Problem 59 - University Physics with Modern Physics

Question

As shown in the figure, sunlight is coming straight down (negative \(z\) -direction) on a solar panel (of length \(L=1.40 \mathrm{~m}\) and width \(W=0.900 \mathrm{~m}\) ) on the Mars rover Spir- it. The amplitude of the electric field in the solar radiation is \(673 \mathrm{~V} / \mathrm{m}\) and is uniform (the radiation has the same amplitude everywhere). If the solar panel has an efficiency of \(18.0 \%\) in converting solar radiation into electrical power, how much average power can the panel generate?

Step-by-Step Solution

Answer

Solution: After performing the calculations, the average power generated by the solar panel is: \(P = \left(\frac{1}{2} \cdot 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.00 \times 10^8 \, \text{m/s} \cdot (673 \, \text{V/m})^2 \right) \cdot 0.180 \cdot (1.40\,\text{m} \cdot 0.900\,\text{m})\) Calculate the value of \(P\), and provide the result in watts.

Step 1: Calculate the intensity of solar radiation

Using the given amplitude of the electric field, \(E = 673 \,\text{V/m}\), we can calculate the intensity of the solar radiation using the formula: \(I = \frac{1}{2} \epsilon_0 c E^2\), where \(\epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m}\) is the permittivity of free space, and \(c = 3.00 \times 10^8 \, \text{m/s}\) is the speed of light. \(I = \frac{1}{2} \cdot 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.00 \times 10^8 \, \text{m/s} \cdot (673 \, \text{V/m})^2\)

Step 2: Calculate the active surface area of the solar panel

Given the dimensions of the solar panel, we can find the active surface area by multiplying its length and width: \(A = L \cdot W = 1.40\,\text{m} \cdot 0.900\,\text{m}\)

Step 3: Calculate the average power generated by the solar panel

To find the average power generated, we have to multiply the intensity of solar radiation, the efficiency of the solar panel, and the active surface area: \(P = I \cdot \eta \cdot A\), where \(\eta = 18.0\% = 0.180\) \(P = \left(\frac{1}{2} \cdot 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.00 \times 10^8 \, \text{m/s} \cdot (673 \, \text{V/m})^2 \right) \cdot 0.180 \cdot (1.40\,\text{m} \cdot 0.900\,\text{m})\) Now, calculate the value of \(P\) to get the average power generated by the solar panel.

Key Concepts

Understanding Solar Radiation Intensity

Solar radiation intensity refers to the power per unit area received from the Sun in the form of electromagnetic radiation. This intensity can significantly affect how much power a solar panel can generate. To find it, we need the formula:

• \( I = \frac{1}{2} \epsilon_0 c E^2 \)

Here:

• \( \epsilon_0 \) is the permittivity of free space, \( 8.85 \times 10^{-12} \, \text{F/m} \)
• \( c \) is the speed of light, \( 3.00 \times 10^8 \, \text{m/s} \)
• \( E \) is the electric field amplitude

This expression tells us how solar energy is spread across an area at a point in space, influencing how efficiently solar panels on devices like the Mars rover can capture this energy.
As your calculations reveal intensity, it becomes easier to connect these insights with practical applications.

Role of Electric Field Amplitude in Solar Panels

Electric field amplitude (\( E \)) is vital for calculating solar radiation intensity. It represents the maximum strength of the electric field component of solar radiation. In our scenario, the amplitude is \( 673 \, \text{V/m} \). This value is crucial because solar panels convert the electromagnetic energy of sunlight into electrical energy, and the electric field plays a central role in this conversion process.

• Higher electric field amplitude means higher potential energy transformations.
• It's directly involved in the equations that help us determine the potential power output of a solar panel.

Thus, knowing the electric field amplitude helps predict the efficiency and potential power output of solar modules, which is essential for planning energy needs and solutions for exploration missions like those on Mars.

Mars Rover Power Output: Meeting Energy Demands

Mars rovers, like Spirit, rely heavily on solar panels for their energy. This power output is pivotal, as it determines the rover's ability to perform scientific tasks and maintain essential functions. The steps for assessing power output include:

• Calculating solar radiation intensity, which provides the baseline solar power available to the panel.
• Determining the solar panel’s efficiency to ascertain how much of the received solar energy can be converted to electrical energy.

Given the values, a solar panel with an efficiency of \( 18\% \) means only a fraction of the received solar radiation converts into usable power.
Knowing the power output helps engineers and scientists ensure that the rover has enough energy to complete its mission tasks and survive in the harsh Martian climate.

Solar Power Calculation: Harnessing Energy Efficiently

To calculate the solar power generated by a panel, we need to understand both the solar radiation intensity that hits the panel and the panel's efficiency. The expression used is:

• \( P = I \cdot \eta \cdot A \)

Where:

• \( I \) is the intensity of solar radiation
• \( \eta \) is the efficiency of the solar panel (expressed in decimals)
• \( A \) is the active surface area of the solar panel

The active surface area is simply the panel's length times its width, allowing us to calculate the total area exposed to sunlight.
Efficiency consideration ensures we understand how much of the sunlight energy is converted into usable electricity.
Proper calculation assists in optimizing energy usage to meet the specific power requirements of devices or missions, like the Mars rover, and can be critical in planning explorative or sustainable energy systems.