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Ch. 23 - Electric Potential
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 22, Problem 103a

Inside a high-voltage lab, engineers have designed a storage container for electrical energy using a nonconducting sphere of radius r2 that contains a concentric spherical cavity of radius r1. The material between r₁ and r₂ carries a uniform charge density ρE ( C/m³). Determine the electric potential V, relative to V = 0 at r = ∞, as a function of the distance r from the center for r > r₂.

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Step 1: Recognize that the problem involves a spherically symmetric charge distribution. For r > r₂, the electric potential V can be determined using Gauss's law to first find the electric field E, and then integrating E to find V.
Step 2: Apply Gauss's law to find the electric field E for r > r₂. The total charge enclosed within a Gaussian surface of radius r (where r > r₂) is given by Q_enc = ρ_E * (4/3)π(r₂³ - r₁³). Use the symmetry of the problem to write the electric field as E = Q_enc / (4πr²ε₀).
Step 3: Substitute Q_enc into the expression for E. This gives E = [ρ_E * (r₂³ - r₁³)] / [3ε₀r²]. This is the electric field for r > r₂.
Step 4: To find the electric potential V, integrate the electric field E with respect to r. The potential difference between two points is given by V(r) - V(∞) = -∫(E dr). Since V(∞) = 0, the potential at a distance r > r₂ is V(r) = -∫[ρ_E * (r₂³ - r₁³) / (3ε₀r²)] dr.
Step 5: Perform the integration. The integral of 1/r² with respect to r is -1/r. Substituting this result, the potential V(r) for r > r₂ becomes V(r) = [ρ_E * (r₂³ - r₁³)] / (3ε₀r). This is the expression for the electric potential as a function of r for r > r₂.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Potential

Electric potential (V) is the amount of electric potential energy per unit charge at a point in an electric field. It is a scalar quantity that indicates the work done to move a unit positive charge from a reference point (usually at infinity) to a specific point in the field. Understanding how to calculate electric potential is crucial for analyzing systems with charge distributions, such as the one described in the question.
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Electric Potential

Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed by that surface. It is expressed mathematically as ∮E·dA = Q_enc/ε₀, where E is the electric field, dA is the differential area vector, Q_enc is the enclosed charge, and ε₀ is the permittivity of free space. This law is essential for determining the electric field in symmetric charge distributions, which is necessary for calculating the electric potential in the given problem.
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Uniform Charge Density

Uniform charge density (ρ_E) refers to a constant distribution of charge per unit volume within a specified region. In this scenario, it indicates that the charge is evenly spread throughout the material between the inner and outer spheres. This concept is important for calculating the total charge within a given volume, which directly influences the electric field and potential in the system.
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Related Practice
Textbook Question

The liquid-drop model of the nucleus suggests that high-energy oscillations of certain nuclei can split (“fission”) a large nucleus into two unequal fragments plus a few neutrons. Using this model, consider the case of a uranium nucleus fissioning into two spherical fragments, one with a charge q₁ = +38e and radius r₁ = 5.5 x 10⁻¹⁵ m, the other with q₂ = + 54e and r₂ = 6.2 x 10⁻¹⁵ m. Calculate the electric potential energy (MeV) of these fragments, assuming that the charge is uniformly distributed throughout the volume of each spherical nucleus and that their surfaces are initially in contact at rest. The electrons surrounding the nuclei can be neglected. This electric potential energy will then be entirely converted to kinetic energy as the fragments repel each other. How does your predicted kinetic energy of the fragments agree with the observed value associated with uranium fission (approximately 200 MeV total)? [ 1 MeV = 10⁶ eV.]

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Textbook Question

A Geiger counter is used to detect charged particles emitted by radioactive nuclei. It consists of a thin, positively charged central wire of radius Rₐ surrounded by a concentric conducting cylinder of radius Rᵦ with an equal negative charge (Fig. 23–57). The charge per unit length on the inner wire is λ (units C/m). The interior space between wire and cylinder is filled with low-pressure inert gas. Charged particles ionize some of these gas atoms; the resulting free electrons are attracted toward the positive central wire. If the radial electric field is strong enough, the freed electrons gain enough energy to ionize other atoms, causing an “avalanche” of electrons to strike the central wire, generating an electric “signal.” Find the expression for the electric field between the wire and the cylinder, and (b) show that the potential difference between Rₐ and Rᵦ is Vₐ - Vᵦ = ( λ / 2π∊₀ ) ln( Rᵦ/Rₐ) .

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Textbook Question

A thin flat disk of radius R₀ carries a total charge Q that is distributed uniformly over its surface. The electric potential at a distance x on the x axis is given by V(x) = Q/ 2π∊₀R₀²[(x² + R²₀) ¹⸍² - x]. (See Example 23–10.) Show that the electric field at a distance x on the x axis is given by E(x) = Q/2π∊₀R₀² ( 1 - ( x / ( x² + R²₀))¹⸍². Make graphs of V(x) and E(x) as a function of x/R₀ for x/R₀ = 0 to 4. (Do the calculations in steps of 0.1.) Use Q = 5.0μC and R₀ = 10 cm for the calculation and graphs.

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Textbook Question

A Van de Graaff generator (Fig. 23–58) can develop a very large potential difference, even millions of volts. Electrons are pulled off the belt by the high voltage pointed electrode (positive) at A, leaving the belt positively charged. (Recall Example 23–5 where we saw that near sharp points the electric field is high and ionization can occur.) The belt carries the positive charge up inside the spherical shell where electrons from the large conducting sphere are attracted over to the pointed conductor at B, leaving the outer surface of the conducting sphere positively charged. As more charge is brought up, the sphere reaches extremely high voltage. Consider a Van de Graaff generator with a sphere of radius 0.20 m. What is the electric potential on the surface of the sphere when electrical breakdown occurs ( E = 3 x 10⁶ V/m) ? Assume V = 0 at r = ∞.

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Textbook Question

A Van de Graaff generator (Fig. 23–58) can develop a very large potential difference, even millions of volts. Electrons are pulled off the belt by the high voltage pointed electrode (positive) at A, leaving the belt positively charged. (Recall Example 23–5 where we saw that near sharp points the electric field is high and ionization can occur.) The belt carries the positive charge up inside the spherical shell where electrons from the large conducting sphere are attracted over to the pointed conductor at B, leaving the outer surface of the conducting sphere positively charged. As more charge is brought up, the sphere reaches extremely high voltage. Consider a Van de Graaff generator with a sphere of radius 0.20 m.

(b) What is the charge on the sphere for the potential found in part (a)

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