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Chapter 23: Problem 13

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-4} \mathrm{kg} .\) It moves from point \(A,\) where the electric potential is \(V_{A}=+200 \mathrm{V},\) to point \(B,\) where the electric potential is \(V_{B}=+800 \mathrm{V} .\) The electric force is the only force acting on the particle. The particle has speed 5.00 \(\mathrm{m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.

Short Answer

Expert verified
Particle slows down and stops, possibly reversing direction.

Step by step solution

01

Understand the Scenario

The problem describes a charged particle moving in an electric potential field. We need to determine the speed of the particle at point B, knowing its initial speed at point A. The change in electric potential will affect the particle's kinetic energy, given that the electric force is the only force acting on it.
02

Use the Energy Principle

Since only electric forces do work, the work done is equal to the change in kinetic energy: \[ W = ext{KE}_B - ext{KE}_A = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \]
03

Relate Work to Electric Potential Energy

The work done by the electric force can also be calculated as the change in electric potential energy:\[ W = q(V_B - V_A) \] where \(q\) is the charge of the particle.
04

Set Up the Energy Equation

Substitute the expressions for work and kinetic energy difference in the equation:\[ q(V_B - V_A) = \frac{1}{2} m v_B^2 - \frac{1}{2} m v_A^2 \]
05

Rearrange to Solve for Final Speed

Rearrange the equation to solve for the final speed \(v_B\):\[ \frac{1}{2} m v_B^2 = q(V_B - V_A) + \frac{1}{2} m v_A^2 \]\[ v_B^2 = \frac{2q(V_B - V_A)}{m} + v_A^2 \]\[ v_B = \sqrt{\frac{2q(V_B - V_A)}{m} + v_A^2} \]
06

Insert the Known Values

Insert the given values: \( q = -5.00 \times 10^{-6} \) C, \( m = 2.00 \times 10^{-4} \) kg, \( V_A = 200 \) V,\( V_B = 800 \) V, \( v_A = 5.00 \) m/s:\[ v_B = \sqrt{\frac{2(-5.00 \times 10^{-6})(800 - 200)}{2.00 \times 10^{-4}} + 5.00^2} \]
07

Calculate Final Speed

Calculate \( v_B \) using the equation:\[ v_B = \sqrt{\frac{-6 \times 10^{-3}}{2 \times 10^{-4}} + 25} \]\[ v_B = \sqrt{-30 + 25} \]Since the term under the square root is negative, the calculation indicates a decrease in kinetic energy.
08

Interpret the Result

The calculation shows that the kinetic energy decreases because the electric potential energy change is negative. Since our calculation leads to a complex number (physical impossibility in this context), the particle slows down and potentially stops somewhere between A and B before switching direction.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that a particle has because of its motion. It depends on two factors: the mass of the particle and its speed. Mathematically, we express kinetic energy using the formula:\[ KE = \frac{1}{2} m v^2 \]where:
  • \( KE \) is the kinetic energy,
  • \( m \) is the mass of the particle, and
  • \( v \) is the speed of the particle.
This relationship means that if a particle moves faster, its kinetic energy increases. Similarly, a heavier particle with the same speed will have more kinetic energy than a lighter one. In our problem, the particle's speed at point \( B \) is affected by the change in kinetic energy as it moves through the electric potential field.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. In our problem, the charge of the particle is given as \(-5.00 \mu \mathrm{C}\), which is equivalent to \(-5.00 \times 10^{-6}\) Coulombs.A few essential points to understand about electric charge:
  • Charges can be positive or negative.
  • Like charges repel each other, while opposite charges attract.
  • Charges interact with electric fields, causing them to move if they are free to do so.
In this exercise, the negative charge moves from one point to another in a region with different electric potential. This movement affects the particle's energy levels. The nature of the charge specifically dictates how it interacts with different potentials, affecting both the direction and magnitude of the force experienced.
Work-Energy Principle
The work-energy principle is a powerful concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. In equation form, this is represented as:\[ \text{Work done} = \Delta KE = KE_B - KE_A \]Where:
  • \( \text{Work done} \) is performed by the forces acting on the object.
  • \( \Delta KE \) represents the change in the object's kinetic energy.
  • \( KE_A \) and \( KE_B \) are the initial and final kinetic energy at points A and B, respectively.
For the given problem, the work done by the electric force is linked with the particle's movement through a potential difference. As the particle travels from a lower to a higher potential area, the change in the electric potential energy affects its kinetic energy. This illustrates how potential energy converts to or from kinetic energy, depending on the nature of the charge and the direction of the electric field. The work-energy principle helps us understand why our calculated final speed \( v_B \) showed a reduction in kinetic energy, indicating that the particle slows down.

