A wave traveling in a lossless, nonmagnetic medium has an electric field amplitude of \( 24.56 \mathrm{~V} / \mathrm{m} \) and an average power density of \( 2.4 \mathrm{~W} / \mathrm{m}^{2} \). Determine the phase velocity of the wave ( 4 marks )

To determine the phase velocity of the wave, we can use the relationship between the electric field amplitude \( E \), the average power density \( S \), and the phase velocity \( v \). The average power density \( S \) in an electromagnetic wave is given by the formula: \[ S = \frac{1}{2} \epsilon_0 c E^2 \] where: - \( S \) is the average power density, - \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.854 \times 10^{-12} \, \text{F/m} \)), - \( c \) is the speed of light in vacuum (\( c \approx 3 \times 10^8 \, \text{m/s} \)), - \( E \) is the electric field amplitude. We can rearrange this formula to solve for the phase velocity \( v \): \[ v = \frac{S}{\frac{1}{2} \epsilon_0 E^2} \] Now, we can substitute the known values into this equation. 1. Given: - \( S = 2.4 \, \text{W/m}^2 \) - \( E = 24.56 \, \text{V/m} \) 2. Calculate \( v \): Let's calculate \( v \) using the above formula. First, we need to calculate \( \frac{1}{2} \epsilon_0 E^2 \): \[ \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \times (8.854 \times 10^{-12}) \times (24.56)^2 \] Now, I will perform the calculations. Calculate the value by following steps: - step0: Calculate: \(\frac{1}{2}\left(8.854e-12\right)\times 24.56^{2}\) - step1: Convert the expressions: \(\frac{1}{2}\left(8.854e-12\right)\left(\frac{614}{25}\right)^{2}\) - step2: Multiply the numbers: \(\frac{2\times 307^{2}}{25^{2}}\times \left(8.854e-12\right)\) - step3: Multiply: \(\frac{2\times 307^{2}\left(8.854e-12\right)}{25^{2}}\) - step4: Multiply: \(\frac{1668961.292e-24\times 307^{2}}{25^{2}}\) - step5: Evaluate the power: \(\frac{1668961.292e-24\times 307^{2}}{625}\) - step6: Simplify: \(\frac{2274731.152445}{625}\) - step7: Rewrite the expression: \(3639.569844\) Calculate or simplify the expression \( 2.4 / 3639.569844 \). Calculate the value by following steps: - step0: Calculate: \(\frac{2.4}{3639.569844}\) - step1: Convert the expressions: \(\frac{\frac{12}{5}}{\frac{909892461}{250000}}\) - step2: Multiply by the reciprocal: \(\frac{12}{5}\times \frac{250000}{909892461}\) - step3: Reduce the numbers: \(4\times \frac{50000}{303297487}\) - step4: Multiply: \(\frac{4\times 50000}{303297487}\) - step5: Multiply: \(\frac{200000}{303297487}\) Now that we have calculated \( \frac{1}{2} \epsilon_0 E^2 \): \[ \frac{1}{2} \epsilon_0 E^2 \approx 3639.57 \, \text{W/m}^2 \] Next, we can find the phase velocity \( v \) using the average power density \( S \): \[ v = \frac{S}{\frac{1}{2} \epsilon_0 E^2} = \frac{2.4}{3639.57} \] Calculating this gives: \[ v \approx 0.000659 \, \text{m/s} \] Thus, the phase velocity of the wave is approximately \( 0.000659 \, \text{m/s} \). However, this value seems unusually low for a wave in a nonmagnetic medium. Let's verify the calculations and ensure that the units and values are consistent. The phase velocity \( v \) in a lossless medium is typically expected to be close to the speed of light \( c \). Let's re-evaluate the calculations to ensure accuracy. 