Calculate the capacitance of the Earth. Treat the Earth as an isolated spherical conductor of radius \(6371 \mathrm{~km}\).

Short Answer

Expert verified
Answer: The approximate capacitance of the Earth is 715.4 pF (picoFarads).

Step by step solution

01

Convert radius to meters

First, we need to convert the given radius from kilometers to meters: $$ 6371 \mathrm{~km} \times \frac{1000 \mathrm{~m}}{1 \mathrm{~km}} = 6,371,000 \mathrm{~m} $$ The radius of the Earth in meters is now \(6,371,000 \mathrm{~m}\).
02

Calculate capacitance

Now, we can plug the values into the capacitance formula: $$ C = 4\pi \epsilon_0 R $$ Where \(\epsilon_0 = 8.854\times 10^{-12} \mathrm{F/m}\) is the vacuum permittivity. Substituting the values, we get: $$ C = 4\pi (8.854\times 10^{-12} \mathrm{F/m}) (6,371,000 \mathrm{~m}) $$
03

Compute the result

Now, we can perform the calculations to find the capacitance: $$ C \approx 4 \times 3.14 \times (8.854\times 10^{-12} \mathrm{F/m}) \times (6,371,000 \mathrm{~m}) $$ $$ C \approx 715.4 \times 10^{-12} \mathrm{F} $$ Thus, the capacitance of the Earth can be approximated as \(715.4 \times 10^{-12} \mathrm{F}\) or \(715.4 \mathrm{pF}\) (picoFarads).

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

Which of the following is proportional to the capacitance of a parallel plate capacitor? a) the charge stored on each conducting plate b) the potential difference between the two plates c) the separation distance between the two plates d) the area of each plate e) all of the above f) none of the above

A parallel plate capacitor with air in the gap between the plates is connected to a \(6.00-\mathrm{V}\) battery. After charging, the energy stored in the capacitor is \(72.0 \mathrm{~nJ}\). Without disconnecting the capacitor from the battery, a dielectric is inserted into the gap and an additional \(317 \mathrm{~nJ}\) of energy flows from the battery to the capacitor. a) What is the dielectric constant of the dielectric? b) If each of the plates has an area of \(50.0 \mathrm{~cm}^{2}\), what is the charge on the positive plate of the capacitor after the dielectric has been inserted? c) What is the magnitude of the electric field between the plates before the dielectric is inserted? d) What is the magnitude of the electric field between the plates after the dielectric is inserted?

A parallel plate capacitor of capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A close-fitting slab of dielectric, with dielectric constant \(\kappa\), is then inserted and fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.

When working on a piece of equipment, electricians and electronics technicians sometimes attach a grounding wire to the equipment even after turning the device off and unplugging it. Why would they do this?

Determine all the values of equivalent capacitance you can create using any combination of three identical capacitors with capacitance \(C\).

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