The rms value of the electric field of the light coming from the sun is 720 NC^-1. The average total

Q: Why don't we use the magnetic field to find the energy density?

You can! The total energy density is also equal to $B_{\text{rms}}^2 / \mu_0$. Since we are only given the electric field, it's more direct to use the formula involving $E_{\text{rms}}$.

Q: Is the energy density constant?

The *instantaneous* energy density fluctuates as the wave propagates, but the *average* energy density over time is constant, which is what we calculated.

Q: Why is the formula $\varepsilon_0 E_{\text{rms}}^2$ and not $\frac{1}{2} \varepsilon_0 E_{\text{rms}}^2$?

The electric energy density alone is $\frac{1}{2} \varepsilon_0 E_{\text{rms}}^2$. Since the magnetic field contributes an equal amount of energy, the *total* average energy density is double that, giving $\varepsilon_0 E_{\text{rms}}^2$.

Q: What is the value of $\varepsilon_0$?

$\varepsilon_0$ is the permittivity of free space, approximately equal to $8.85 \times 10^{-12} \text{ F/m}$ (or $\text{C}^2\text{N}^{-1}\text{m}^{-2}$).

Q: How do I quickly square 720 without a calculator?

You can break it down: $720 = 72 \times 10$. $72^2 = (70+2)^2 = 4900 + 280 + 4 = 5184$. So, $720^2 = 518400$.

Question

The rms value of the electric field of the light coming from the sun is $720 	ext{ NC}^{-1}$. The average total energy density of the electromagnetic wave is

Options:

(A) $4.58 	imes 10^{-6} 	ext{ Jm}^{-3}$

(B) $6.37 	imes 10^{-9} 	ext{ Jm}^{-3}$

(C) $81.35 	imes 10^{-12} 	ext{ Jm}^{-3}$

(D) $3.3 	imes 10^{-3} 	ext{ Jm}^{-3}$

Sigma Percentile · Accepted Answer

### The Energy of Sunlight

Imagine standing outside on a bright, sunny day. The warmth you feel on your skin is energy delivered across millions of kilometers of empty space. This energy travels in the form of **electromagnetic waves**—a synchronized dance of oscillating electric and magnetic fields.

In this problem, we are given the root mean square (rms) value of the electric field of sunlight, which is $E_{	ext{rms}} = 720 	ext{ NC}^{-1}$. Our goal is to find the **average total energy density** of this wave.

### The Master Equation

An electromagnetic wave carries energy in both its electric and magnetic fields. Remarkably, the energy is shared equally between the two. The energy density (energy per unit volume) of the electric field is given by $\frac{1}{2}\varepsilon_0 E_{	ext{rms}}^2$, and the magnetic field contributes an identical amount.

Therefore, the **total average energy density** $u$ is simply the sum of both, which simplifies beautifully to:
$$u = \varepsilon_0 E_{	ext{rms}}^2$$

Here, $\varepsilon_0$ is the permittivity of free space, a fundamental constant of nature approximately equal to $8.85 	imes 10^{-12} 	ext{ C}^2	ext{N}^{-1}	ext{m}^{-2}$.

### Executing the Calculation

Now, we just need to plug in our values and carefully compute the result. Let's substitute the knowns into our master equation:
$$u = (8.85 	imes 10^{-12}) 	imes (720)^2$$

First, let's handle the square of the electric field. Squaring $720$ gives us $518400$, which we can write in scientific notation as $5.184 	imes 10^5$.

Now, our equation looks like this:
$$u = 8.85 	imes 10^{-12} 	imes 5.184 	imes 10^5$$

Multiplying the numerical parts ($8.85 	imes 5.184$) gives approximately $45.87$. Combining the powers of ten ($10^{-12} 	imes 10^5$) leaves us with $10^{-7}$.

So, we have:
$$u \approx 45.87 	imes 10^{-7} 	ext{ Jm}^{-3}$$

To match standard scientific notation and our given options, we shift the decimal point one place to the left, which increases the exponent by one:
$$u = 4.58 	imes 10^{-6} 	ext{ Jm}^{-3}$$

This perfectly matches option (a). The next time you feel the sun's warmth, you'll know exactly how to calculate the energy density of the light hitting you!

Animated Solution for Physics - Electromagnetic Waves: The rms value of the electric field of the light coming from the sun is $720 NC^{- 1}$ . The average total energy density of the electromagnetic wave is

Select Answer:

Visualized Solution

$E_{rms}$ of EM Wave

Total Average Energy Density

Substituting Values

Squaring the Electric Field

Final Calculation

Final Answer

The Way Forward

The Sigma Insight: Characteristics of Electromagnetic Waves

The Energy of Sunlight

The Master Equation

Executing the Calculation

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