Question

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

Question: Calculate the Poynting vector at 0.30 m, 0.32 m, and 1.00 m from a monochromatic point source of light emitting a power of 1.5 W uniformly in all directions. Answer: Using the Poynting vector formula, we get the following values for each distance: 1. At 0.30 m: $$Poynting \thinspace vector = \frac{1.5}{4 \pi (0.30)^2} = 13.283 W/m^2$$ 2. At 0.32 m: $$Poynting \thinspace vector = \frac{1.5}{4 \pi (0.32)^2} = 12.317 W/m^2$$ 3. At 1.00 m: $$Poynting \thinspace vector = \frac{1.5}{4 \pi (1.00)^2} = 0.119 W/m^2$$ As we can see, the Poynting vector decreases as the distance from the source increases.

Step-by-step solution

Compute the Poynting vector at 0.30 m from the source

We will use the formula to calculate the Poynting vector for the given distance. Distance (r) = 0.30 m Power = 1.5 W $$Poynting \thinspace vector = \frac{1.5}{4 \pi (0.30)^2}$$ Calculate the value of the Poynting vector for the given values.

Compute the Poynting vector at 0.32 m from the source

Next, we will use the same formula to calculate the Poynting vector for the given distance. Distance (r) = 0.32 m Power = 1.5 W $$Poynting \thinspace vector = \frac{1.5}{4 \pi (0.32)^2}$$ Calculate the value of the Poynting vector for the given values.

Compute the Poynting vector at 1.00 m from the source

Lastly, we will use the same formula to find the Poynting vector at the given distance. Distance (r) = 1.00 m Power = 1.5 W $$Poynting \thinspace vector = \frac{1.5}{4 \pi (1.00)^2}$$ Calculate the value of the Poynting vector for the given values. After calculating the Poynting vector for each location, compare the values to see how the Poynting vector changes with distance from the source.