The electric field in an electromagnetic wave | vedclass

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Question

(C) The electric field is given by $E = E_0 \sin \omega(t - x/c)$,where $E_0 = 50 \, NC^{-1}$.The energy density $u$ of an electromagnetic wave is giv

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(C) The electric field is given by $E = E_0 \sin \omega(t - x/c)$,where $E_0 = 50 \, NC^{-1}$.The energy density $u$ of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E_0^2$.The total energy $U$ in a volume $V$ is $U = u \cdot V = \frac{1}{2} \epsilon_0 E_0^2 V$.Given $U = 5.5 	imes 10^{-12} \, J$ and $\epsilon_0 = 8.8 	imes 10^{-12} \, C^2 N^{-1} m^{-2}$,we have:$5.5 	imes 10^{-12} = \frac{1}{2} 	imes (8.8 	imes 10^{-12}) 	imes (50)^2 	imes V$.$5.5 = 0.5 	imes 8.8 	imes 2500 	imes V$.$5.5 = 4.4 	imes 2500 	imes V$.$5.5 = 11000 	imes V$.$V = \frac{5.5}{11000} = 0.0005 \, m^3$.Converting $m^3$ to $cm^3$:$V = 0.0005 	imes (100 \, cm)^3 = 0.0005 	imes 10^6 \, cm^3 = 500 \, cm^3$.

The electric field in an electromagnetic wave is given by $E = (50 \, NC^{-1}) \sin \omega(t - x/c)$. The energy contained in a cylinder of volume $V$ is $5.5 \times 10^{-12} \, J$. The value of $V$ is $...... \, cm^3$ (given $\epsilon_0 = 8.8 \times 10^{-12} \, C^2 N^{-1} m^{-2}$).

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