Problem 2195 - JEE Physics

Question

When a plane electromagnetic wave travels in vacuum, the average electric energy density is given by \(\left(E_{0}\right.\) is the amplitude of the electric field) (A) \((1 / 4) \varepsilon_{0} E_{0}^{2}\) (B) \((1 / 2) \varepsilon_{0} E_{0}^{2}\) (C) \(2 e_{\mathrm{o}} \mathrm{E}_{\mathrm{O}}^{2}\) (D) \(4 \varepsilon_{0} E_{0}^{2}\)

Step-by-Step Solution

Answer

The short answer based on the given step-by-step solution is: The average electric energy density when a plane electromagnetic wave travels in a vacuum is given by the expression \((1 / 2) \varepsilon_{0} E_{0}^{2}\), which corresponds to choice (B).

Step 1: Recall the expression for the electric energy density in a vacuum

The electric energy density, represented by letter \(u\), is given by the formula: \[u = \frac{1}{2} \varepsilon_0 E^2\] Where \(\varepsilon_0\) is the vacuum permittivity constant and \(E\) is the electric field amplitude at a certain point.

Step 2: Adapt the formula for the average electric energy density

Since the problem asks for the average electric energy density, we need to use an average value for the electric field amplitude. We'll use \(E_0\) as the average electric field amplitude. Now, substitute \(E_0\) for \(E\) in the \(u\) expression: \[u = \frac{1}{2} \varepsilon_0 E_0^2\]

Step 3: Identify the correct answer based on the derived formula

From Step 2, we found that the formula for the average electric energy density in a vacuum is: \[u = \frac{1}{2} \varepsilon_0 E_0^2\] Comparing this formula with the given choices, we find that the correct answer is: (B) \((1 / 2) \varepsilon_{0} E_{0}^{2}\)

Key Concepts

Electric Energy Density

Electric energy density is a concept that helps us understand how energy is stored in an electric field. Imagine an invisible field that fills the space around charged objects; this field holds energy proportional to its strength. When we're talking about a plane electromagnetic wave traveling in a vacuum, the electric field plays a crucial role in carrying energy.
To calculate how much energy is stored per unit volume in this field, we use a specific formula. The formula for the electric energy density, denoted by the symbol \(u\), is \(u = \frac{1}{2} \varepsilon_0 E^2\). Here, \(\varepsilon_0\) represents the vacuum permittivity, which is a constant, and \(E\) is the electric field amplitude.
Understanding this concept is essential, especially when you're dealing with electromagnetic waves. It’s important because it tells us how efficiently energy can be transferred through space in the form of electromagnetic waves. Remember, the stronger the field, the more energy you have per volume!

Vacuum Permittivity

Vacuum permittivity, symbolized by \(\varepsilon_0\), is a fundamental physical constant that describes how well an electric field can penetrate a vacuum. It forms part of the foundations of electromagnetism and is essential in the calculations of electric energy density.
Vacuum permittivity has a specific value, approximately \(8.854 \times 10^{-12} \text{ F/m (farads per meter)}\), and this constant is significant in determining how much force an electric field can exert over a distance. In simpler terms, it represents the ability of the vacuum itself to allow electric field lines to spread out.

• This value plays a crucial role in the formulas related to electromagnetism, influencing electric forces and the propagation of electromagnetic waves.
• When you're dealing with equations like \(u = \frac{1}{2} \varepsilon_0 E^2\) for electric energy density, understanding \(\varepsilon_0\) helps us appreciate why energy spreads out the way it does in vacuum.

Grasping vacuum permittivity means understanding how space itself affects electric interactions.

Electric Field Amplitude

Electric field amplitude represents the peak strength of an electric field within an electromagnetic wave. This component is vital in our calculations and understanding of electromagnetic energy transfer.
In the context of electromagnetic waves, when we describe the amplitude, we're referring to the maximum value of the electric field, denoted by \(E_0\). Think of it like the height of a wave on water, indicating how powerful the wave is. This amplitude helps determine the energy density of the wave with the formula \(u = \frac{1}{2} \varepsilon_0 E_0^2\).

• Amplitude can vary depending on the source of the electromagnetic wave.
• The higher the amplitude, the more energy is being transferred by the wave.
• It's useful in understanding how different magnitudes affect interactions in physics and engineering applications.

By grasping how electric field amplitude works, we can better predict and analyze the behavior of electromagnetic waves across different scenarios.