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A particle with a charge of \(+4.20 \mathrm{nC}\) is in a uniform electric field \(\vec{E}\) directed in the negative \(x\) direction. It is released from rest, and 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 was the change in electric potential over the distance that the charge moved? (c) What is the magnitude of \(\overrightarrow{\boldsymbol{E}} ?\) (d) What was the change in potential energy of the charge?

Short Answer

Expert verified
(a) Work done: \(1.50 \times 10^{-6} \text{ J}\). (b) Change in potential: \(357.14 \text{ V}\). (c) Magnitude of \(\vec{E}\): \(5952.33 \text{ V/m}\). (d) Potential energy change: \(1.50 \times 10^{-6} \text{ J}\).

Step by step solution

01

Understanding Electric Work Done

The work done by the electric field on a charge is equal to the change in kinetic energy of the particle. Since the kinetic energy is given as final and started from rest, the work done by the electric field is equal to this kinetic energy.\[ W = 1.50 \times 10^{-6} \text{ J} \]
02

Calculating Change in Electric Potential

The relationship between work done and change in electric potential \( \Delta V \) over a distance \( d \) is given by:\[ W = q \Delta V \]Rearranging for \( \Delta V \):\[ \Delta V = \frac{W}{q} = \frac{1.50 \times 10^{-6} \text{ J}}{4.20 \times 10^{-9} \text{ C}} \approx 357.14 \text{ V} \]
03

Finding Magnitude of Electric Field

The electric field \( E \) is related to the electric potential difference \( \Delta V \) by the equation:\[ \Delta V = E \cdot d \]Rearranging for \( E \):\[ E = \frac{\Delta V}{d} = \frac{357.14 \text{ V}}{0.06 \text{ m}} \approx 5952.33 \text{ V/m} \]
04

Determining Change in Potential Energy

The change in potential energy \( \Delta U \) when moving through an electric field is:\[ \Delta U = q \Delta V = 4.20 \times 10^{-9} \text{ C} \times 357.14 \text{ V} = 1.50 \times 10^{-6} \text{ J} \]

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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 associated with the motion of an object. In the world of physics, it's a crucial concept because it allows us to understand how energy transforms from one form to another. For an object moving at a certain velocity, the kinetic energy is calculated using the formula: \[ KE = \frac{1}{2}mv^2 \]In this exercise, the particle starts from rest, meaning its initial kinetic energy is zero. As it moves 6.00 cm in the electric field, it gains kinetic energy amounting to \(1.50 \times 10^{-6} \text{ J}\). This increase is a direct result of the work done by the electric force on the particle, demonstrating the profound relationship between work and kinetic energy.
  • Increases when an object speeds up
  • Directly affected by changes in velocity
Electric Potential
Electric potential reflects the potential energy per unit charge at a particular location in an electric field. Think of it as the electric version of gravitational potential energy. When a charge moves in an electric field, especially over a certain distance, it undergoes a change in electric potential. In this scenario, the work done on the charge translates into a potential difference or voltage \( \Delta V\). The electric potential helps us quantify how much energy a charge can gain or lose when moving between two points. The relationship between the work done \( W \) and change in electric potential is given by:\[ W = q \Delta V \] Rearranging the formula allows us to solve for the change in electric potential:\[ \Delta V = \frac{W}{q} \]Upon plugging in the given values, we find that \( \Delta V \approx 357.14 \text{ V} \).
  • Provides insight into energy changes in a field
  • Measured in volts, representing energy per charge unit
Work Done
Work done in the context of physics is the energy transferred to or from an object via a force acting over a displacement. For electric fields, the work done by the field on a charge results in a change in the charge's kinetic energy. Here, since the particle started from rest, the entire kinetic energy of \(1.50 \times 10^{-6} \text{ J}\) is due to the work done by the electric force. This demonstrates the principle:\[ W = \Delta KE \] which tells us that work done equals the change in kinetic energy. This relationship emphasizes the interchangeable nature of work and energy within physics. In an electric field setting, if we know the work done, we immediately can deduce changes in kinetic energy and vice versa.
  • Energy transfer via force
  • Indicator of energy transformation across systems
Potential Energy
Electric potential energy is the energy stored in a system of charges due to their positions within an electric field. This energy is particularly important when charges are displaced, as in our exercise. Here, a charged particle shifts within the field, leading to changes in its potential energy.The formula for potential energy change in an electric field is similar to that for electric work and is given by:\[ \Delta U = q \Delta V \] For the particle in question, \[ \Delta U = 4.20 \times 10^{-9} \text{ C} \times 357.14 \text{ V} = 1.50 \times 10^{-6} \text{ J} \] This calculation mirrors the outcome of the work done because potential and kinetic energies are interlinked through conservation of energy.
  • Energy based on charge configuration
  • Subject to changes as charges relocate

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Most popular questions from this chapter

Two oppositely charged identical insulating spheres, each \(50.0 \mathrm{~cm}\) in diameter and carrying a uniform charge of magnitude \(175 \mu \mathrm{C},\) are placed \(1.00 \mathrm{~m}\) apart center to center (Figure 18.48 ). (a) If a voltmeter is connected between the nearest points \((a\) and \(b\) ) on their surfaces, what will it read? (b) Which point, \(a\) or \(b\), is at the higher potential? How can you know this without any calculations?

The paper dielectric in a paper-and-foil capacitor is \(0.0800 \mathrm{~mm}\) thick. Its dielectric constant is \(2.50,\) and its dielectric strength is \(50.0 \mathrm{MV} / \mathrm{m} .\) Assume that the geometry is that of a parallel- plate capacitor, with the metal foil serving as the plates. (a) What area of each plate is required for a \(0.200 \mu \mathrm{F}\) capacitor? (b) If the electric field in the paper is not to exceed one-half the dielectric strength, what is the maximum potential difference that can be applied across the capacitor?

A parallel-plate air capacitor has a capacitance of \(920 \mathrm{pF}\). The charge on each plate is \(2.55 \mu \mathrm{C}\). (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is doubled? (c) How much work is required to double the separation?

In the Bohr model of the hydrogen atom, a single electron revolves around a single proton in a circle of radius \(r .\) Assume that the proton remains at rest. (a) By equating the electric force to the electron mass times its acceleration, derive an expression for the electron's speed. (b) Obtain an expression for the electron's kinetic energy, and show that its magnitude is just half that of the electric potential energy. (c) Obtain an expression for the total energy, and evaluate it using \(r=5.29 \times 10^{-11} \mathrm{~m}\). Give your numerical result in joules and in electronvolts.

You are working on an electronics project that requires a variety of capacitors, but you have only five capacitors available, each with a capacitance of \(100 \mathrm{nF}\). (a) What is the maximum capacitance that you can produce with these five capacitors? (b) What is the minimum capacitance, other than zero, that you can produce?

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