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Chapter 18: Problem 2

Two very large charged parallel metal plates are 10.0 \(\mathrm{cm}\) apart and produce a uniform electric field of \(2.80 \times 10^{6} \mathrm{N} / \mathrm{C}\) between them. A proton is fired perpendicular to these plates with an initial speed of 5.20 \(\mathrm{km} / \mathrm{s}\) , starting at the middle of the negative plate and going toward the positive plate. How much work has the electric field done on this proton by the time it reaches the positive plate?

Short Answer

Expert verified
The work done is \( 4.48 \times 10^{-14} \text{ J} \).

Step by step solution

01

Understanding Work Done in an Electric Field

The work done by an electric field on a charged particle is equal to the change in potential energy as the particle moves within the field. It is given by the formula: \[ W = qEd \] where \( W \) is the work done, \( q \) is the charge of the particle, \( E \) is the electric field strength, and \( d \) is the displacement of the charge along the direction of the field.
02

Identify the Charge and Distance

The charge of a proton \( q \) is \( 1.6 \times 10^{-19} \) C. The distance \( d \) the proton travels perpendicular to the plates is equal to the distance between the plates, which is given as 10.0 cm or 0.10 m.
03

Apply the Work Formula

Substitute the values into the work formula: \( W = qEd \). Given: - \( q = 1.6 \times 10^{-19} \) C - \( E = 2.80 \times 10^6 \) N/C - \( d = 0.10 \) m Thus, \[ W = (1.6 \times 10^{-19} \text{ C}) \times (2.80 \times 10^6 \text{ N/C}) \times 0.10 \text{ m} \]
04

Calculate the Work Done

Calculate the numerical result for the work done: \[ W = 1.6 \times 2.8 \times 0.10 \times 10^{-13} \] \[ W = 4.48 \times 10^{-14} \text{ J} \] Thus, the work done by the electric field on the proton is \( 4.48 \times 10^{-14} \text{ Joules} \).

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Key Concepts

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

Work Done by Electric Field
When an electric field acts on a charged particle, it exerts a force on it. This force can do work on the particle as it moves through the field. Work done by an electric field is calculated using the formula: \[ W = qEd \] where:
  • \(W\) is the work done, measured in Joules (J).
  • \(q\) is the charge of the particle, in Coulombs (C).
  • \(E\) is the electric field strength in Newtons per Coulomb (N/C).
  • \(d\) is the displacement the charge moves parallel to the electric field, in meters (m).
The concept of work here is similar to mechanical work. As the proton in the question moves toward the positive plate, the electric field is working on it. This calculated work represents the energy transferred from the electric field to the proton as it moves through the field.
Charged Particles
Charged particles, like protons and electrons, experience forces when placed in electric fields. A proton, which is positively charged, naturally moves from areas of high electric potential to low electric potential. This is because the uniform electric field exerts a force \[ F = qE \] where \(F\) is the force on the proton, \(q\) is the charge of the proton, and \(E\) is the field strength. Some characteristics of charged particles include:
  • Protons have a positive charge of \(1.6 \times 10^{-19}\) C.
  • Electrons have the same magnitude of charge as protons, but negative.
  • Charged particles will move under the influence of electric fields, accelerating due to the force exerted on them.
Understanding the behavior of charged particles in electric fields is crucial for many applications, from electronic circuits to understanding molecular interactions.
Electric Potential Energy
Electric potential energy is the energy that a charged particle possesses due to its position in an electric field. In the case of uniform fields, such as those found between two parallel plates, the potential energy ( \[ U \] ) is related to the work done by the field. The change in electric potential energy as a charge moves is given by: \[ \Delta U = -W \] Where \(W\) is the work done by the electric field. The negative sign arises because the work done by the field reduces the particle's potential energy. Some key points include:
  • Electric potential energy decreases as a positive charge moves toward a negative plate.
  • As a proton moves in the direction of the electric field, it loses potential energy but gains kinetic energy.
  • The change in potential energy equals the work done by the field.
Understanding electric potential energy is vital in explaining how charged particles interact in electric fields.
Parallel Plate Capacitor
Parallel plate capacitors consist of two large conductive plates separated by an insulator, and they store electric potential energy in the uniform electric field between the plates. Key characteristics of parallel plate capacitors include:
  • The electric field \(E\) is uniform and directed from the positive plate to the negative plate.
  • The potential difference \( V \) between the plates is given by \(V = Ed\) where \(d\) is the separation between the plates.
  • The capacitance \( C \) of the capacitor is determined by the area of the plates \( A \), the separation \( d \), and the permittivity of the insulator between the plates \( \epsilon \)
  • The formula for capacitance is \( C = \frac{\epsilon A}{d} \).
The separation and the uniform field make parallel plate capacitors excellent for understanding basic electric field principles, including field strength, potential difference, and the energy stored in the electric field.

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Most popular questions from this chapter

A 5.80\(\mu\) F parallel-plate air capacitor has a plate separation of 5.00 mm and is charged to a potential difference of 400 \(\mathrm{V}\) . Calculate the energy density in the region between the plates, in units of \(\mathrm{J} / \mathrm{m}^{3} .\)

Dipole. A dipole is located on a sheet of paper. (a) In the plane of that paper, carefully sketch the electric field lines for this dipole. (b) Use your field lines in part (a) to sketch the equipotential curves where the equipotential surfaces intersect the paper.

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{675}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?

Two stationary point charges of \(+3.00 \mathrm{nC}\) and \(+2.00 \mathrm{nC}\) are separated by a distance of 50.0 \(\mathrm{cm} .\) An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy for the electron when it is (a) at the midpoint and (b) 10.0 \(\mathrm{cm}\) from the \(+3.00 \mathrm{nC}\) charge?

Electrical sensitivity of sharks. Certain sharks can detect an electric field as weak as 1.0\(\mu \mathrm{V} / \mathrm{m} .\) To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5 \(\mathrm{V}\) A battery across theseplates, how far apart would the plates have to be?
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