The electric field in an electromagnetic wave

Question:

The electric field in an electromagnetic wave is given by $\mathrm{E}=\left(50 \mathrm{NC}^{-1}\right) \sin \omega(\mathrm{t}-\mathrm{x} / \mathrm{c})$

The energy contained in a cylinder of volume $V$ is $5.5 \times 10^{-12} \mathrm{~J}$. The value of $\mathrm{V}$ is $\mathrm{cm}^{3}$.

Solution:

$\mathrm{E}=50 \sin \left(\omega \mathrm{t}-\frac{\omega}{\mathrm{c}} \cdot \mathrm{x}\right)$

Energy density $=\frac{1}{2} \in_{0} \mathrm{E}_{0}^{2}$

Energy density $=\frac{1}{2} \in_{0} \mathrm{E}_{0}^{2}$

$\frac{1}{2} 8.8 \times 10^{-12} \times 2500 \mathrm{~V}=5.5 \times 10^{-12}$

$\frac{1}{2} 8.8 \times 10^{-12} \times 2500 \mathrm{~V}=5.5 \times 10^{-12}$

$\mathrm{V}=\frac{5.5 \times 2}{2500 \times 8.8}=.0005 \mathrm{~m}^{3}$

$=.0005 \times 10^{6}(\mathrm{c} . \mathrm{m})^{3}$

$=500(\mathrm{c} . \mathrm{m})^{3}$

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