Problem 2174 - JEE Physics

Question

An observer is at \(2 \mathrm{~m}\) from an isotropic point source of light emitting \(40 \mathrm{w}\) power. The rm.s value of electric due to the source at the position of the observer is (A) \(5.77 \times 10^{-8} \mathrm{Vm}^{-1}\) (B) \(17.3 \mathrm{Vm}^{-1}\) (C) \(57.7 \times 10^{-8} \mathrm{Vm}^{-1}\) (D) \(1.73 \mathrm{Vm}^{-1}\)

Step-by-Step Solution

Answer

The RMS value of the electric field due to the source at the position of the observer is \(5.77 \times 10^{-8} Vm^{-1}\).

Step 1: Determine the energy density of the wave at the observer's position

Using the isotropic point source nature of the light, we can determine the power radiated per unit solid angle. The total power of the source is 40 W, and since it is isotropic (radiating equally in all directions), the power per unit solid angle can be found using the formula: Power per unit solid angle = \(\frac{Total Power}{4 \pi}\)

Step 2: Calculate the intensity of the wave

The intensity of the wave (energy per unit area per unit time) can be found by dividing the power per unit solid angle by the surface area of the sphere with a radius equivalent to the distance between the observer and the source (2 m). The formula is: Intensity = \(\frac{Power per unit solid angle}{4 \pi r^2}\)

Step 3: Determine the energy density

The energy density (energy per unit volume) of the wave is given by the ratio of intensity to the speed of light (c). The formula is: Energy density = \(\frac{Intensity}{c}\)

Step 4: Calculate the RMS value of the electric field

Using Maxwell's equations, we can determine the RMS value of the electric field from the energy density. The formula is: Electric Field RMS = \(\sqrt{2 \times \frac{Energy Density}{\epsilon_0}}\) Where \(\epsilon_0\) is the vacuum permittivity.

Step 5: Compute the values

Using the given data and formulas above, we can now compute the answers. Power per unit solid angle = \(\frac{40 W}{4 \pi} = 10\pi W/sr\) Intensity = \(\frac{10\pi W / sr}{4 \pi (2m)^2} = \frac{5}{4} \frac{W}{m^2}\) Energy density = \(\frac{\frac{5}{4} \frac{W}{m^2}}{3 \times 10^{8} m/s} = 1.25 \times 10^{-9} \frac{J}{m^3}\) Electric Field RMS = \(\sqrt{2 \times \frac{1.25 \times 10^{-9}\frac{J}{m^3}}{8.85\times10^{-12}F/m}} = 5.77 \times 10^{-8} Vm^{-1}\) So, the RMS value of the electric field due to the source at the position of the observer is \(5.77 \times 10^{-8} Vm^{-1}\). Therefore, the correct option is (A).

Key Concepts

Isotropic Point Source

In physics, an **isotropic point source** refers to a source that emits energy equally in all directions. Think of it as a tiny sphere, from which light or sound spreads out in every direction with the same intensity. This property is especially useful in situations where uniformity of the transmission is important.

The key idea is how this uniform emission allows us to simplify calculations of power and intensity. Since the total power is distributed equally over the entire surface area of a sphere, we can describe its behavior mathematically as:

• Power per unit solid angle: \(\frac{Total \ Power}{4\pi}\)

Where 4π represents the total solid angle around the source measured in steradians. For an observer at a distance, this simplification helps us to rely solely on geometry and basic physical laws to predict outcomes like intensity and field strength.
Understanding isotropic sources can greatly aid in visualizing how energy disperses in space, easing the path to calculating other linked quantities in electromagnetic theory.

Electric Field RMS

The **Electric Field RMS** (Root Mean Square) value is a crucial measurement in electromagnetics. It describes the effective value of a varying electric field, akin to a kind of average field that can do equivalent work as the time-varying one.

The RMS value of the electric field can be determined from the energy density using Maxwell's equations. The formula that connects these is:

• Electric Field RMS: \(\sqrt{2 \times \frac{Energy \ Density}{\epsilon_0}}\)

Here, \(\epsilon_0\) is the vacuum permittivity, a constant that describes how electric fields interact in a vacuum. RMS values are important because they allow for straightforward comparisons and simplifications when discussing fields that oscillate, such as light waves.
This concept makes it possible to use simpler arithmetic in the analysis and presentation of data, particularly in fields like electronics and signal processing where AC currents are prevalent.

Wave Intensity

**Wave intensity** quantifies how much energy a wave carries as it travels through space. It is defined as the energy passing per unit area per unit time and provides a measure of how strong or weak a wave is at a certain point.

To calculate wave intensity for an isotropic point source, we use:

• Intensity: \(\frac{Power \ per \ unit \ solid \ angle}{4 \pi r^2}\)

Here, \(r\) is the distance from the source. Intensity decreases as you move further from the source, due to the geometric spreading of the wave. The further you are, the more the wave's energy is spread out over a larger area.
Wave intensity is a key concept in many applications, such as optics and telecommunications, where understanding how waves propagate and diminish over distance can impact design and efficiency.

Energy Density

**Energy density** describes how much energy is present in a particular volume of space, conveyed by an electromagnetic wave. It's akin to the energy "density" within a certain area, explaining how much energy is packed into the volume that the wave occupies.

For electromagnetic waves, energy density relates directly to wave intensity:

• Energy density: \(\frac{Intensity}{c}\)

Here, \(c\) is the speed of light. This relationship arises because energy travels through space at the speed of light in a vacuum. The higher the intensity, the more energy there is in every small patch of space, which means a higher energy density.
Understanding energy density is essential for designing and evaluating systems that transmit or interact with electromagnetic energy, like antennas and other communication systems, where optimizing how energy is delivered and absorbed can be crucial for system performance.