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Chapter 19: Problem 57

Two hollow metal spheres are concentric with each other. The inner sphere has a radius of \(0.1500 \mathrm{~m}\) and a potential of \(85.0 \mathrm{~V}\). The radius of the outer sphere is \(0.1520 \mathrm{~m}\) and its potential is \(82.0 \mathrm{~V}\). If the region between the spheres is filled with Teflon, find the electric energy contained in this space.

Short Answer

Expert verified
The electric energy contained in the space is approximately \(3.13 \times 10^{-10} \, \text{J}\).

Step by step solution

01

Understand the Problem

Two hollow metal spheres are concentric, with different potentials on the inner and outer spheres. The goal is to find the electric energy stored in the Teflon-filled space between them.
02

Recall Relevant Formulas

The electric energy \( U \)stored in a capacitor is given by \( U = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the potential difference.
03

Determine Capacitance of Spheres

For two concentric spheres, the capacitance \( C \)is calculated as \(C = \frac{4\pi\varepsilon_r\varepsilon_0}{\frac{1}{R_{\text{inner}}} - \frac{1}{R_{\text{outer}}}},\)where \( R_{\text{inner}} = 0.1500 \, \text{m} \) and \( R_{\text{outer}} = 0.1520 \, \text{m} \), \( \varepsilon_0 \) is the permittivity of free space \(\approx 8.85 \times 10^{-12} \, \text{F/m}\), and \( \varepsilon_r = 2.1 \) is the relative permittivity of Teflon.
04

Calculate Capacitance

Substitute the values into the formula from Step 3: \(C = \frac{4\pi \times 2.1 \times 8.85 \times 10^{-12}}{\frac{1}{0.1500} - \frac{1}{0.1520}} \approx 6.95 \times 10^{-11} \, \text{F}.\)
05

Calculate Potential Difference

The potential difference \( V \) between the spheres is the difference in their potentials: \( V = 85.0 \, \text{V} - 82.0 \, \text{V} = 3.0 \, \text{V}. \)
06

Calculate Electric Energy

Substitute \( C \)and\( V \)into the energy formula: \[U = \frac{1}{2} \times 6.95 \times 10^{-11} \times (3.0)^2 \approx 3.13 \times 10^{-10} \, \text{J}.\]

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Key Concepts

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

Capacitance
Capacitance is a measure of a system's ability to store electric charge. In our problem, we have two concentric metal spheres acting like a capacitor. The inner sphere with a smaller radius and higher potential stores energy due to the electric field between the spheres.

The formula used to calculate the capacitance (\(C\)) of two concentric spheres is:
  • \(C = \frac{4\pi\varepsilon_r\varepsilon_0}{\frac{1}{R_{\text{inner}}} - \frac{1}{R_{\text{outer}}}}\)
Where:
  • \(\varepsilon_r\) represents the relative permittivity of the medium between them, and in this case, Teflon.
  • \(R_{\text{inner}}\) and \(R_{\text{outer}}\) are the radii of the inner and outer spheres, respectively.
  • \(\varepsilon_0\) is the permittivity of free space.
This formula helps us find how much electric charge these spheres can hold for a given electric potential difference between them.
Electric Potential
Electric potential, often referred to as voltage, measures the potential energy per unit charge at a point in an electric field. It's like the electric 'height' of a point relative to a reference point, often the ground.

In our scenario, the inner sphere has a higher electric potential than the outer sphere, which creates a potential difference across the space between them. This potential difference drives the flow of charge, similar to how height differences in a ski slope create movement downhill.
  • The potential difference \( V \) is calculated as:
    \( V = 85.0 \, \text{V} - 82.0 \, \text{V} = 3.0 \, \text{V} \)
This potential difference is vital for calculating the stored electric energy and creating the electric field in the Teflon-filled space.
Permittivity
Permittivity is a measure of how an electric field affects, and is affected by, a dielectric medium. It's a crucial factor in determining a material's ability to transmit electric force.

In this exercise, two types of permittivity are considered:
  • The permittivity of free space \(\varepsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m}\).
  • The relative permittivity \(\varepsilon_r\) of Teflon, which is \(2.1\).
The total permittivity \(\varepsilon\) used in calculations is the product of these values, \(\varepsilon = \varepsilon_r \times \varepsilon_0\). This determines how much electric field the Teflon can sustain in the space between concentric spheres.
Teflon
Teflon is a synthetic polymer known for its insulating properties. It's commonly used as a dielectric material in capacitors due to its high relative permittivity and chemical inertness.

In our problem, Teflon fills the space between the two metal spheres, enhancing the system's capacitance by allowing it to store more electric energy than it would with air or a vacuum.
  • Teflon鈥檚 relative permittivity (\(\varepsilon_r\)) is \(2.1\), higher than air (\(\varepsilon_r \approx 1\)), meaning it can sustain a stronger electric field.
  • This property makes Teflon ideal for increasing the electric energy storage in capacitors used in various electronic applications.
Thus, Teflon not only provides physical separation but also electrically benefits the capacitor system.

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Most popular questions from this chapter

An axon is the relatively long tail-like part of a neuron, or nerve cell. The outer surface of the axon membrane (dielectric constant \(=5,\) thickness \(=1 \times 10^{-8} \mathrm{~m}\) ) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that an axon can be treated like a parallel plate capacitor with a plate area of \(5 \times 10^{-6} \mathrm{~m}^{2},\) what is its capacitance?

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about \(0.070 \mathrm{~V}\) exists across the membrane. The thickness of the cell membrane is \(8.0 \times 10^{-9} \mathrm{~m}\). What is the magnitude of the electric field in the membrane?

An electron is released at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing), (a) As the electron gains kinetic energy, does its electric potential energy increase or decrease? Why? (b) The difference in the electron's electric potential energy between the positive and negative plates is EPE positive \(-\mathrm{EPE}_{\text {negative. }}\) How is this difference related to the charge on the electron \((-e)\) and to the difference \(V\) positive \(-V_{\text {negative in the electric potential between }}\) the plates? (c) How is the potential difference \(V\) positive \(-V_{\text {negative }}\) related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate? The plates of a parallel plate capacitor are separated by a distance of \(1.2 \mathrm{~cm}\), and the electric field within the capacitor has a magnitude of \(2.1 \times 10^{6} \mathrm{~V} / \mathrm{m}\). An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?

Suppose that the electric potential outside a living cell is higher than that inside the cell by \(0.070 \mathrm{~V}\). How much work is done by the electric force when a sodium ion (charge \(=+e\) ) moves from the outside to the inside?

Refer to Interactive Solution 19.45 at and Multiple-Concept Example 10 for help in solving this problem. An empty capacitor has a capacitance of \(3.2 \mu \mathrm{F}\) and is connected to a 12-V battery. A dielectric material \((\kappa=4.5)\) is inserted between the plates of this capacitor. What is the magnitude of the surface charge on the dielectric that is adjacent to either plate of the capacitor? (Hint: The surface charge is equal to the difference in the charge on the plates with and without the dielectric.)
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