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Chapter 31: Problem 35

A monochromatic point source of light emits \(1.5 \mathrm{~W}\) of electromagnetic power uniformly in all directions. Find the Poynting vector at a point situated at each of the following locations: a) \(0.30 \mathrm{~m}\) from the source b) \(0.32 \mathrm{~m}\) from the source c) \(1.00 \mathrm{~m}\) from the source

Short Answer

Expert verified
Question: Calculate the Poynting vector (energy flux) at points with different distances from a point source of light that emits 1.5 W of electromagnetic power uniformly in all directions: a) 0.30m, b) 0.32m, and c) 1.00m. Answer: a) At 0.30m, the Poynting vector is 1.326 W/m². b) At 0.32m, the Poynting vector is 1.161 W/m². c) At 1.00m, the Poynting vector is 0.119 W/m².

Step by step solution

01

Find the area at 0.30m distance

First, we need to find the surface area for a sphere whose radius is 0.30m. We'll use the formula for the surface area of a sphere: $$A = 4\pi r^2$$ With \(r = 0.30 \mathrm{~m}\): $$A_{0.30m} = 4\pi (0.30)^2 = 4\pi (0.09) = 1.131\mathrm{~m²}$$
02

Calculate the Poynting vector at 0.30m distance

Now, we'll use the formula for the Poynting vector to find the energy flux at 0.30m distance from the source: $$\vec{S}=\frac{P}{A}$$ With \(P = 1.5 \mathrm{~W}\) and \(A = 1.131\mathrm{~m²}\): $$\vec{S}_{0.30m}=\frac{1.5}{1.131} = 1.326\mathrm{~W/m²}$$
03

Find the area at 0.32m distance

As we did in Step 1, we need to find the surface area for a sphere whose radius is 0.32m: $$A_{0.32m} = 4\pi (0.32)^2 = 4\pi (0.1024) = 1.292\mathrm{~m²}$$
04

Calculate the Poynting vector at 0.32m distance

Now, we'll use the formula for the Poynting vector to find the energy flux at 0.32m distance from the source: $$\vec{S}_{0.32m}=\frac{P}{A}$$ With \(A = 1.292\mathrm{~m²}\): $$\vec{S}_{0.32m}=\frac{1.5}{1.292} = 1.161\mathrm{~W/m²}$$
05

Find the area at 1.00m distance

Finally, we need to find the surface area for a sphere whose radius is 1.00m: $$A_{1.00m} = 4\pi (1.00)^2 = 4\pi = 12.566\mathrm{~m²}$$
06

Calculate the Poynting vector at 1.00m distance

We'll use the formula for the Poynting vector to find the energy flux at 1.00m distance from the source: $$\vec{S}_{1.00m}=\frac{P}{A}$$ With \(A = 12.566\mathrm{~m²}\): $$\vec{S}_{1.00m}=\frac{1.5}{12.566} = 0.119\mathrm{~W/m²}$$ To summarize, the Poynting vector for the given distances is: a) \(0.30\mathrm{~m}\) : \(\vec{S}_{0.30m}=1.326\mathrm{~W/m²}\) b) \(0.32\mathrm{~m}\) : \(\vec{S}_{0.32m}=1.161\mathrm{~W/m²}\) c) \(1.00\mathrm{~m}\) : \(\vec{S}_{1.00m}=0.119\mathrm{~W/m²}\)

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