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Chapter 23: Problem 35

Electron Released from Rest An electron is released from rest in a uniform electric field of magnitude \(2.00 \times 10^{4} \mathrm{~N} / \mathrm{C}\). Calculate the acceleration of the electron. (Ignore gravitation).

Short Answer

Expert verified
The acceleration of the electron is 3.51 脳 10^15 m/s虏.

Step by step solution

01

- Identify the given values

Given in the problem: the magnitude of the electric field is E = 2.00 脳 10^4 N/C. An electron is released, so initial velocity is zero.
02

- Write down the formula for the force acting on an electron in an electric field

The force (F) acting on an electron in an electric field (E) can be calculated using the formula: F = eE where e is the elementary charge with a value of approximately 1.60 脳 10^-19 C.
03

- Calculate the force acting on the electron

Using the values given: F = (1.60 脳 10^-19 C) 脳 (2.00 脳 10^4 N/C) F = 3.20 脳 10^-15 N.
04

- Use Newton's second law to find the acceleration

According to Newton's second law, the acceleration (a) can be calculated using the formula: a = F/m where m is the mass of the electron, approximately 9.11 脳 10^-31 kg.
05

- Substitute the values and calculate the acceleration

Using the values calculated and given: a = (3.20 脳 10^-15 N) / (9.11 脳 10^-31 kg) a = 3.51 脳 10^15 m/s虏.

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Key Concepts

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

electric field
An electric field is a region around a charged particle where a force would be experienced by other charges. The strength of an electric field is measured in newtons per coulomb (N/C). For example, in this exercise, the electric field has a magnitude of 2.00 脳 10^4 N/C.
The direction of the force within an electric field depends on the nature of the charge. A positive charge will be pushed away from another positive charge and attracted toward a negative one. Conversely, a negative charge like an electron will move toward a positive charge.
To calculate the force exerted on a charge in an electric field, use the formula: \[ F = eE \] Here, \(F\) is the force, \(e\) is the charge of the electron (approximately 1.60 脳 10^-19 C), and \(E\) is the electric field strength.
Newton's second law
Newton's second law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. This can be formulated as: \[ F = ma \] In this equation, \(F\) stands for the force, \(m\) is the mass, and \(a\) is the acceleration.
In simpler terms, if you know the force acting on an object and its mass, you can easily calculate its acceleration.
For an electron in an electric field, you substitute the force calculated from the previous step into Newton's second law to find acceleration. Because the electron's mass is known (approximately 9.11 脳 10^-31 kg), you can determine the acceleration by rearranging the equation: \[ a = \frac{F}{m} \]
acceleration calculation
Finally, you need to calculate the acceleration of the electron. From the previous steps, we have calculated the force on the electron as 3.20 脳 10^-15 N. Using the mass of the electron (9.11 脳 10^-31 kg) and Newton's second law, we substitute these values into the equation \[ a = \frac{F}{m} \].
So: \[ a = \frac{3.20 \times 10^{-15} \text{ N}}{9.11 \times 10^{-31} \text{ kg}} = 3.51 \times 10^{15} \text{ m/s}^2 \]
This calculation shows the electron's acceleration when released from rest in the given electric field.
  • Ensure you always use the correct units.
  • Keep track of significant figures.
  • Understanding each step helps to solve similar physics problems easily.
This approach will make concepts clearer and improve your ability to solve related problems in physics.

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Most popular questions from this chapter

Dipole in a Field An electric dipole, consisting of charges of magnitude \(1.50 \mathrm{nC}\) separated by \(6.20 \mu \mathrm{m}\), is in an electric field of strength \(1100 \mathrm{~N} / \mathrm{C}\). (a) What is the magnitude of the electric dipole moment? (b) What is the difference between the potential energies corresponding to dipole orientations parallel to and antiparallel to the field?

Charged Cloud A charged cloud system produces an electric field in the air near Earth's surface. A particle of charge \(-2.0 \times\) \(10^{-9} \mathrm{C}\) is acted on by a downward electrostatic force of \(3.0 \times\) \(10^{-6} \mathrm{~N}\) when placed in this field. (a) What is the magnitude of the electric field? (b) What are the magnitude and direction of the electrostatic force exerted on a proton placed in this field? (c) What is the gravitational force on the proton? (d) What is the ratio of the magnitude of the electrostatic force to the magnitude of the gravitational force in this case?

Two Point Charges Two point charges \(q_{A}=2.1 \times 10^{-8} \mathrm{C}\) and \(q_{B}=-4.0 q_{A}\) are fixed in place \(50 \mathrm{~cm}\) apart. Find the point along the straight line passing through the two charges at which the electric field is zero.

Uniform Upward Field In Fig. \(23-49\), a uniform, upwarddirected electric field \(\vec{E}\) of magnitude \(2.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\) has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length \(L=10.0 \mathrm{~cm}\) and separation \(d=2.00 \mathrm{~cm} .\) An electron is then shot between the plates from the left edge of the lower plate. The initial velocity \(\vec{v}_{1}\) of the electron makes an angle \(\theta=45.0^{\circ}\) with the lower plate and has a magnitude of \(6.00 \times 10^{6} \mathrm{~m} / \mathrm{s}\). (a) Will the electron strike one of the plates? (b) If so, which plate and how far horizontally from the left edge will the electron strike?

Field Retards Motion An electron with a speed of \(5.00 \times\) \(10^{8} \mathrm{~cm} / \mathrm{s}\) enters an electric field of magnitude \(1.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\), traveling along the field lines in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily and (b) how much time will have elapsed? (c) If the region with the electric field is only \(8.00 \mathrm{~mm}\) long (too short for the electron to stop within it), what fraction of the electron's initial kinetic energy will be lost in that region?
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