An electromagnetic wave of frequency 1 times | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given: Frequency of the electromagnetic wave $f = 1.0 	imes 10^{14} \,Hz$.Amplitude of the electric field $E_0 = 4 \,Vm^{-1}$.Permittivity of fre

Vedclass · Accepted Answer

$70.4 	imes 10^{-12} \,Jm^{-3}$

Vedclass · Answer

(C) Given: Frequency of the electromagnetic wave $f = 1.0 	imes 10^{14} \,Hz$.Amplitude of the electric field $E_0 = 4 \,Vm^{-1}$.Permittivity of free space $\varepsilon_0 = 8.8 	imes 10^{-12} \,C^2 \,N^{-1} \,m^{-2}$.The energy density $u_E$ of an electric field is given by the formula:$u_E = \frac{1}{2} \varepsilon_0 E_0^2$Substituting the given values into the formula:$u_E = \frac{1}{2} 	imes (8.8 	imes 10^{-12}) 	imes (4)^2$$u_E = \frac{1}{2} 	imes 8.8 	imes 10^{-12} 	imes 16$$u_E = 4.4 	imes 16 	imes 10^{-12}$$u_E = 70.4 	imes 10^{-12} \,Jm^{-3}$Therefore,the energy density of the electric field is $70.4 	imes 10^{-12} \,Jm^{-3}$.

An electromagnetic wave of frequency $1 \times 10^{14} \,Hz$ is propagating along the $z$-axis. The amplitude of the electric field is $4 \,Vm^{-1}$. What is the energy density of the electric field? (Permittivity of free space $\varepsilon_0 = 8.8 \times 10^{-12} \,C^2 \,N^{-1} \,m^{-2}$)

Similar Questions

An electromagnetic wave of frequency $3 \, GHz$ enters a dielectric medium of relative electric permittivity $2.25$ from vacuum. The wavelength of this wave in that medium will be $....... \times 10^{-2} \, cm$.

Energy stored in electromagnetic oscillations is in the form of

The electric field of a plane electromagnetic wave is given by $\vec E = E_0 \hat i \cos(kz) \cos(\omega t)$. The corresponding magnetic field $\vec B$ is then given by:

If ${\mu_r}$ is the relative permeability and $K$ is the dielectric constant of a medium,its refractive index $n$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module