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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 them?

Short Answer

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Question: Determine the fractional difference in capacitances between a spherical capacitor and a parallel plate capacitor, given the radii of the concentric conducting shells of the spherical capacitor as \(r_1\) and \(r_2\), and the separation between the plates of the parallel plate capacitor as \(d = r_2 - r_1\). Answer: The fractional difference in capacitances between the spherical capacitor and the parallel plate capacitor is \(\frac{r_2 - r_1}{r_1}\).

Step by step solution

01

Express the capacitance of a spherical capacitor

The capacitance of a spherical capacitor is given by the formula: $$C_s = 4 \pi \epsilon_0 \frac{r_1 r_2}{r_2 - r_1}$$ where \(\epsilon_0\) is the vacuum permittivity.
02

Express the capacitance of a parallel plate capacitor

The capacitance of a parallel plate capacitor is given by the formula: $$C_p = \frac{\epsilon_0 A}{d}$$ where \(A\) is the area of the plates and \(d\) is the separation between them. In our case, \(A = 4 \pi r_1^2\) (the surface area of the inner sphere) and \(d = r_2 - r_1\).
03

Calculate the capacitance of the parallel plate capacitor

Plugging in the values for \(A\) and \(d\), we get: $$C_p = \frac{\epsilon_0 (4 \pi r_1^2)}{r_2 - r_1}$$
04

Determine the fractional difference in capacitances

Now, we need to find the fractional difference between the capacitances of the spherical capacitor and the parallel plate capacitor. We can calculate this as: $$\text{Fractional difference} = \frac{C_s - C_p}{C_p}$$ Plugging in the values for \(C_s\) and \(C_p\), we get: $$\text{Fractional difference} = \frac{4 \pi \epsilon_0 \frac{r_1 r_2}{r_2 - r_1} - \frac{\epsilon_0 (4 \pi r_1^2)}{r_2 - r_1}}{\frac{\epsilon_0 (4 \pi r_1^2)}{r_2 - r_1}}$$ Now, we can simplify the expression: $$\text{Fractional difference} = \frac{4 \pi r_1 r_2 - 4 \pi r_1^2}{4 \pi r_1^2}$$ $$\text{Fractional difference} = \frac{4 \pi r_1 (r_2 - r_1)}{4 \pi r_1^2}$$ Cancelling out the common terms, we get the final expression for the fractional difference: $$\text{Fractional difference} = \frac{r_2 - r_1}{r_1}$$

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Most popular questions from this chapter

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 supplying a potential difference \(V=9.00 \mathrm{~V}\) is connected to the plates. What is the charge distribution, \(\sigma,\) on the left plate? What are the capacitance, \(C^{\prime},\) and the 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?

The space between the plates of an isolated parallel plate capacitor is filled with a slab of dielectric material. The magnitude of the charge \(Q\) on each plate is kept constant. If the dielectric material is removed the energy stored in the capacitor a) increases. c) decreases. b) stays the same. d) may increase or decrease.

Calculate the maximum surface charge distribution that can be maintained on any surface surrounded by dry air.

A parallel plate capacitor with capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A slab of dielectric, with dielectric constant \(\kappa\), is then inserted into and completely fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.

Thermocoax is a type of coaxial cable used for high-frequency filtering in cryogenic quantum computing experiments. Its stainless steel shield has an inner diameter of \(0.350 \mathrm{~mm}\), and its Nichrome conductor has a diameter of \(0.170 \mathrm{~mm} .\) Nichrome is used because its resistance doesn't change much in going from room temperature to near absolute zero. The insulating dielectric is magnesium oxide \((\mathrm{MgO})\), which has a dielectric constant of \(9.70 .\) Calculate the capacitance per meter of Thermocoax.

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