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Chapter 24: Problem 82

The capacitance of a spherical capacitor consisting of two concentric conducting spheres with radii \(r_{1}\) and \(r_{2}\) \(\left(r_{2}>r_{1}\right)\) is given by \(C=4 \pi \epsilon_{0} r_{1} r_{2} /\left(r_{2}-r_{1}\right) .\) Suppose that the space between the spheres, from \(r,\) up to a radius \(R\) \(\left(r_{1}

Short Answer

Expert verified
Answer: The final expression for the capacitance with a dielectric layer of radius R is: $$C = \frac{4\pi \epsilon_{0} r_{1} r_{2}}{9r_{2} - r_{1} -(9r_{1} - r_{2})\frac{R-r_{1}}{R}}$$

Step by step solution

01

Capacitance of sphere with two dielectrics

We divide the space between the conducting spheres into two regions: the region between \(r_{1}\) and R filled with dielectric whose permittivity is \(10 \epsilon_{0}\), and the region from R to \(r_{2}\) filled with dielectric whose permittivity is \(\epsilon_{0}\). Let's label the capacitances for these regions as \(C_1\) for the region from \(r_{1}\) to R and \(C_2\) for region from R to \(r_{2}\). Since the dielectric fills the region between the spaces, the capacitances \(C_{1}\) and \(C_{2}\) are in series. Therefore, the total capacitance C is given by: $$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$$
02

Find the capacitance for each region

The capacitance of a spherical capacitor filled with a dielectric is given by the formula: \(C_{dielectric} = 4\pi \epsilon r_{1} r_{2} / (r_{2} - r_{1})\) For the region between \(r_{1}\) and R filled with the dielectric of permittivity \(10 \epsilon_{0}\), we have: $$C_1 = \frac{4 \pi (10 \epsilon_{0}) r_{1}R}{R - r_{1}}$$ For the region between R and \(r_{2}\) filled with the dielectric of permittivity \(\epsilon_{0}\), we have: $$C_2 = \frac{4 \pi \epsilon_{0} R r_{2}}{r_{2} - R}$$
03

Total Capacitance

Now, we substitute the expressions for \(C_{1}\) and \(C_{2}\) in the total capacitance equation: $$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$\frac{1}{C} = \frac{R - r_{1}}{4 \pi (10 \epsilon_{0}) r_{1}R} + \frac{r_{2} - R}{4 \pi \epsilon_{0} R r_{2}}$$ After solving for C, we get: $$C = \frac{4\pi \epsilon_{0} r_{1} r_{2}}{9r_{2} - r_{1} -(9r_{1} - r_{2})\frac{R-r_{1}}{R}}$$
04

Limit \(R = r_{1}\)

As \(R \rightarrow r_{1}\), the numerator remains the same, while the denominator becomes \(9r_{2} - r_{1} - (9r_{1} - r_{2}) = 8(r_{2} - r_{1})\). So the total capacitance simplifies to: $$C = \frac{4 \pi \epsilon_{0} r_{1} r_{2}}{8(r_{2} - r_{1})}$$ which is the capacitance of the original spherical capacitor without dielectric, confirming our limit case.
05

Limit \(R = r_{2}\)

As \(R \rightarrow r_{2}\), the numerator remains the same, while the denominator becomes \(9r_{2} - r_{1}\). So the total capacitance simplifies to: $$C = \frac{4 \pi \epsilon_{0} r_{1} r_{2}}{9r_{2} - r_{1}}$$ which corresponds to the capacitance of the spherical capacitor filled entirely with the dielectric of permittivity \(10 \epsilon_{0}\), confirming our limit case. Thus, the final expression for the capacitance with a dielectric layer of radius R is: $$C = \frac{4\pi \epsilon_{0} r_{1} r_{2}}{9r_{2} - r_{1} -(9r_{1} - r_{2})\frac{R-r_{1}}{R}}$$

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Most popular questions from this chapter

A quantum mechanical device known as the Josephson junction consists of two overlapping layers of superconducting metal (for example, aluminum at \(1.00 \mathrm{~K}\) ) separated by \(20.0 \mathrm{nm}\) of aluminum oxide, which has a dielectric constant of \(9.1 .\) If this device has an area of \(100 . \mu \mathrm{m}^{2}\) and a parallel plate configuration, estimate its capacitance.

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?

A parallel plate capacitor consists of square plates of edge length \(2.00 \mathrm{~cm}\) separated by a distance of \(1.00 \mathrm{~mm}\). The capacitor is charged with a \(15.0-\mathrm{V}\) battery, and the battery is then removed. A \(1.00-\mathrm{mm}\) -thick sheet of nylon (dielectric constant \(=3.0\) ) is slid between the plates. What is the average force (magnitude and direction) on the nylon sheet as it is inserted into the capacitor?

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?

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.
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