What is the radius of an isolated spherical conductor that has a capacitance of \(1.00 \mathrm{~F} ?\)

Short Answer

Expert verified
Answer: The radius of the isolated spherical conductor is approximately \(8.92 \times 10^6 \mathrm{m}\).

Step by step solution

01

Recall the Formula for the Capacitance of a Spherical Conductor

The formula for capacitance (C) of an isolated spherical conductor is given by: C = \(4\pi\epsilon_0 R\) Where \(\epsilon_0\) is the permittivity of free space (\(8.854 \times 10^{-12} \mathrm{F/m}\)) and R represents the radius of the sphere.
02

Rearrange the Formula to Solve for the Radius

In order to find the radius (R), we will rearrange the formula above to have R by itself on one side: R = \(\frac{C}{4\pi\epsilon_0}\)
03

Substitute the given Capacitance Value and Calculate the Radius

The given capacitance value is \(1.00 \mathrm{F}\), so now we will substitute this value into the formula to find the radius: R = \(\frac{1.00 \mathrm{F}}{4\pi(8.854 \times 10^{-12} \mathrm{F/m})}\) Now, we will perform the calculation to find the radius. R = \(\frac{1.00}{4\pi(8.854 \times 10^{-12})} \approx 8.92 \times 10^6 \mathrm{m}\) Thus, the radius of the isolated spherical conductor is approximately \(8.92 \times 10^6 \mathrm{m}\).

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

A parallel plate capacitor has a capacitance of \(120 .\) pF and a plate area of \(100 . \mathrm{cm}^{2}\). The space between the plates is filled with mica whose dielectric constant is \(5.40 .\) The plates of the capacitor are kept at \(50.0 \mathrm{~V}\) a) What is the strength of the electric field in the mica? b) What is the amount of free charge on the plates? c) What is the amount of charge induced on the mica?

A parallel plate capacitor is constructed from two plates of different areas. If this capacitor is initially uncharged and then connected to a battery, how will the amount of charge on the big plate compare to the amount of charge on the small plate?

An \(8.00-\mu F\) capacitor is fully charged by a \(240 .-V\) battery, which is then disconnected. Next, the capacitor is connected to an initially uncharged capacitor of capacitance \(C,\) and the potential difference across it is found to be \(80.0 \mathrm{~V}\) What is \(C ?\) How much energy ends up being stored in the second capacitor?

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 \(4.00 \cdot 10^{3}-n F\) parallel plate capacitor is connected to a \(12.0-\mathrm{V}\) battery and charged. a) What is the charge \(Q\) on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor? The \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor is then disconnected from the \(12.0-\mathrm{V}\) battery and used to charge three uncharged capacitors, a \(100 .-n F\) capacitor, a \(200 .-\mathrm{nF}\) capacitor, and a \(300 .-\mathrm{nF}\) capacitor, connected in series. c) After charging, what is the potential difference across each of the four capacitors? d) How much of the electrical energy stored in the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor was transferred to the other three capacitors?

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