1. The average power density \( S \) is given as \( 2.4 \, \text{W/m}^2 \). 2. The electric field amplitude \( E \) is \( 24.56 \, \text{V/m} \). 3. The permittivity of free space \( \epsilon_0 \) is \( 8.854 \times 10^{-12} \, \text{F/m} \). Let's recalculate \( \frac{1}{2} \epsilon_0 E^2 \) and then \( v \) again. Calculate the value by following steps: - step0: Calculate: \(\frac{1}{2}\left(8.854e-12\right)\times 24.56^{2}\) - step1: Convert the expressions: \(\frac{1}{2}\left(8.854e-12\right)\left(\frac{614}{25}\right)^{2}\) - step2: Multiply the numbers: \(\frac{2\times 307^{2}}{25^{2}}\times \left(8.854e-12\right)\) - step3: Multiply: \(\frac{2\times 307^{2}\left(8.854e-12\right)}{25^{2}}\) - step4: Multiply: \(\frac{1668961.292e-24\times 307^{2}}{25^{2}}\) - step5: Evaluate the power: \(\frac{1668961.292e-24\times 307^{2}}{625}\) - step6: Simplify: \(\frac{2274731.152445}{625}\) - step7: Rewrite the expression: \(3639.569844\) Calculate or simplify the expression \( 2.4 / (1/2 * (8.854e-12) * (24.56^2)) \). Calculate the value by following steps: - step0: Calculate: \(\frac{2.4}{\left(\frac{1}{2}\left(8.854e-12\right)\times 24.56^{2}\right)}\) - step1: Remove the parentheses: \(\frac{2.4}{\frac{1}{2}\left(8.854e-12\right)\times 24.56^{2}}\) - step2: Convert the expressions: \(\frac{2.4}{\frac{1}{2}\left(8.854e-12\right)\left(\frac{614}{25}\right)^{2}}\) - step3: Multiply the terms: \(\frac{2.4}{\frac{1668961.292e-24\times 307^{2}}{25^{2}}}\) - step4: Convert the expressions: \(\frac{\frac{12}{5}}{\frac{1668961.292e-24\times 307^{2}}{25^{2}}}\) - step5: Evaluate the power: \(\frac{\frac{12}{5}}{\frac{1668961.292e-24\times 307^{2}}{625}}\) - step6: Multiply by the reciprocal: \(\frac{12}{5}\times \frac{625}{1668961.292e-24\times 307^{2}}\) - step7: Reduce the numbers: \(12\times \frac{125}{1668961.292e-24\times 307^{2}}\) - step8: Multiply: \(\frac{12\times 125}{1668961.292e-24\times 307^{2}}\) - step9: Multiply: \(\frac{1500}{1668961.292e-24\times 307^{2}}\) - step10: Simplify: \(\frac{1500}{2274731.152445}\) - step11: Rewrite the expression: \(0.000659\) The calculations have been verified, and the results remain consistent: 1. The value of \( \frac{1}{2} \epsilon_0 E^2 \) is approximately \( 3639.57 \, \text{W/m}^2 \). 2. The phase velocity \( v \) calculated as \( \frac{2.4}{3639.57} \) is approximately \( 0.000659 \, \text{m/s} \). However, this value seems incorrect for a wave in a nonmagnetic medium. The phase velocity should be close to the speed of light in vacuum, which is approximately \( 3 \times 10^8 \, \text{m/s} \). Let's re-evaluate the relationship between the average power density \( S \) and the electric field amplitude \( E \) to ensure we are using the correct formula. The correct formula for the average power density in terms of the electric field amplitude is: \[ S = \frac{1}{2} \epsilon_0 c E^2 \] From this, we can express the phase velocity \( v \) as: \[ v = \frac{c}{\sqrt{\epsilon_0 \mu_0}} = c \] In a lossless, nonmagnetic medium, the phase velocity \( v \) is equal to the speed of light \( c \). Thus, the phase velocity of the wave is: \[ v \approx 3 \times 10^8 \, \text{m/s} \] This is the expected result for a wave traveling in a lossless, nonmagnetic medium.