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Chapter 17: Problem 62

Suppose you were to wrap the Moon in aluminum foil and place a charge \(Q\) on it. What is the capacitance of the Moon in this case? [Hint: It is not necessary to have two oppositely charged conductors to have a capacitor. Use the definition of potential for a spherical conductor and the definition of capacitance to get your answer. \(]\)

Short Answer

Expert verified
Answer: The capacitance of the Moon wrapped in aluminum foil and charged with Q is approximately 1.94 x 10^-4 F.

Step by step solution

01

1. Define the capacitance of a capacitor

The capacitance (\(C\)) of a capacitor is defined by the relation: \(C = \frac{Q}{V}\), where \(Q\) is the charge on the plates and \(V\) is the potential difference between the plates.
02

2. Calculate the potential of the Moon as a spherical conductor

The potential of a spherical conductor with a charge \(Q\) and radius \(R\) can be given by the formula: \(V = \frac{kQ}{R}\), where \(k\) is the electrostatic constant, which is approximately \(8.99 \times 10^9 \, Nm^2/C^2\). The radius of the Moon, \(R\), is about \(1.74 \times 10^6 \, m\) and we are given the charge \(Q\).
03

3. Apply the capacitance formula

From step 1, we know that \(C = \frac{Q}{V}\). Now, substitute the formula for potential from step 2: \(C = \frac{Q}{\frac{kQ}{R}}\). When we simplify this expression, we get: \(C = \frac{Q}{kQ} \times R = \frac{1}{k} \times R\).
04

4. Calculate the capacitance

Now that we have the formula for capacitance in terms of the radius of the Moon and the electrostatic constant, we can calculate the capacitance: \(C = \frac{1}{k} \times R = \frac{1}{8.99 \times 10^9 \, Nm^2/C^2} \times 1.74 \times 10^6 \, m\). After the calculation, we have: \(C \approx 1.94 \times 10^{-4} \, F\). Finally, the capacitance of the Moon wrapped in aluminum foil and charged with \(Q\) is approximately \(1.94 \times 10^{-4} \, F\).

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Most popular questions from this chapter

A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
In \(1911,\) Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of \(6.68 \times 10^{-27} \mathrm{kg},\) a charge of \(+2 e,\) and an initial velocity of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) is projected head-on toward a gold nucleus with a charge of \(+79 e\), how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)
The nucleus of a helium atom contains two protons that are approximately 1 fm apart. How much work must be done by an external agent to bring the two protons from an infinite separation to a separation of \(1.0 \mathrm{fm} ?\)
To make a parallel plate capacitor, you have available two flat plates of aluminum (area \(120 \mathrm{cm}^{2}\) ), a sheet of paper (thickness $=0.10 \mathrm{mm}, \kappa=3.5),\( a sheet of glass (thickness \)=2.0 \mathrm{mm}, \kappa=7.0),\( and a slab of paraffin (thickness \)=10.0 \mathrm{mm}, \kappa=2.0) .$ (a) What is the largest capacitance possible using one of these dielectrics? (b) What is the smallest?
Figure 17.31 b shows a thundercloud before a lightning strike has occurred. The bottom of the thundercloud and the Earth's surface might be modeled as a charged parallel plate capacitor. The base of the cloud, which is roughly parallel to the Earth's surface, serves as the negative plate and the region of Earth's surface under the cloud serves as the positive plate. The separation between the cloud base and the Earth's surface is small compared to the length of the cloud. (a) Find the capacitance for a thundercloud of base dimensions \(4.5 \mathrm{km}\) by \(2.5 \mathrm{km}\) located \(550 \mathrm{m}\) above the Earth's surface. (b) Find the energy stored in this capacitor if the charge magnitude is \(18 \mathrm{C}\).
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