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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 88

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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Step 1: Identify the forces acting on the balls. Both balls experience gravitational force, which is given by \( F_g = m g \), where \( m = 0.550 \; \text{kg} \) and \( g = 9.8 \; \text{m/s}^2 \). Additionally, the charged balls experience an electric force due to the electric field, given by \( F_e = q E \), where \( q \) is the charge of the ball and \( E = 150 \; \text{V/m} \) is the electric field strength.
Step 2: Write the total mechanical energy for each ball. The total energy includes gravitational potential energy, electric potential energy, and kinetic energy. For the positively charged ball, the total energy is \( E_1 = m g h + q_1 E h + \frac{1}{2} m v_1^2 \), and for the negatively charged ball, the total energy is \( E_2 = m g h + q_2 E h + \frac{1}{2} m v_2^2 \), where \( h = 2.00 \; \text{m} \) is the initial height and \( v_1 \) and \( v_2 \) are the final speeds of the balls.
Step 3: Apply conservation of energy. At the ground level, the potential energies (gravitational and electric) are zero, so the total energy is purely kinetic. For the positively charged ball, \( m g h + q_1 E h = \frac{1}{2} m v_1^2 \). For the negatively charged ball, \( m g h + q_2 E h = \frac{1}{2} m v_2^2 \).
Step 4: Solve for the final speeds of the balls. Rearrange the equations to isolate \( v_1 \) and \( v_2 \): \( v_1 = \sqrt{2 g h + \frac{2 q_1 E h}{m}} \) and \( v_2 = \sqrt{2 g h + \frac{2 q_2 E h}{m}} \). Substitute the given values for \( g \), \( h \), \( q_1 \), \( q_2 \), \( E \), and \( m \) into these equations.
Step 5: Determine the difference in speeds. The difference in speeds is \( \Delta v = v_1 - v_2 \). Substitute the expressions for \( v_1 \) and \( v_2 \) to find \( \Delta v = \sqrt{2 g h + \frac{2 q_1 E h}{m}} - \sqrt{2 g h + \frac{2 q_2 E h}{m}} \). Simplify this expression to calculate the difference in speeds.

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Key Concepts

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

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this context, the potential energy of the balls at the height of 2.00 m is converted into kinetic energy as they fall. The total mechanical energy remains constant, allowing us to equate the initial potential energy to the final kinetic energy to find the speeds of the balls just before they hit the ground.
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Electric Field

An electric field is a region around a charged object where other charged objects experience a force. The strength of the electric field near the Earth's surface is approximately 150 V/m, directed downward. This field affects the motion of charged objects, such as the positively and negatively charged balls in this scenario, influencing their acceleration and final speeds as they fall.
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Kinetic Energy and Speed Relationship

Kinetic energy (KE) is the energy of an object due to its motion, defined by the equation KE = 0.5 * m * v², where m is mass and v is velocity. As the balls fall, their potential energy is converted into kinetic energy, which determines their speed upon impact. The difference in charge between the two balls will affect their acceleration due to the electric field, leading to different final speeds despite having the same mass.
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Related Practice
Textbook Question

An analog voltage signal can vary from 0 V to 5.00 V, and it is to be converted to an 8-bit binary representation. What binary number would best represent 3.47 volts?

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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

Two point charges are fixed 4.0 cm apart from each other. Their charges are Q₁ = Q₂ = 6.5 μC and their masses are m₁ = 2.5 mg and m₂ = 3.5 mg. If Q₁ is released from rest, what will be its speed after a very long time?

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Textbook Question

In a photocell, ultraviolet (UV) light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. See Fig. 23–52. If the plate voltage is -3.02 V (compared to the barium) when the fastest electrons are stopped, what was the speed of these electrons when they were emitted?

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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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