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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 12.5 cm, and the outer sphere has radius 14.8 cm. A potential difference of 120 V is applied to the capacitor. (a) What is the energy density at \(r\) = 12.6 cm, just outside the inner sphere? (b) What is the energy density at \(r\) = 14.7 cm, just inside the outer sphere? (c) For a parallel-plate capacitor the energy density is uniform in the region between the plates, except near the edges of the plates. Is this also true for a spherical capacitor?

Short Answer

Expert verified
(a) Energy density at 12.6 cm: Higher. (b) Energy density at 14.7 cm: Lower. (c) No, it's not uniform. It decreases with increasing radius.

Step by step solution

01

Understand Energy Density Formula

The energy density \( u \) in the electric field is given by the formula \( u = \frac{1}{2} \varepsilon_0 E^2 \), where \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \)), and \( E \) is the electric field strength.
02

Electric Field in a Spherical Capacitor

For a spherical capacitor, the electric field \( E \) at a distance \( r \) from the center is given by \( E = \frac{Q}{4 \pi \varepsilon_0 r^2} \), where \( Q \) is the charge on the inner sphere. The potential difference \( V \) between the spheres is related to the charge and the radii by \( V = \int_{r_1}^{r_2} E \, dr = \frac{Q}{4 \pi \varepsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \).
03

Solve for Charge Q

Rearrange the formula for the potential difference to solve for \( Q \): \( Q = V \cdot 4 \pi \varepsilon_0 \left(\frac{1}{r_1} - \frac{1}{r_2}\right)^{-1} \). Substitute \( V = 120 \, \text{V} \), \( r_1 = 0.125 \, \text{m} \), and \( r_2 = 0.148 \, \text{m} \) to find \( Q \).
04

Find Electric Field at r = 12.6 cm

Using the calculated charge \( Q \), substitute \( r = 0.126 \, \text{m} \) into the electric field formula \( E = \frac{Q}{4 \pi \varepsilon_0 r^2} \) and compute \( E \).
05

Calculate Energy Density at r = 12.6 cm

Substitute \( E \) from step 4 into the energy density formula \( u = \frac{1}{2} \varepsilon_0 E^2 \) to find the energy density at \( r = 12.6 \, \text{cm} \).
06

Find Electric Field at r = 14.7 cm

Using the same charge \( Q \) from step 3, substitute \( r = 0.147 \, \text{m} \) into the electric field formula to find \( E \) at this distance.
07

Calculate Energy Density at r = 14.7 cm

Substitute \( E \) from step 6 into the energy density formula \( u = \frac{1}{2} \varepsilon_0 E^2 \) to find the energy density at \( r = 14.7 \, \text{cm} \).
08

Compare Energy Densities

Analyze the calculated energy densities to discuss whether the energy density is uniform between the shells, comparing it to the behavior in a parallel-plate capacitor.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Density
Energy density refers to the amount of energy stored in a given system or region per unit volume. For our spherical capacitor, the energy density at any point within the electric field is given by the formula: \[ u = \frac{1}{2} \varepsilon_0 E^2 \]where \( u \) is the energy density, \( \varepsilon_0 \) is the permittivity of free space, and \( E \) is the electric field strength. The energy density tells us how much energy is concentrated in a certain space, which is crucial for understanding capacitors' efficiency.
In spherical capacitors, unlike parallel-plate capacitors, the energy density is not uniform because the electric field varies with the distance from the center. This variation is due to the spherical geometry, which affects how the electric field disperses over space.
Electric Field
The electric field \( E \) in a spherical capacitor is the force field created by the charge on the inner sphere. It influences how charged particles move between the shells. For a spherical capacitor, the electric field at a distance \( r \) from the center is determined by \[ E = \frac{Q}{4 \pi \varepsilon_0 r^2} \]where \( Q \) is the charge on the inner sphere.
Here's a quick breakdown of what happens:
  • The electric field decreases as you move away from the inner sphere because it is inversely proportional to \( r^2 \).
  • This decrease means the electric field is stronger closer to the inner sphere, which affects the energy density calculation.
Understanding this helps us predict how the physical properties like voltage and energy are distributed inside the capacitor.
Potential Difference
The potential difference, often referred to as voltage, is key in determining how much energy can be stored in a capacitor. In our spherical capacitor, the potential difference \( V \) between the two shells is calculated by integrating the electric field across the radial distance between the shells:\[ V = \int_{r_1}^{r_2} E \, dr = \frac{Q}{4 \pi \varepsilon_0} \left(\frac{1}{r_1} - \frac{1}{r_2}\right) \]It quantifies the work done to move a unit charge from one shell to the other.
  • The potential difference is a crucial factor because it determines the electric field's magnitude and direction.
  • In practical scenarios, a higher potential difference usually means the capacitor can store more energy, assuming the physical dimensions are constant.
For our given spherical capacitor, the potential difference is set to 120 volts, which influences all further calculations of electric field and energy density.
Spherical Shells
Spherical shells describe the geometry and structure of our capacitor. The capacitor consists of two conducting shells:
  • The inner shell (with radius 12.5 cm)
  • The outer shell (with radius 14.8 cm)
The space between these shells is a vacuum, and this separation is crucial for how the capacitor functions.
A key point to remember about spherical capacitors:
  • They differ from parallel-plate capacitors mainly due to the shape and arrangement of the electrodes.
  • This geometry affects not only the electric field distribution but also the energy density and capacitance.
Understanding the structure of spherical shells helps in grasping how the electric field and potential difference are managed within the system.
Permittivity of Free Space
The permittivity of free space, denoted by \( \varepsilon_0 \), is a fundamental physical constant that represents the ability of a vacuum to permit electric field lines. Its value is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \cdot \text{m}^2 \).
This constant is essential in electrostatics because:
  • It determines the ability of the vacuum (or any other material) to resist the formation of an electric field.
  • It appears in fundamental equations used to calculate electric fields, forces, and energy densities.
In our context of a spherical capacitor, \( \varepsilon_0 \) helps determine both the electric field \( E \) and subsequently the energy density \( u \). Understanding this constant sheds light on why capacitors behave a certain way in vacuum conditions.

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Most popular questions from this chapter

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00\(\times\) 10\(^6\) V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is \(E = 3.20\times10^5\space V/m\). When the space is filled with dielectric, the electric field is \(E = 2.50\times10^5\space V/m\). (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

A spherical capacitor is formed from two concentric, spherical, conducting shells separated by vacuum. The inner sphere has radius 15.0 cm and the capacitance is 116 pF. (a) What is the radius of the outer sphere? (b) If the potential difference between the two spheres is 220 V, what is the magnitude of charge on each sphere?

A parallel-plate air capacitor has a capacitance of 920 pF. The charge on each plate is 3.90 \(\mu\)C. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference if the plate separation is doubled? (c) How much work is required to double the separation?

A parallel-plate vacuum capacitor has 8.38 J of energy stored in it. The separation between the plates is 2.30 mm. If the separation is decreased to 1.15 mm, what is the energy stored (a) if the capacitor is disconnected from the potential source so the charge on the plates remains constant, and (b) if the capacitor remains connected to the potential source so the potential difference between the plates remains constant?

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