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 \(8.92 \times 10^6 \mathrm{m}\).

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Most popular questions from this chapter

A large parallel plate capacitor with plates that are square with side length \(1.00 \mathrm{~cm}\) and are separated by a distance of \(1.00 \mathrm{~mm}\) is dropped and damaged. Half of the areas of the two plates are pushed closer together to a distance of \(0.500 \mathrm{~mm}\). What is the capacitance of the damaged capacitor?

Two concentric metal spheres are found to have a potential difference of \(900 . \mathrm{V}\) when a charge of \(6.726 \cdot 10^{-8} \mathrm{C}\) is applied to them. The radius of the outer sphere is \(0.210 \mathrm{~m}\). What is the radius of the inner sphere?

A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as \(625 \mathrm{~mA} \mathrm{~h}\), meaning that much charge can be delivered at approximately \(1.5 \mathrm{~V}\).) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of \(1.0 \mathrm{~mm}\) and with air filling the space between the plates. a) Assuming that the potential difference across the capacitor is \(1.5 \mathrm{~V},\) what must the area of each plate be? b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be? c) Is either capacitor a practical replacement for the AAA batterv?

A parallel plate capacitor has square plates of side \(L=\) \(10.0 \mathrm{~cm}\) and a distance \(d=1.00 \mathrm{~cm}\) between the plates. Of the space between the plates, \(\frac{1}{3}\) is filled with a dielectric with dielectric constant \(\kappa_{1}=20.0 .\) The remaining \(\frac{4}{5}\) of the space is filled with a different dielectric with \(\kappa_{2}=5.00 .\) Find the capacitance of the capacitor.

The Earth is held together by its own gravity. But it is also a charge-bearing conductor. a) The Earth can be regarded as a conducting sphere of radius \(6371 \mathrm{~km},\) with electric field \(\vec{E}=(-150 . \mathrm{V} / \mathrm{m}) \hat{r}\) at its surface, where \(\hat{r}\) is a unit vector directed radially outward. Calculate the total electrostatic potential energy associated with the Earth's electric charge and field. b) The Earth has gravitational potential energy, akin to the electrostatic potential energy. Calculate this energy, treating the Earth as a uniform solid sphere. (Hint: \(d U=-(G m / r) d m\). c) Use the results of parts (a) and (b) to address this question: To what extent do electrostatic forces affect the structure of the Earth?

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