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

Giancoli Douglas 5th edition

Ch. 23 - Electric Potential

Problem 101

Chapter 22, Problem 101

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.

Verified step by step guidance

Step 1: Start by recalling the relationship between the electric potential V(x) and the electric field E(x). The electric field is the negative gradient of the potential, which in one dimension is given by E(x) = -dV(x)/dx. This means we need to differentiate the given expression for V(x) with respect to x.

Step 2: Write down the given expression for V(x): V(x) = (Q / (2π∊₀R₀²)) * [(x² + R₀²)^(1/2) - x]. Differentiate this expression term by term with respect to x. Use the chain rule for the term involving (x² + R₀²)^(1/2).

Step 3: Compute the derivative of the first term: d/dx [(x² + R₀²)^(1/2)] = (1/2)(x² + R₀²)^(-1/2) * 2x = x / (x² + R₀²)^(1/2). Then compute the derivative of the second term: d/dx [-x] = -1.

Step 4: Substitute the derivatives back into the expression for E(x). Combine terms to simplify the result. You should find that E(x) = (Q / (2π∊₀R₀²)) * [1 - (x / (x² + R₀²)^(1/2))].

Step 5: To create the graphs of V(x) and E(x) as a function of x/R₀, substitute the given values Q = 5.0 μC and R₀ = 10 cm into the expressions for V(x) and E(x). Calculate V(x) and E(x) for x/R₀ ranging from 0 to 4 in steps of 0.1. Plot V(x) and E(x) on separate graphs, with the x-axis labeled as x/R₀ and the y-axis labeled as V(x) or E(x), respectively.

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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, denoted as 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 in bringing a unit positive charge from infinity to that point without any acceleration. The formula provided in the question shows how the potential varies with distance x from the charged disk, reflecting the influence of both the total charge Q and the geometry of the disk.

Electric Field

The electric field, represented as E, is a vector field that describes the force experienced by a unit positive charge placed in the field. It is defined as the negative gradient of the electric potential, indicating how the potential changes with position. The equation given for E(x) shows how the electric field strength varies with distance x from the disk, highlighting the relationship between electric potential and electric field in electrostatics.

Graphing Functions

Graphing functions involves plotting the relationship between two variables, in this case, electric potential V(x) and electric field E(x) as functions of the normalized distance x/R₀. This visual representation helps in understanding how these quantities change with distance, allowing for easier interpretation of their behavior and interactions. The specified range and step size for x/R₀ will provide a detailed view of the trends in both V and E.

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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

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 ρ_E ( 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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Textbook Question

Near the surface of the Earth there is an electric field of about 150 V/m which points downward. Two identical balls with mass m = 0.550 kg are dropped from a height of 2.00 m, but one of the balls is positively charged with q₁ = 650 μC, and the second is negatively charged with q₂ = -650 μC. Use conservation of energy to determine the difference in the speeds of the two balls when they hit the ground. (Neglect air resistance.)

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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 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

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

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