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?

Short Answer

Expert verified
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 by step solution

01

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. \(I = \frac{1}{2} \cdot 8.85 \times 10^{-12} \, \text{F/m} \cdot 3.00 \times 10^8 \, \text{m/s} \cdot (673 \, \text{V/m})^2\)
02

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: \(A = L \cdot W = 1.40\,\text{m} \cdot 0.900\,\text{m}\)
03

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Three FM radio stations covering the same geographical area broadcast at frequencies \(91.1,91.3,\) and \(91.5 \mathrm{MHz},\) respectively. What is the maximum allowable wavelength width of the band-pass filter in a radio receiver so that the FM station 91.3 can be played free of interference from FM 91.1 or FM 91.5? Use \(c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\), and calculate the wavelength to an uncertainty of \(1 \mathrm{~mm} .\)

Suppose an RLC circuit in resonance is used to produce a radio wave of wavelength \(150 \mathrm{~m}\). If the circuit has a 2.0 -pF capacitor, what size inductor is used?

Calculate the average value of the Poynting vector, \(S_{\text {ave }}\) for an electromagnetic wave having an electric field of amplitude \(100 . \mathrm{V} / \mathrm{m}\) a) What is the average energy density of this wave? b) How large is the amplitude of the magnetic field?

Silica aerogel, an extremely porous, thermally insulating material made of silica, has a density of \(1.00 \mathrm{mg} / \mathrm{cm}^{3}\). A thin circular slice of aerogel has a diameter of \(2.00 \mathrm{~mm}\) and a thickness of \(0.10 \mathrm{~mm}\). a) What is the weight of the aerogel slice (in newtons)? b) What is the intensity and radiation pressure of a \(5.00-\mathrm{mW}\) laser beam of diameter \(2.00 \mathrm{~mm}\) on the sample? c) How many \(5.00-\mathrm{mW}\) lasers with a beam diameter of \(2.00 \mathrm{~mm}\) would be needed to make the slice float in the Earth's gravitational field? Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

A \(14.9-\mu F\) capacitor, a \(24.3-\mathrm{k} \Omega\) resistor, a switch, and a 25.-V battery are connected in series. What is the rate of change of the electric field between the plates of the capacitor at \(t=0.3621 \mathrm{~s}\) after the switch is closed? The area of the plates is \(1.00 \mathrm{~cm}^{2}\) .

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free