What is the average magnitude of the Poynting vector 5.00 miles from a radio transmitter broadcasting isotropically (equally in all directions) with an average power of 250 kW?

$$r = 5.00 \text{ miles} \times 1609.34 \frac{\text{m}}{\text{mile}} = 8046.7 \text{ m}$$ $$P = 250 \text{ kW} = 250 \times 10^3 \text{ W}$$

$$I = \frac{P}{A} = \frac{250 \times 10^3 \text{ W}}{8.138 \times 10^8 \text{ m}^2} = 3.07 \times 10^{-4} \frac{\text{W}}{\text{m}^2}$$

The average magnitude of the Poynting vector is $\boxed{\mathbf{3.07 \times 10^{-4} \frac{W}{m^2}}}$.

2. A parallel plate capacitor has circular plates with radius of 7.0cm. If the electric field between the plates is changing at a rate of $2 \times 10^{12} \text{ V/m/s}$. Find the displacement current.

The displacement current ($I_d$) is given by the formula $I_d = \epsilon_0 A \frac{dE}{dt}$.

$$r = 7.0 \text{ cm} = 0.07 \text{ m}$$ $$A = \pi r^2 = \pi (0.07 \text{ m})^2 = 0.01539 \text{ m}^2$$

$$\frac{dE}{dt} = 2 \times 10^{12} \frac{\text{V}}{\text{m} \cdot \text{s}}$$ $$\epsilon_0 = 8.854 \times 10^{-12} \frac{\text{F}}{\text{m}}$$ $$I_d = (8.854 \times 10^{-12} \frac{\text{F}}{\text{m}}) (0.01539 \text{ m}^2) (2 \times 10^{12} \frac{\text{V}}{\text{m} \cdot \text{s}})$$

$$I_d = 0.272 \text{ A}$$

The displacement current is $\boxed{\mathbf{0.272 \text{ A}}}$.

3. The electric field amplitude of a laser beam is $6.0 \times 10^8 \text{ V/m}$. Calculate its average and maximum intensity.

The maximum intensity ($I_{max}$) and average intensity ($I_{avg}$) are related to the electric field amplitude ($E_0$).

$$I_{max} = c \epsilon_0 E_0^2$$

Where $c = 3.00 \times 10^8 \text{ m/s}$ and $\epsilon_0 = 8.854 \times 10^{-12} \text{ F/m}$.

$$I_{max} = (3.00 \times 10^8 \frac{\text{m}}{\text{s}}) (8.854 \times 10^{-12} \frac{\text{F}}{\text{m}}) (6.0 \times 10^8 \frac{\text{V}}{\text{m}})^2$$ $$I_{max} = (3.00 \times 10^8) (8.854 \times 10^{-12}) (3.6 \times 10^{17}) \frac{\text{W}}{\text{m}^2}$$ $$I_{max} = 9.56 \times 10^{14} \frac{\text{W}}{\text{m}^2}$$

For a sinusoidal electromagnetic wave, the average intensity is half of the maximum intensity.

$$I_{avg} = \frac{1}{2} I_{max}$$ $$I_{avg} = \frac{1}{2} (9.56 \times 10^{14} \frac{\text{W}}{\text{m}^2}) = 4.78 \times 10^{14} \frac{\text{W}}{\text{m}^2}$$

The maximum intensity is $\boxed{\mathbf{9.56 \times 10^{14} \frac{W}{m^2}}}$ and the average intensity is $\boxed{\mathbf{4.78 \times 10^{14} \frac{W}{m^2}}}$.

4. A microwave oven has an average intensity of $8.33 \times 10^3 \text{ W/m}^2$. Find the peak electric field strength ($E_0$).

The average intensity ($I_{avg}$) is related to the peak electric field strength ($E_0$) by the formula $I_{avg} = \frac{1}{2} c \epsilon_0 E_0^2$.

$$E_0^2 = \frac{2 I_{avg}}{c \epsilon_0}$$ $$E_0 = \sqrt{\frac{2 I_{avg}}{c \epsilon_0}}$$

$$I_{avg} = 8.33 \times 10^3 \frac{\text{W}}{\text{m}^2}$$ $$c = 3.00 \times 10^8 \frac{\text{m}}{\text{s}}$$ $$\epsilon_0 = 8.854 \times 10^{-12} \frac{\text{F}}{\text{m}}$$ $$E_0 = \sqrt{\frac{2 \times (8.33 \times 10^3 \frac{\text{W}}{\text{m}^2})}{(3.00 \times 10^8 \frac{\text{m}}{\text{s}}) (8.854 \times 10^{-12} \frac{\text{F}}{\text{m}})}}$$ $$E_0 = \sqrt{\frac{1.666 \times 10^4}{2.6562 \times 10^{-3}}} \frac{\text{V}}{\text{m}}$$ $$E_0 = \sqrt{6.272 \times 10^6} \frac{\text{V}}{\text{m}}$$

$$E_0 = 2504.4 \frac{\text{V}}{\text{m}}$$

The peak electric field strength is $\boxed{\mathbf{2504.4 \frac{V}{m}}}$.

5. A 100W light bulb radiates energy uniformly in all directions. What is the intensity 2.0m away?

The intensity ($I$) is the power ($P$) distributed over the surface area ($A$) of a sphere at a given distance ($r$).

$$I = \frac{P}{A} = \frac{100 \text{ W}}{50.265 \text{ m}^2} = 1.99 \frac{\text{W}}{\text{m}^2}$$

The intensity 2.0m away is $\boxed{\mathbf{1.99 \frac{W}{m^2}}}$.