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Chapter 24: Problem 28

A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between plates?

Short Answer

Expert verified
Answer: The fractional difference in the capacitances of the spherical capacitor and the parallel plate capacitor is given by \(\frac{r_2 - r_1}{r_1}\), where \(r_1\) and \(r_2\) are the radii of the inner and outer spheres of the spherical capacitor, respectively.

Step by step solution

01

Capacitance of a Spherical Capacitor

To calculate the capacitance of a spherical capacitor made from two thin concentric conducting shells with radii \(r_{1}\) and \(r_{2}\), we use the following formula: \(C_\text{spherical} = 4 \pi \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1}\) where \(\varepsilon_0\) is the vacuum permittivity.
02

Capacitance of a Parallel Plate Capacitor

To calculate the capacitance of a parallel plate capacitor with the same area as the inner sphere and the same separation \(d = r_2 - r_1\) between the plates, we use the following formula: \(C_\text{parallel} = \frac{\varepsilon_0 A}{d}\) where \(A\) is the area of the plates which is equal to the surface area of the inner sphere: \(A = 4 \pi r_1^2\) Therefore, the capacitance of the parallel plate capacitor is: \(C_\text{parallel} = \frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}\)
03

Find the Difference in Capacitances

We will now find the difference between the capacitances of the spherical and parallel plate capacitors: \(\Delta C = C_\text{spherical} - C_\text{parallel}\) Substituting the values, we get: \(\Delta C = 4 \pi \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1} - \frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}\)
04

Calculate the Fractional Difference

Finally, we will calculate the fractional difference between the capacitances. The fractional difference is given by: \(\text{Fractional Difference} = \frac{\Delta C}{C_\text{parallel}}\) Substituting the values, we get: \(\text{Fractional Difference} = \frac{4 \pi \varepsilon_0 \frac{r_1 r_2}{r_2 - r_1} - \frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}}{\frac{\varepsilon_0 \cdot 4 \pi r_1^2}{r_2 - r_1}}\) Simplifying, we get: \(\text{Fractional Difference} = \frac{r_2 - r_1}{r_1}\) This is the fractional difference in the capacitances of the spherical capacitor and the parallel plate capacitor.

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Most popular questions from this chapter

The Earth can be thought of as a spherical capacitor. If the net charge on the Earth is \(-7.8 \cdot 10^{5} \mathrm{C},\) find \((\) a) the capacitance of the Earth and (b) the electric potential energy stored on the Earth's surface

A parallel plate capacitor with a plate area of \(12.0 \mathrm{~cm}^{2}\) and air in the space between the plates, which are separated by \(1.50 \mathrm{~mm},\) is connected to a \(9.00-\mathrm{V}\) battery. If the plates are pulled back so that the separation increases to \(2.75 \mathrm{~mm}\) how much work is done?

Two parallel plate capacitors, \(C_{1}\) and \(C_{2},\) are connected in series to a \(96.0-\mathrm{V}\) battery. Both capacitors have plates with an area of \(1.00 \mathrm{~cm}^{2}\) and a separation of \(0.100 \mathrm{~mm} ;\) \(C_{1}\) has air between its plates, and \(C_{2}\) has that space filled with porcelain (dielectric constant of 7.0 and dielectric strength of \(5.70 \mathrm{kV} / \mathrm{mm}\) ). a) After charging, what are the charges on each capacitor? b) What is the total energy stored in the two capacitors? c) What is the electric field between the plates of \(C_{2} ?\)

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

A capacitor consists of two parallel plates, but one of them can move relative to the other as shown in the figure. Air fills the space between the plates, and the capacitance is \(32.0 \mathrm{pF}\) when the separation between plates is \(d=0.500 \mathrm{~cm} .\) a) A battery with potential difference \(V=9.0 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma\), on the left plate? What are the capacitance, \(C^{\prime},\) and charge distribution, \(\sigma^{\prime},\) when \(d\) is changed to \(0.250 \mathrm{~cm} ?\) b) With \(d=0.500 \mathrm{~cm}\), the battery is disconnected from the plates. The plates are then moved so that \(d=0.250 \mathrm{~cm}\) What is the potential difference \(V^{\prime},\) between the plates?
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