Question

An EM wave of frequency 8 x $10^{14}$ Hz having amplitude of electric field 2 volt/m. The total energy density associated with wave is x x $10^{-12} J/m^2$. The x is

Answer 1

17.7

Answer 2

The total energy density associated with an electromagnetic wave is given by the formula:

$u_{avg} = \frac{1}{2} \epsilon_0 E_0^2$

Where:

• $u_{avg}$ is the average total energy density.
• $\epsilon_0$ is the permittivity of free space, approximately $8.85 \times 10^{-12}$ F/m.
• $E_0$ is the amplitude of the electric field.

Given values:

• $E_0 = 2$ V/m
• $\epsilon_0 = 8.85 \times 10^{-12}$ F/m

Substitute the values into the formula: $u_{avg} = \frac{1}{2} \times (8.85 \times 10^{-12} \text{ F/m}) \times (2 \text{ V/m})^2$ $u_{avg} = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4$ $u_{avg} = 8.85 \times 10^{-12} \times 2$ $u_{avg} = 17.7 \times 10^{-12} \text{ J/m}^3$

The problem states that the total energy density is $x \times 10^{-12} J/m^3$. Comparing our calculated value with the given format: $x \times 10^{-12} J/m^3 = 17.7 \times 10^{-12} J/m^3$

Therefore, $x = 17.7$. The frequency $8 \times 10^{14}$ Hz is not required for this calculation as the energy density depends only on the amplitude of the electric field and the permittivity of the medium.