}$. Imagine a parallel plate capacitor immersed in sea water and driven by an alternating voltage source $V(t)=V_0 \sin (2 \pi v t)$. What fraction of the conduction current density is the displacement current density?
Suppose distance between the parallel plates is $d$ and applied voltage $V_{(t)}=V_0 2 \pi v t$.
Thus, electric field
$$\begin{aligned} & E=\frac{V_0}{d} \sin (2 \pi v t) \\ & \text { Now using Ohm's law, } \\ & J_c=\frac{1}{\rho} \frac{V_0}{d} \sin (2 \pi v t) \\ & \Rightarrow \quad=\frac{V_0}{\rho d} \sin (2 \pi v t)=J_0^c \sin 2 \pi v t \\ \text{Here,}\quad & J_0^c=\frac{V_0}{\rho d} \end{aligned}$$
$$\begin{aligned} &\text { Now the displacement current density is given as }\\ &\begin{aligned} J_d & =\varepsilon \frac{\delta E}{d t}=\frac{\varepsilon \delta}{d t} \quad\left[\frac{V_0}{d} \sin (2 \pi v t)\right]\\ & =\frac{\varepsilon 2 \pi v V_0}{d} \cos (2 \pi v t) \\ \Rightarrow \quad & =J_0^d \cos (2 \pi v t) \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { where, } \quad J_0^d & =\frac{2 \pi V \varepsilon V_0}{d} \\ \Rightarrow \quad \frac{J_0^d}{J_0^c} & =\frac{2 \pi v \varepsilon V_0}{d} \cdot \frac{\rho d}{V_0}=2 \pi \vee \varepsilon \rho \\ & =2 \pi \times 80 \varepsilon_0 \vee \times 0.25=4 \pi \varepsilon_0 \vee \times 10 \\ & =\frac{10 v}{9 \times 10^9}=\frac{4}{9} \end{aligned}$$