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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

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 \(C_{2} ?\)

A parallel plate capacitor of capacitance \(C\) has plates of area \(A\) with distance \(d\) between them. When the capacitor is connected to a battery of potential difference \(V\), it has a charge of magnitude \(Q\) on its plates. While the capacitor is connected to the battery, the distance between the plates is decreased by a factor of \(3 .\) The magnitude of the charge on the plates and the capacitance are then a) \(\frac{1}{3} Q\) and \(\frac{1}{3} C\). c) \(3 Q\) and \(3 C\). b) \(\frac{1}{3} Q\) and \(3 C\). d) \(3 Q\) and \(\frac{1}{3} C\).

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.

A dielectric slab with thickness \(d\) and dielectric constant \(\kappa=2.31\) is inserted in a parallel place capacitor that has been charged by a \(110 .-\mathrm{V}\) battery and having area \(A=\) \(100, \mathrm{~cm}^{2},\) and separation distance \(d=2.50 \mathrm{~cm}\). a) Find the capacitance, \(C,\) the potential difference, \(V,\) the electric field, \(E,\) the total charge stored on the capacitor \(Q\), and electric potential energy stored in the capacitor, \(U\), before the dielectric material is inserted. b) Find \(C, V, E, Q,\) and \(U\) when the dielectric slab has been inserted and the battery is still connected. c) Find \(C, V, E, Q\) and \(U\) when the dielectric slab is in place and the battery is disconnected.

Two capacitors, with capacitances \(C_{1}\) and \(C_{2},\) are connected in series. A potential difference, \(V_{0}\), is applied across the combination of capacitors. Find the potential differences \(V_{1}\) and \(V_{2}\) across the individual capacitors, in terms of \(V_{0}\), \(C_{1},\) and \(C_{2}\).
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