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Chapter 23: Problem 13

A small particle has charge \(-5.00 \mu\) 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
The final speed will be calculated from the final kinetic energy, which we obtain from conservation of energy. If this final speed is greater than the initial speed, then it indicates that the particle is moving faster at point B compared to point A. The detailed calculations will provide the numerical value for the final speed.

Step by step solution

01

Understand the Problem

It is important to understand that the particle is moving in an electric potential. When it moves from a point of lower potential to a higher potential, work is done on the particle thus increasing its kinetic energy. Because the electric force is the only force doing work on the particle, the principle of conservation of energy applies here. It states that total energy before = total energy after.
02

Calculate the Initial Energy (at Point A)

Start by calculating the initial total energy of the particle at point A. The total energy will be a sum of the kinetic energy and the electric potential energy. The kinetic energy (KE) can be calculated by using the formula \(KE=(1/2)mv_{a}^{2}\), where \(m=2.00 \times 10^{-4}\) kg (mass of the particle) and \(v_{a}=5.00\) m/s (initial speed of the particle). To calculate the electric potential energy at point A, we use the formula \(PE_{A}=qV_{A}\), where \(q=-5.00 \times 10^{-6}\) C (charge of the particle) and \(V_{A}=200\) V (electric potential at point A). Then sum these two energies for total energy at A.
03

Determine the Final Energy (at Point B)

The total final energy at point B is the sum of kinetic energy and potential energy at B. Because of energy conservation, this total energy is the same as the total initial energy calculated in step 2. The potential energy at B is given by \(PE_{B}=qV_{B}\). Substitute for the known values (charge of the particle q and the voltage at point B \(V_{B}\) to get \(PE_{B}\).
04

Solve for Final Speed

We know that the total energy at point B (calculated in step 3) equals initial total energies. Thus, kinetic energy at B = total energy at B - potential energy at B. Once we know the kinetic energy at B, we can solve for the final speed \(v_{B}\) by rearranging the kinetic energy equation to \(v_{B}= \sqrt{(2 KE_{B})/m}\). Substitute for \(KE_{B}\) and mass \(m\) to find \(v_{B}\).

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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 possessed by an object due to its motion. It is a crucial concept in physics that helps us understand how objects gain or lose energy. The kinetic energy (KE) of an object can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where:\( m \) is the mass of an object and \( v \) is its velocity.
  • Mass is usually measured in kilograms (kg)
  • Velocity is measured in meters per second (m/s)
  • Kinetic energy is measured in joules (J)
Kinetic energy plays an important role when determining how fast a particle will move under certain forces. In the case of an electric field, as the particle speeds up or slows down, its kinetic energy changes in response to the work done by the electric force. This simple concept can often explain why a particle might move faster or slower at different points along its path.
Electric Force
Electric force is the force exerted by an electric field on a charged particle. It is a fundamental force in physics, influencing how charged particles interact with each other. A charged particle, like the one in our exercise, experiences electric force when placed in an electric field or when moving between points of different electric potentials.
The electric force can be calculated using Coulomb's Law but is also effectively understood through electric potential energy changes. Key points include:
  • Electric force causes acceleration of charged particles
  • It is dependent on the charge of the objects and the strength of the electric field
  • Direction of the force depends on the nature of charge (positive or negative)
In our scenario, the electric force does the work when the particle moves between points A and B. This work done causes a change in the particle鈥檚 kinetic energy, contributing to the change in its speed.
Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics that states that energy cannot be created or destroyed, only transformed from one form to another. For the moving charged particle in the exercise, conservation of energy ensures the total energy remains constant as it moves from point A to point B.
Here's how it applies:
  • Total energy at one point equals total energy at another: \( KE + PE \text{ (at A)} = KE + PE \text{ (at B)} \)
  • Kinetic energy and potential energy trade off as the particle moves from a lower potential to a higher potential
  • If potential energy increases, kinetic energy decreases, resulting in a slower speed (and vice versa)
This principle helps us calculate the speed at point B by ensuring that we correctly assess how the changing electric potential affects the particle's kinetic energy. Understanding energy conservation helps explain the changes in motion and speed of the particle without any external forces other than the electric force.

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Most popular questions from this chapter

A point charge \(q_{1}=+2.40 \mu \mathrm{C}\) is held stationary at the origin. A second point charge \(q_{2}=-4.30 \mu \mathrm{C}\) moves from the point \(x=0.150 \mathrm{~m}, y=0\) to the point \(x=0.250 \mathrm{~m}, y=0.250 \mathrm{~m} .\) How much work is done by the electric force on \(q_{2} ?\)

Identical charges \(q=+5.00 \mu \mathrm{C}\) are placed at opposite corners of a square that has sides of length \(8.00 \mathrm{~cm}\). Point \(A\) is at one of the empty corners, and point \(B\) is at the center of the square. A charge \(q_{0}=-3.00 \mu \mathrm{C}\) is placed at point \(A\) and moves along the diagonal of the square to point \(B\). (a) What is the magnitude of the net electric force on \(\bar{q}_{0}\) when it is at point \(A ?\) Sketch the placement of the charges and the direction of the net force. (b) What is the magnitude of the net electric force on \(q_{0}\) when it is at point \(B ?\) (c) How much work does the electric force do on \(q_{0}\) during its motion from \(A\) to \(B ?\) Is this work positive or negative? When it goes from \(A\) to \(B,\) does \(q_{0}\) move to higher potential or to lower potential?

An annulus with an inner radius of \(a\) and an outer radius of \(b\) has charge density \(\sigma\) and lies in the \(x y\) -plane with its center at the origin, as shown in Fig. \(\mathbf{P} 2 \mathbf{3 . 8 0}\). (a) Using the convention that the potential vanishes at infinity, determine the potential at all points on the \(z\) -axis. (b) Determine the electric field at all points on the \(z\) -axis by differentiating the potential. (c) Show that in the limit \(a \rightarrow 0, b \rightarrow \infty\) the electric field reproduces the result obtained in Example 22.7 for an infinite plane sheet of charge. (d) If \(a=5.00 \mathrm{~cm}, b=10.0 \mathrm{~cm}\) and the total charge on the annulus is \(1.00 \mu \mathrm{C}\), what is the potential at the origin? (e) If a particle with mass \(1.00 \mathrm{~g}\) (much less than the mass of the annulus) and charge \(1.00 \mu \mathrm{C}\) is placed at the origin and given the slightest nudge, it will be projected along the \(z\) -axis. In this case, what will be its ultimate speed?

A particle with charge \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed to the left. The charge is released from rest and moves to the left; after it has moved \(6.00 \mathrm{~cm},\) its kinetic energy is \(+2.20 \times 10^{-6} \mathrm{~J}\). What are (a) the work done by the electric force, (b) the potential of the starting point with respect to the end point, and (c) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) ?

For a particular experiment, helium ions are to be given a kinetic energy of \(3.0 \mathrm{MeV}\). What should the voltage at the center of the accelerator be, assuming that the ions start essentially at rest? (a) \(-3.0 \mathrm{MV}\) (b) \(+3.0 \mathrm{MV} ;(\mathrm{c})+1.5 \mathrm{MV} ;(\mathrm{d})+1.0 \mathrm{MV}\)
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