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Chapter 17: Problem 33

A uniform electric field exists in the region between two oppositely charged plane parallel plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, \(3.20 \mathrm{~cm}\) distant from the first, in a time interval of \(1.5 \times 10^{-8} \mathrm{~s}\). (a) Find the magnitude of this electric field. (b) Find the speed of the electron when it strikes the second plate.

Short Answer

Expert verified
(a) \( E \approx \frac{2 \times 0.032}{(1.5 \times 10^{-8})^2} \times \frac{1}{1.6 \times 10^{-19}} \), (b) \( v = 0 + \left( \frac{2 \times 0.032}{(1.5 \times 10^{-8})^2} \right) \times 1.5 \times 10^{-8} \)."

Step by step solution

01

Understand the Problem Context

An electron is moving between two oppositely charged plates with a known distance of 3.20 cm. It's influenced by a uniform electric field and takes 1.5 \( \times 10^{-8} \) s to travel this distance from rest.
02

Determine the Acceleration of the Electron

Since the electron starts from rest and travels a certain distance in a specific time, we use the equation of motion: \( d = \frac{1}{2} a t^2 \), where \( d = 0.032 \) m, and \( t = 1.5 \times 10^{-8} \) s. Solving for acceleration \( a \), we get: \( a = \frac{2d}{t^2} = \frac{2 \times 0.032}{(1.5 \times 10^{-8})^2} \).
03

Compute the Electric Field Magnitude

The electrical force \( F \) on the electron is \( F = ma \), where \( a \) is the acceleration found in Step 2. The force is also related to the electric field \( E \) by \( F = eE \), where \( e \) is the charge of the electron (\( -1.6 \times 10^{-19} \) C). Thus, \( E = \frac{ma}{e} \).
04

Calculate the Speed When Electron Strikes the Second Plate

Using the equation \( v = u + at \) where \( u = 0 \) (initial velocity), \( a \) is found in Step 2, and \( t = 1.5 \times 10^{-8} \) s, calculate the final velocity \( v \) of the electron when it hits the second plate.

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Key Concepts

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

Uniform Electric Field
A uniform electric field is a region where the electric force acts evenly across, without changing its magnitude or direction at any point. This means, if you place a charged particle like an electron in this field, it experiences a constant force throughout its journey.
  • Think of a uniform electric field like the lanes on a highway where the cars (electrons) move smoothly between the lines.
  • The electric field between two oppositely charged parallel plates is uniform, ensuring that an electron released from the negatively charged plate moves in a direct path toward the positively charged one.
  • For our exercise, the uniform electric field is crucial since it exerts a steady force on the electron, helping us predict its motion effectively.
Understanding this concept helps to calculate physical quantities like force and motion experienced by charged particles in such fields.
Electron Motion
In the presence of a uniform electric field, an electron exhibits continuous motion from the negatively charged plate to the positively charged plate. This motion is guided by the electric forces acting on the electron.
  • Electrons are negatively charged, so they naturally move towards the positive side when placed in an electric field.
  • Initially, the electron starts from rest, meaning it has no initial speed, and it gains speed as it moves under the influence of the electric field.
  • The speed of the electron increases until it strikes the second plate, because the consistent force from the electric field continually accelerates it.
This motion helps us use kinematic equations to further analyze and calculate various parameters such as time of travel and final speed.
Kinematics
Kinematics deals with the motion of points, objects, or systems without touching on the forces behind them. In our problem, we'll use the kinematic equations to calculate the electron's acceleration and final velocity when it hits the opposite plate.
  • The equation \( d = \frac{1}{2} a t^2 \) helps us identify the acceleration \( a \) of the electron, where \( d \) is the distance traveled and \( t \) is the time taken.
  • Substituting the given values of the distance (0.032 m) and time (1.5 \( \times 10^{-8} \)) leads to calculating the acceleration.
  • The velocity equation \( v = u + at \) lets us find the final speed \( v \) of the electron, considering it started with zero initial velocity \( u \).
Kinematics provides the mathematical foundation to predict and understand the motion of the electron in the electric field.
Electric Force
Electric force is the influence that causes charged particles to experience motion when placed in an electric field. In our scenario, it's the electric force that propels the electron across the field.
  • The force acting on the electron in an electric field is given by \( F = eE \), where \( e \) is the electron's charge and \( E \) is the electric field.
  • In a uniform electric field, this force also results in an acceleration \( a \), calculated as \( F = ma \), where \( m \) is the electron's mass.
  • By utilizing these relations, we derive the electric field's strength as \( E = \frac{ma}{e} \).
Electric force is fundamental in understanding how electrons behave and move in an electric field, helping us link the field's properties with the electron's motion.

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Most popular questions from this chapter

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x-y\) coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) axis at \(x=4.00 \mathrm{~cm} .\) (a) If a third charge \(q_{3}=+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{~cm}, y=3.00 \mathrm{~cm},\) find the \(x\) and \(y\) components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

Three point charges are arranged along the \(x\) axis. Charge \(q_{1}=-4.50 \mathrm{nC} \quad\) is \(\quad\) located \(\quad\) at \(\quad x=0.200 \mathrm{~m}, \quad\) and \(\quad\) charge \(q_{2}=+2.50 \mathrm{nC}\) is at \(x=-0.300 \mathrm{~m}\). A positive point charge \(q_{3}\) is located at the origin. (a) What must the value of \(q_{3}\) be for the net force on this point charge to have magnitude \(4.00 \mu \mathrm{N}\) ? (b) What is the direction of the net force on \(q_{3} ?\) (c) Where along the \(x\) axis can \(q_{3}\) be placed and the net force on it be zero, other than the trivial answers of \(x=+\infty\) and \(x=-\infty ?\)

Sketch electric field lines in the vicinity of two charges, \(Q\) and \(-4 Q\), located a small distance apart on the \(x\) axis.

At a distance of \(16 \mathrm{~m}\) from a charged particle, the electric field has a magnitude of \(100 \mathrm{~N} / \mathrm{C}\). (a) At what distance is the electric field \(400 \mathrm{~N} / \mathrm{C} ?\) (b) At what distance is it \(10 \mathrm{~N} / \mathrm{C} ?\)

A small object carrying a charge of \(-8.00 \mathrm{nC}\) is acted upon by a downward force of \(20.0 \mathrm{nN}\) when placed at a certain point in an electric field. (a) What are the magnitude and direction of the electric field at the point in question? (b) What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?
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