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Chapter 17: Problem 64

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling \(4.50 \mathrm{~m}\) in the first \(3.00 \mu \mathrm{s}\) after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignoring the effects of gravity? Justify your answer quantitatively.

Short Answer

Expert verified
(a) Magnitude: \( 5.69 \times 10^{3} \text{ N/C} \); Direction: Downwards. (b) Yes, ignore gravity as electric force is much larger.

Step by step solution

01

Identify Known Values

Given that the electron travels a distance of 4.50 m in 3.00 渭s from rest, we need to identify the initial velocity \( u \), time \( t \), and displacement \( s \) for use in kinematics and force equation. Here \( u = 0 \), \( t = 3.00 \times 10^{-6} \) s, and \( s = 4.50 \) m.
02

Find Acceleration Using Kinematics

We use the kinematic equation: \[ s = ut + \frac{1}{2} a t^2 \]Substituting the known values:\[ 4.50 = 0 + \frac{1}{2} a (3.00 \times 10^{-6})^2 \]Solving for \( a \), we find:\[ a = \frac{2 \times 4.50}{(3.00 \times 10^{-6})^2} \]\[ a = 1.00 \times 10^{13} \text{ m/s}^2 \]
03

Calculate Electric Field Magnitude

The force on the electron due to the electric field can be found using \( F = ma \), where \( m = 9.11 \times 10^{-31} \) kg is the mass of the electron. The force is also given by \( F = eE \), where \( e = 1.60 \times 10^{-19} \) C:\[ ma = eE \]\[ E = \frac{ma}{e} \]Substituting the values:\[ E = \frac{9.11 \times 10^{-31} \times 1.00 \times 10^{13}}{1.60 \times 10^{-19}} \]\[ E \approx 5.69 \times 10^{3} \, \text{N/C} \]
04

Determine Direction of Electric Field

Since the electron is negatively charged and it accelerates upwards, the direction of the electric field must be downwards to exert an upward force on the electron.
05

Evaluate the Effect of Gravity

The gravitational force on the electron is \( F_g = mg \), where \( g = 9.81 \text{ m/s}^2 \):\[ F_g = 9.11 \times 10^{-31} \times 9.81 \]\[ F_g \approx 8.93 \times 10^{-30} \text{ N} \]Compare this to the electric force \( F_e = ma \):\[ F_e = 9.11 \times 10^{-31} \times 1.00 \times 10^{13} = 9.11 \times 10^{-18} \text{ N} \]Since \( F_e \) is much larger than \( F_g \), we are justified in ignoring the effects of gravity.

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Key Concepts

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

Electron Motion
When an electron is introduced into a uniform electric field, it experiences an electric force that causes it to accelerate. The electron begins from rest, meaning it starts with an initial velocity of zero. As it moves through the electric field, the acceleration impacts how quickly it reaches a final velocity. Because electrons carry a negative charge, they move in the direction opposite to the field lines.

The formula to find acceleration using kinematics is given by:
  • \[ s = ut + \frac{1}{2} a t^2 \] where \( s \) is the displacement, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time taken.
In this exercise, the electron covers a distance of 4.50 m in just 3 microseconds. This quick movement highlights the intense influence of the electric field.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces causing the motion. In the case of an electron moving vertically upwards, kinematics provides a framework to calculate its acceleration.

We can use the following kinematic equation:
  • \[ s = ut + \frac{1}{2} a t^2 \]
Since the electron starts from rest (\( u = 0 \)), the equation simplifies to:
  • \[ s = \frac{1}{2} a t^2 \]
By solving for acceleration (\( a \)), we can determinelate that the acceleration of the electron due to the electric field is incredibly significant. Once acceleration is known, it facilitates the calculation of the electric field's magnitude.
Gravitational Effects
Gravitational effects denote the attraction force between objects with mass, defined by the gravitational constant and the masses involved. However, in physics problems involving electrons in electric fields, gravity is often too small to substantially impact the outcome.

To evaluate the significance, we compute the gravitational force \( F_g \) using
  • \[ F_g = mg \]
where \( m \) is the electron's mass and \( g \) is the acceleration due to gravity.

Comparing this to the electric force \( F_e \), calculated as
  • \[ F_e = ma \]
shows that the electric force is magnitudes larger than the gravitational force, thus justifying the omission of gravitational influences.
Physics Problem Solving
Solving physics problems, like those involving electric fields and electron motion, requires a structured approach. Identifying known values, applying the right equations, and understanding the physics principles behind the formulas are key.

Key steps include:
  • Define the problem: Identify what is known and what needs finding.
  • Select the relevant physics principle: Determine the applicable laws related to electric forces, such as Coulomb's law or Newton's second law.
  • Apply kinematic equations: Use these to calculate an object's motion due to forces acting on it, like acceleration.
  • Justify approximations: Assess when it is appropriate to ignore certain forces, like gravitational effects, based on their relative magnitude.
Being thorough with these steps not only clarifies the problem-solving path but also ensures accuracy in determining outcomes, such as the direction and magnitude of an electric field.

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Most popular questions from this chapter

A charge of \(-3.00 \mathrm{nC}\) is placed at the origin of an \(x-y\) coordinate system, and a charge of \(2.00 \mathrm{nC}\) is placed on the \(y\) axis at \(y=4.00 \mathrm{~cm} .\) (a) If a third charge, of \(5.00 \mathrm{nC},\) is now placed at the point \(x=3.00 \mathrm{~cm}, y=4.00 \mathrm{~cm},\) 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.

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

Two very large parallel sheets of the same size carry equal magnitudes of charge spread uniformly over them, as shown in Figure \(17.58 .\) In each of the cases that follow, sketch the net pattern of electric field lines in the region between the sheets, but far from their edges. (Hint: First sketch the field lines due to each sheet, and then add these fields to get the net field.) (a) The top sheet is positive and the bottom sheet is negative, as shown, (b) both sheets are positive, (c) both sheets are negative.

Which of the following is true about \(\vec{E}\) inside a negatively charged sphere as described here? A. It points from the center of the sphere to the surface and is largest at the center. B. It points from the surface to the center of the sphere and is largest at the surface. C. It is zero. D. It is constant but is not zero.

Point charges of \(3.00 \mathrm{nC}\) are situated at each of three corners of a square whose side is \(0.200 \mathrm{~m}\). What are the magnitude and direction of the resultant force on a point charge of \(-1.00 \mu \mathrm{C}\) if it is placed (a) at the center of the square, (b) at the vacant corner of the square?
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