The electric field in an electromagnetic wave is given as

$$\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}$$

where $\omega$ and $c$ are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of $5 \times 10^{-4} \mathrm{~m}^{3}$ will be

(Given $\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2}$ )

<p>To find the energy contained in a volume of an electromagnetic wave, we need to calculate the energy density and then multiply it by the volume.</p> <p>For an electromagnetic wave, the energy density $u$ is given by:</p> <p>$u = \frac{1}{2} \varepsilon_0 E^2$</p> <p>where $\varepsilon_0$ is the vacuum permittivity and $E$ is the electric field amplitude.</p> <p>First, let&#39;s find the amplitude of the electric field. In the given equation:</p> <p>$$\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}$$</p> <p>The amplitude $E$ is $20 \mathrm{~N/C}$.</p> <p>Now, we can find the energy density:</p> <p>$$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} \cdot (20 \mathrm{~N/C})^2$$</p> <p>$u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot 400$</p> <p>$u = 1.77 \times 10^{-9} \mathrm{J/m^3}$</p> <p>The energy contained in a volume of $5 \times 10^{-4} \mathrm{m^3}$ will be:</p> <p>$E_\text{total} = u \cdot V$</p> <p>$$E_\text{total} = 1.77 \times 10^{-9} \mathrm{J/m^3} \cdot 5 \times 10^{-4} \mathrm{m^3}$$</p> <p>$E_\text{total} = 8.85 \times 10^{-13} \mathrm{J}$</p>