The electric field in an electromagnetic wave | vedclass

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

Question

(C) The electric field is given by $\vec{E} = E_0 \sin \omega (t - \frac{x}{c}) \hat{j}$,where $E_0 = 20 	ext{ V/m}$.The average energy density $(u_{

Vedclass · Accepted Answer

$8.85$

Vedclass · Answer

(C) The electric field is given by $\vec{E} = E_0 \sin \omega (t - \frac{x}{c}) \hat{j}$,where $E_0 = 20 	ext{ V/m}$.The average energy density $(u_{avg})$ of an electromagnetic wave is given by $u_{avg} = \frac{1}{2} \varepsilon_0 E_0^2$.The total energy $(U)$ stored in a volume $(V)$ is $U = u_{avg} 	imes V$.Substituting the given values:$U = \frac{1}{2} 	imes (8.85 	imes 10^{-12} 	ext{ C}^2/	ext{Nm}^2) 	imes (20 	ext{ V/m})^2 	imes (5 	imes 10^{-4} 	ext{ m}^3)$.$U = \frac{1}{2} 	imes 8.85 	imes 10^{-12} 	imes 400 	imes 5 	imes 10^{-4} 	ext{ J}$.$U = 8.85 	imes 10^{-12} 	imes 200 	imes 5 	imes 10^{-4} 	ext{ J}$.$U = 8.85 	imes 10^{-12} 	imes 1000 	imes 10^{-4} 	ext{ J}$.$U = 8.85 	imes 10^{-12} 	imes 10^{-1} 	ext{ J} = 8.85 	imes 10^{-13} 	ext{ J}$.Thus,the value is $8.85 	imes 10^{-13} 	ext{ J}$.

Similar Questions

Which one of the following statements is correct?

When an $EM$ wave propagates through a vacuum,then:

The velocity of electromagnetic radiation in a medium of permittivity $\varepsilon_0$ and permeability $\mu_0$ is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module