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Most popular questions from this chapter

The Millikan Oil-Drop Experiment. The charge of an electron was first measured by the American physicist Robert Millikan during \(1909-1913 .\) In his experiment, oil is sprayed in very fine drops (around \(10^{-4} \mathrm{mm}\) in diameter) into the space between two parallel horizontal plates separated by a distance \(d . \mathrm{A}\) potential difference \(V_{A B}\) is maintained between the parallel plates, causing a downward electric field between them. Some of the oil drops acquire a negative charge because of frictional effects or because of ionization of the surrounding air by \(x\) rays or radioactivity. The drops are observed through a microscope. (a) Show that an oil drop of radius \(r\) at rest between the plates will remain at rest if the magnitude of its charge is $$q=\frac{4 \pi}{3} \frac{\rho r^{3} g d}{V_{A B}}$$ where \(\rho\) is the density of the oil. (Ignore the buoyant force of the air.) By adjusting \(V_{A B}\) to keep a given drop at rest, the charge on that drop can be determined, provided its radius is known. (b) Millikan's ooil drops were much too small to measure their radii directly. Instead, Millikan determined \(r\) by cutting off the electric field and measuring the terminal speed \(v_{t}\) of the drop as it fell. (We discussed the concept of terminal speed in Section \(5.3 . )\) The viscous force \(F\) on a sphere of radius \(r\) moving with speed \(v\) through a fluid with viscosity \(\eta\) is given by Stokes's law: \(F=6 \pi \eta r v .\) When the drop is falling at \(v_{1}\) , the viscous force just balances the weight \(w=m g\) of the drop. Show that the magnitude of the charge on the drop is $$q=18 \pi \frac{d}{V_{A B}} \sqrt{\frac{\eta^{3} v_{t}^{3}}{2 \rho g}}$$ Within the limits of their experimental error, every one of the thousands of drops that Millikan and his coworkers measured had a charge equal to some small integer multiple of a basic charge \(e\) . That is, they found drops with charges of \(\pm 2 e, \quad \pm 5 e,\) and so on, but none with values such as 0.76\(e\) or 2.49\(e .\) A drop with charge \(-e\) has acquired one extra electron; if its charge is \(-2 e,\) it has acquired two extra electrons, and so on. (c) A charged oil drop in a Millikan oil-drop apparatus is observed to fall 1.00 mm at constant speed in 39.3 s if \(V_{A B}=0 .\) The same drop can be held at rest between two plates separated by 1.00 \(\mathrm{mm}\) if \(V_{A B}=9.16 \mathrm{V} .\) How many excess electrons has the drop acquired, and what is the radius of the drop? The viscosity of air is \(1.81 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2},\) and the density of the oil is 824 \(\mathrm{kg} / \mathrm{m}^{3} .\)

A very long wire carries a uniform linear charge density \(\lambda\). Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed 2.50 \(\mathrm{cm}\) from the wire and the other probe is 1.00 \(\mathrm{cm}\) farther from the wire, the meter reads 575 \(\mathrm{V}\) (a) What is \(\lambda ?\) (b) If you now place one probe at 3.50 \(\mathrm{cm}\) from the wire and the other probe 1.00 \(\mathrm{cm}\) farther away, will the voltmeter read 575 \(\mathrm{V}\) ? If not, will it read more or less than 575 \(\mathrm{V}\) ? Why? (c) If you place both probes 3.50 \(\mathrm{cm}\) from the wire but 17.0 \(\mathrm{cm}\) from each other, what will the voltmeter read?

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conduct- ing wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

A proton and an alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

A small metal sphere, carrying a net charge of \(q_{1}=\) \(-2.80 \mu \mathrm{C},\) is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_{2}=-7.80 \mu \mathrm{C} \quad\) and \(\operatorname{mass}\) \(1.50 \mathrm{g},\) is projected toward \(q_{1}\). When the two spheres are 0.800 \(\mathrm{m}\) apart, \(q_{2}\) is moving toward \(q_{1}\) with speed 22.0 \(\mathrm{m} / \mathrm{s}\) (Fig. E23.5). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_{2}\) when the spheres are 0.400 \(\mathrm{m}\) apart? (b) How close does \(q_{2}\) get to \(q_{1}\) ?
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