Table of contents

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

26. Capacitors & Dielectrics

Parallel Plate Capacitors

Physics

26. Capacitors & Dielectrics

Parallel Plate Capacitors

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Multiple Choice

One gallon of typical gasoline has 132 MJ of energy. You wish to store that much energy in a parallel-plate capacitor for an electric car. Assume you charge the capacitor using a $120 V$ power source. You can fill the capacitor with any of the materials listed in Table 21.3, but the spacing between plates has to be 0.080 mm. What is the minimum surface area required for the capacitor? This may help to understand challenges in engineering electric cars.

2.8 \times 10^{8} m^{2}

2.0 \times 10^{9} m^{2}

5.5 \times 10^{9} m^{2}

6.6 \times 10^{10} m^{2}

1.7 \times 10^{11} m^{2}

5.5 \times 10^{8} m^{2}

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Verified step by step guidance

First, understand that the energy stored in a capacitor is given by the formula: $ E = \frac{1}{2} C V^2 $, where $ E $ is the energy, $ C $ is the capacitance, and $ V $ is the voltage.

Rearrange the formula to solve for capacitance $ C $: $ C = \frac{2E}{V^2} $. Substitute $ E = 132 \text{ MJ} = 132 \times 10^6 \text{ J} $ and $ V = 120 \text{ V} $ into the equation.

The capacitance $ C $ of a parallel-plate capacitor is also given by $ C = \frac{\varepsilon A}{d} $, where $ \varepsilon $ is the permittivity of the dielectric material, $ A $ is the surface area of one of the plates, and $ d $ is the separation between the plates.

Rearrange this formula to solve for the surface area $ A $: $ A = \frac{C \cdot d}{\varepsilon} $. Substitute the value of $ C $ from the previous step, $ d = 0.080 \text{ mm} = 0.080 \times 10^{-3} \text{ m} $, and choose an appropriate $ \varepsilon $ from Table 21.3.

Calculate the minimum surface area $ A $ using the values obtained. This will give you the minimum surface area required for the capacitor to store the same amount of energy as one gallon of gasoline.