Animated Solution for Physics - Electromagnetic Waves: The electric field intensity produced by the radiation coming from a $100\text{ W}$ bulb at a distance of $3\text{ m}$ is $E$. The electric field intensity produced by the radiation coming from $60\text{ W}$ at the same distance is $\sqrt{\frac{x}{5}}E$, where the value of $x$ is ........... .

Imagine you are standing in a dark room, and someone turns on a light bulb. The light you see is an electromagnetic wave, carrying energy through space. The brightness or intensity of this light depends on two main factors: how powerful the bulb is, and how far away you are standing.

In this problem, we are exploring the relationship between the power of a light source and the electric field intensity it produces at a specific distance. Let's break down the physics behind this illuminating concept.

We are given two different light bulbs. The first bulb has a power of $P_1=100\text{ W}$, and it produces an electric field of magnitude $E$ at a distance of $r=3\text{ m}$. The second bulb has a lower power of $P_2=60\text{ W}$. We need to find the electric field $E_2$ produced by this second bulb at the exact same distance of $3\text{ m}$.

To connect power and electric field, we need to look at the intensity of the electromagnetic wave. Intensity ($I$) is defined as the power ($P$) radiated per unit area ($A$). Since a bulb acts like a point source, it radiates energy spherically in all directions. Therefore, the area it covers at a distance $r$ is the surface area of a sphere, $4\pi r^2$.

But intensity can also be expressed in terms of the peak electric field ($E_0$) of the electromagnetic wave. According to electromagnetic theory, the average intensity is given by:

Here, ${ϵ}_0$ is the permittivity of free space, and $c$ is the speed of light. By equating these two expressions for intensity, we can uncover the hidden relationship between power and electric field:

Since we are comparing two bulbs at the same distance $r$, all the terms like $4\pi r^2$, $\frac{1}{2}$, ${ϵ}_0$, and $c$ are constants. This simplifies our equation beautifully, revealing a direct proportionality:

This tells us that the electric field is directly proportional to the square root of the power. We can now set up a ratio for our two bulbs:

Now, it's just a matter of plugging in the numbers. We know $P_1=100\text{ W}$, $P_2=60\text{ W}$, and $E_1=E$.

Rearranging this to solve for $E_2$, we get:

The problem states that the new electric field is $\sqrt{\frac{x}{5}}E$. By comparing our derived expression with the given one, it is crystal clear that the value of $x$ is exactly 3.