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Chapter 19: Problem 59

The potential difference between the plates of a capacitor is \(175 \mathrm{~V}\). Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant that the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.

Short Answer

Expert verified
Proton's initial speed is \(\sqrt{\frac{2 \cdot e \cdot 87.5}{m_p}} \).

Step by step solution

01

Understand the System

The system involves a capacitor with a potential difference of \(175\,\mathrm{V}\). A proton and an electron are released midway between the plates. The electron starts from rest, while the proton has an initial speed towards the negative plate.
02

Determine the Work-Energy Principle

Apply the work-energy principle for the electron. The change in electric potential energy is transformed into kinetic energy:\[ \Delta K_e = q_e \Delta V \] where \(q_e\) is the charge of the electron and \(\Delta V = 87.5\,\mathrm{V}\) (half of the total voltage since the electron travels half the distance).
03

Calculate Electron's Kinetic Energy

The change in kinetic energy for the electron is given by:\[ \frac{1}{2} m_e v_e^2 = e \cdot 87.5 \] where \(e = 1.6 \times 10^{-19}\,\mathrm{C}\) and \(m_e = 9.11 \times 10^{-31}\,\mathrm{kg}\).
04

Solve for Electron's Final Velocity

Solving for \(v_e\):\[ v_e = \sqrt{\frac{2 \cdot 1.6 \times 10^{-19} \cdot 87.5}{9.11 \times 10^{-31}}} \]
05

Determine Time for Electron to Reach Plate

Calculate the time \(t\) it takes for the electron to reach the positive plate:\[ t = \frac{d}{v_e} \] where \(d\) is half the distance between the plates. Assume \(d = \frac{1}{2}\) of plate separation.
06

Solve for Proton's Initial Speed

For the proton, use the time \(t\) for it to reach the negative plate. From kinetic and potential energy considerations, we find,\[ \frac{1}{2} m_p (v_{p0})^2 + e \cdot 87.5 = \frac{1}{2} m_p (v_{pf})^2 \]Now, solving within \(t\):\[ v_{pf} = 0 \text{ (since it stops at the plate)} \]Gives, \[ v_{p0} = \sqrt{\frac{2 \cdot e \cdot 87.5}{m_p}} \]Solve using \(e = 1.6 \times 10^{-19} \mathrm{C}\) and \(m_p = 1.67 \times 10^{-27} \mathrm{kg}\).

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Key Concepts

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

Work-Energy Principle
The work-energy principle is a fundamental concept in physics that explains how work done on an object is related to its energy changes. In simple terms, when work is done on an object, it can change the object's kinetic energy or potential energy.
In this problem, we see this principle applied to charged particles within an electric field, which is the space between the plates of a capacitor.
The work done by the electric field changes the kinetic energy of the particles.
  • The work-energy principle can be expressed as: \[ \Delta K = q \Delta V \]where \( \Delta K \) is the change in kinetic energy, \( q \) is the charge of the particle, and \( \Delta V \) is the change in electric potential (voltage).
This helps us understand how the energy dynamics between the particle and the electric field determine its motion.
Potential Difference
Potential difference, often referred to as voltage, is the difference in electric potential energy between two points in a circuit or electric field.
This is a crucial concept in understanding the behavior of charged particles in an electric field.
  • The potential difference creates a force that influences charged particles, thus affecting their motion and energy.
  • In the given problem, the potential difference between the capacitor plates is \( 175 \mathrm{~V} \). A proton and an electron experience forces due to this voltage, impacting their movement.
Understanding this concept allows us to see how different particles react to electric fields, which is key in determining their paths and speeds.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion.
In physics, it is a core concept used to quantify the energy transferred during interactions like collisions or movement under a force.
  • The formula for kinetic energy is:\[ K = \frac{1}{2} m v^2 \]where \( m \) is the mass of the object and \( v \) is its velocity.
  • In the context of this problem, we consider the kinetic energy changes of both the proton and the electron as they move due to the electric forces acting on them.
Through the work-energy principle, we understand how potential energy from the electric field transforms into kinetic energy.
Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field.
This concept is integral to understanding how charged particles behave in an electric field change.
  • There are two types of charge: positive and negative. Electrons have a negative charge, while protons have a positive charge.
  • The amount of charge affects how particles move in an electric field. In this problem, the electron and proton have equal and opposite charges, influencing how each is affected by the capacitor's field.
Grasping this concept helps explain why particles move and accelerate differently when exposed to electric potential differences.

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Most popular questions from this chapter

A capacitor stores \(5.3 \times 10^{-5} \mathrm{C}\) of charge when connected to a 6.0 -V battery. How much charge does the capacitor store when connected to a \(9.0-\mathrm{V}\) battery?

An electron is released at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing), (a) As the electron gains kinetic energy, does its electric potential energy increase or decrease? Why? (b) The difference in the electron's electric potential energy between the positive and negative plates is EPE positive \(-\mathrm{EPE}_{\text {negative. }}\) How is this difference related to the charge on the electron \((-e)\) and to the difference \(V\) positive \(-V_{\text {negative in the electric potential between }}\) the plates? (c) How is the potential difference \(V\) positive \(-V_{\text {negative }}\) related to the electric field within the capacitor and the displacement of the positive plate relative to the negative plate? The plates of a parallel plate capacitor are separated by a distance of \(1.2 \mathrm{~cm}\), and the electric field within the capacitor has a magnitude of \(2.1 \times 10^{6} \mathrm{~V} / \mathrm{m}\). An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?

Four identical charges \((+2.0 \mu \mathrm{C}\) each \()\) are brought from infinity and fixed to a straight line. The charges are located \(0.40 \mathrm{~m}\) apart. Determine the electric potential energy of this group.

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{~V} ?\)

Refer to Interactive Solution 19.39 at for one approach to this problem. The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an \(850-\mu \mathrm{F}\) capacitor is \(280 \mathrm{~V}\). (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for \(3.9 \times 10^{-3} \mathrm{~s},\) find the effective power or "wattage" of the flash.
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