/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 Suppose you were to wrap the Moo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 \(1.94 \times 10^{-4} \, F\).

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 91影视!

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

A point charge \(q=+3.0\) nC moves through a potential difference $\Delta V=V_{\mathrm{f}}-V_{\mathrm{i}}=+25 \mathrm{V} .$ What is the change in the electric potential energy?
As an electron moves through a region of space, its speed decreases from $8.50 \times 10^{6} \mathrm{m} / \mathrm{s}\( to \)2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ The electric force is the only force acting on the electron. (a) Did the electron move to a higher potential or a lower potential? (b) Across what potential difference did the electron travel?
A parallel plate capacitor is composed of two square plates, $10.0 \mathrm{cm}\( on a side, separated by an air gap of \)0.75 \mathrm{mm} .$ (a) What is the charge on this capacitor when there is a potential difference of \(150 \mathrm{V}\) between the plates? (b) What energy is stored in this capacitor?
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 capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.
See all solutions

Recommended explanations on Physics Textbooks

Physics of Motion

Read Explanation

Translational Dynamics

Read Explanation

Medical Physics

Read Explanation

Force

Read Explanation

Space Physics

Read Explanation

Magnetism

Read Explanation
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.