Due to the presence of an em-wave whose elect | vedclass

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Question

(B) The energy density of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E^2$. The total energy $U$ contained in a cylinder of length

Vedclass · Accepted Answer

$200 \sin(\omega t - kx) \ NC^{-1}$

Vedclass · Answer

(B) The energy density of an electromagnetic wave is given by $u = \frac{1}{2} \epsilon_0 E^2$. The total energy $U$ contained in a cylinder of length $L$ and radius $R$ is $U = u 	imes V = \frac{1}{2} \epsilon_0 E^2 	imes (\pi R^2 L)$. Given that the energy $U$ remains the same for both cylinders,we have $U_1 = U_2$. $\frac{1}{2} \epsilon_0 E_1^2 \pi R_1^2 L_1 = \frac{1}{2} \epsilon_0 E_2^2 \pi R_2^2 L_2$. Since $L_1 = L_2$ and $R_2 = \frac{R_1}{2}$,the equation simplifies to $E_1^2 R_1^2 = E_2^2 (\frac{R_1}{2})^2$. $E_1^2 R_1^2 = E_2^2 \frac{R_1^2}{4}$. $E_2^2 = 4 E_1^2$,which implies $E_2 = 2 E_1$. Given $E_1 = 100 \ NC^{-1}$,we get $E_2 = 2 	imes 100 = 200 \ NC^{-1}$. Thus,the electric field is $200 \sin(\omega t - kx) \ NC^{-1}$.

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