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Ch 25: The Electric Potential
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 25: The Electric PotentialProblem 45
Chapter 25, Problem 45

An arrangement of source charges produces the electric potential V=5000x2 along the x-axis, where V is in volts and x is in meters. What is the maximum speed of a 1.0 g, 10 nC charged particle that moves in this potential with turning points at ±8.0 cm?

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Identify the given quantities: The electric potential is given as \( V = 5000x^2 \) (in volts), the charge of the particle is \( q = 10 \text{ nC} = 10 \times 10^{-9} \text{ C} \), the mass of the particle is \( m = 1.0 \text{ g} = 0.001 \text{ kg} \), and the turning points are at \( x = \pm 0.08 \text{ m} \).
Understand the relationship between electric potential energy and kinetic energy: The total mechanical energy of the particle is conserved. At the turning points, the particle's kinetic energy is zero, and all the energy is stored as electric potential energy. The maximum speed occurs when all the potential energy difference is converted into kinetic energy.
Write the expression for the electric potential energy: The electric potential energy \( U \) is given by \( U = qV \). Substituting \( V = 5000x^2 \), we get \( U = q \cdot 5000x^2 \).
Determine the change in potential energy between the turning points and the origin: At the turning points \( x = \pm 0.08 \text{ m} \), the potential energy is \( U_{\text{turning}} = q \cdot 5000 \cdot (0.08)^2 \). At the origin \( x = 0 \), the potential energy is \( U_{\text{origin}} = q \cdot 5000 \cdot 0^2 = 0 \). The change in potential energy is \( \Delta U = U_{\text{turning}} - U_{\text{origin}} = q \cdot 5000 \cdot (0.08)^2 \).
Use energy conservation to find the maximum speed: The change in potential energy \( \Delta U \) is equal to the maximum kinetic energy \( K_{\text{max}} \). Kinetic energy is given by \( K = \frac{1}{2}mv^2 \). Set \( \Delta U = \frac{1}{2}mv_{\text{max}}^2 \) and solve for \( v_{\text{max}} \): \( v_{\text{max}} = \sqrt{\frac{2 \cdot \Delta U}{m}} \). Substitute \( \Delta U = q \cdot 5000 \cdot (0.08)^2 \) and \( m = 0.001 \text{ kg} \) to find \( v_{\text{max}} \).

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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 measured in volts (V) and indicates how much work would be done to move a charge from a reference point to a specific point in the field. In this question, the potential is given as a function of position along the x-axis, which influences the motion of the charged particle.
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Conservation of Energy

The principle of conservation of energy states that the total energy in a closed system remains constant. For a charged particle in an electric potential, the total mechanical energy is the sum of its kinetic energy and electric potential energy. As the particle moves through the potential, its speed will vary, but the total energy will remain constant, allowing us to calculate its maximum speed at turning points.
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Kinetic Energy

Kinetic energy (KE) is the energy that an object possesses due to its motion, calculated using the formula KE = 0.5mv^2, where m is the mass and v is the velocity of the object. In this scenario, the kinetic energy of the charged particle will be at its maximum when it is at the turning points, where the electric potential energy is at its minimum, allowing us to determine the maximum speed of the particle.
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Related Practice
Textbook Question

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Textbook Question

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Textbook Question

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Textbook Question

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Textbook Question

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Textbook Question

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