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Chapter 23: Problem 14

A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{\boldsymbol{E}}\) directed to the left. It is released from rest and moves to the left; after it has moved \(6.00 \mathrm{cm},\) its kinetic energy is found to be \(+1.50 \times 10^{-6} \mathrm{J.}\) (a) What work was done by the electric force? (b) What is the potential of the starting point with respect to the end point? (c) What is the magnitude of \(\vec{\boldsymbol{E}}\)?

Short Answer

Expert verified
(a) Work done: \(1.50 \times 10^{-6} \mathrm{J}\). (b) Potential difference: +357.14 V. (c) Magnitude of \(\vec{E}\): 5952.33 V/m.

Step by step solution

01

Calculate the Work Done by the Electric Force

According to the work-energy theorem, the work done by the electric force equals the change in kinetic energy of the particle. Initially, the kinetic energy is 0 because the particle is released from rest. The final kinetic energy is given as \( 1.50 \times 10^{-6} \mathrm{J} \). Hence, the work done \( W \) is simply the final kinetic energy: \[ W = 1.50 \times 10^{-6} \mathrm{J}. \]
02

Determine the Potential Difference

The work done by the electric force is also related to the potential difference \( \Delta V \) and the charge \( q \) by the equation \( W = q \Delta V \). Solving for \( \Delta V \), we have: \[ \Delta V = \frac{W}{q} = \frac{1.50 \times 10^{-6} \, \mathrm{J}}{4.20 \times 10^{-9} \, \mathrm{C}}. \] Calculating gives: \[ \Delta V = 357.14 \, \mathrm{V}. \] The potential at the starting point relative to the end point is +357.14 V.
03

Find the Magnitude of the Electric Field

The electric potential difference is also related to the electric field \( E \) and the distance \( d \) by \( \Delta V = E \times d \), where \( d = 0.06 \, \mathrm{m} \). Solving for \( E \), we get: \[ E = \frac{\Delta V}{d} = \frac{357.14 \, \mathrm{V}}{0.06 \, \mathrm{m}}. \] Calculating gives: \[ E = 5952.33 \, \mathrm{V/m}. \] The magnitude of the electric field is \( 5952.33 \, \mathrm{V/m}. \)

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Key Concepts

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

Electric Potential
Electric potential is a crucial concept in understanding how charged particles experience forces and movement in electric fields. Think of electric potential as the energy landscape for charges 鈥� it dictates how much work is needed to move a charge within the field.
Electric potential is measured in volts and tells us how much potential energy a unit charge would have at a certain point. This becomes particularly important when dealing with electric fields.
  • The work done on a charge by an electric field is related to the change in electric potential, also known as the potential difference or voltage.
  • The formula that relates work done to electric potential and charge is given by: \[ W = q \, \Delta V \]where \( W \) is work, \( q \) is the charge, and \( \Delta V \) is the potential difference.
  • In the described exercise, we found a potential difference of 357.14 V by calculating how much work the electric field performed while moving the charge.
Understanding electric potential helps us predict how charges will move and interact within electric fields.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It's a key concept when analyzing the dynamics of particles, especially when they are propelled by forces within an electric field.
When a charged particle is released in an electric field, it experiences a force that accelerates it, converting the potential energy into kinetic energy. This change in energy is a direct consequence of the work done by the electric field.
  • The kinetic energy of an object is defined as:\[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass and \( v \) is the velocity of the object.
  • In electric fields, the initial kinetic energy is often zero as particles start from rest and the electric field does the work to provide them with kinetic energy.
  • In the given problem, the particle's kinetic energy became \( 1.50 \times 10^{-6} \) J after moving through the field, which was equivalent to the work done by the electric force.
By understanding kinetic energy, we can better comprehend how energy is transferred and transformed in electric fields, leading to particle motion.
Electromagnetic Theory
Electromagnetic theory provides the framework for understanding electric and magnetic fields and their interactions with matter. It encompasses principles such as electric forces, potential, and energy transformations.
An electric field is a region around a charged object, where other charges experience a force. It's a vector field 鈥� meaning it has both magnitude and direction 鈥� and is vital to explaining how electric forces cause motion in charged particles.
  • The electric field strength \( E \) relates the force experienced by a test charge in the field through the equation:\[ F = qE \]where \( F \) is the force and \( q \) is the magnitude of the charge.
  • This exercise relied on the relationship between electric potential difference, electric field strength, and distance, given by:\[ \Delta V = E \times d \]
  • Calculating the electric field \( E \) in this scenario, we found it to be \( 5952.33 \, \text{V/m} \), using the known potential difference and distance the particle traveled.
Mastery of electromagnetic theory helps us predict not only how particles behave in fields but also deeper phenomena such as electromagnetic waves and their applications.

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Most popular questions from this chapter

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{cm} . \mathrm{A}\) small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within 8.00 \(\mathrm{cm}\) of the surface of the large sphere?

A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. A thin wire lies on the axis of a hollow metal cylinder and is insulated from it (Fig. P23.64). A large potential difference is established between the wire and the outer cylinder, with the wire at higher potential; this sets up a strong electric field directed radially out- ward. When ionizing radiation enters the device, it ionizes a few air molecules. The free electrons produced are accelerated by the electric field toward the wire and, on the way there, ionize many more air molecules. Thus a current pulse is produced that can be detected by appropriate electronic circuitry and converted to an audible "click." Suppose the radius of the central wire is 145\(\mu \mathrm{m}\) and the radius of the hollow cylinder is 1.80 \(\mathrm{cm} .\) What potential difference between the wire and the cylinder produces an electric field of \(2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) at a distance of 1.20 \(\mathrm{cm}\) from the axis of the wire? (The wire and cylinder are both very long in comparison to their radii, so the results of Problem 23.63 apply.)

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 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 small sphere with mass \(5.00 \times 10^{-7} \mathrm{kg}\) and charge \(+3.00 \mu C\) is released from rest a distance of 0.400 \(\mathrm{m}\) above a large horizontal insulating sheet of charge that has uniform surface charge density \(\sigma=+8.00 \mathrm{pC} / \mathrm{m}^{2} .\) Using energy methods, calculate the speed of the sphere when it is 0.100 \(\mathrm{m}\) above the sheet of charge?
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