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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
The magnitude of electric field is calculated to be \(E\). The direction is downward due to the negative charge of the electron. As the force due to gravity is substantially smaller than the electric force, it is considered negligible.

Step by step solution

01

Determine the acceleration of the electron

The task mentions an electron's journey upwards in a given time for a predetermined distance. Use these variables to determine acceleration with the kinematic equation \(a = \frac{2 \cdot (d - u \cdot t)}{t^2}\), where \(u\) is the initial velocity (given as zero), \(d\) is the distance, and \(t\) is the time. Substituting these values, we find the acceleration \(a\).
02

Calculate the Electric Field

With the acceleration calculated, the electric field \(E\) can be found by using the equation \(E = \frac{F}{q}\), where \(F\) denotes the force and \(q\) denotes the charge. For an electron, the force is aligned with acceleration; hence \(F = m \cdot a\), where \(m\) is mass of an electron, and \(a\) is the acceleration computed in Step 1. Since \(q = -e\), where \(e\) is the elementary charge, substitute these values into the equation. \(E = \frac{m * a}{-e}\) will give us the electric field. The negative sign signifies the direction of the field, which is downwards.
03

Evaluate Gravity Effects

We need to evaluate whether gravity had any noticeable effect on the electron's motion. Compute the force of gravity using \(F_{g} = m \cdot g\), where \(g\) is acceleration due to gravity. Compare \(F_{g}\) to the electric force \(F\). If \(F_{g}\) is substantially smaller than \(F\), one can argue that gravity's effect on the electron's motion is negligible.

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Key Concepts

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

Electron Motion
An electron is a tiny subatomic particle carrying a negative charge. When an electron is placed in a uniform electric field, it experiences a force due to the field, which causes it to accelerate. The direction of this acceleration is determined by the direction of the electric field and the charge on the electron. Since electrons are negatively charged, they move in the direction opposite to the electric field lines.

In our exercise, the electron is initially at rest, meaning its starting velocity is zero. When released in an electric field, it accelerates upwards over a certain distance and time. Understanding this motion involves concepts of charge and force:
  • An electric field exerts a force on charged particles.
  • This force is calculated using the product of the charge and the field strength: F = qE.
  • For an electron, with charge \(q = -e\), direction of force will be opposite to that of field.
As the problem states a vertical upward motion over a specific period, we can derive information about the electron's path from this acceleration.
Kinematic Equations
Kinematic equations are useful tools in physics for describing motion. They relate different physical quantities such as displacement, time, and acceleration. In the context of our exercise, the motion of the electron is described through these equations.

The relevant kinematic equation for this problem is: \[ a = \frac{2 \cdot (d - u \cdot t)}{t^2} \] Where:
  • \(a\) is the acceleration
  • \(d\) is the distance traveled
  • \(u\) is the initial velocity (zero in this scenario)
  • \(t\) is the time taken
By plugging in the time and distance values, we find the acceleration the electron undergoes. This acceleration, resulting from the electric force, is crucial in establishing the magnitude of the electric field acting on the electron.
Effects of Gravity
In typical problems involving objects in motion near Earth's surface, gravity is a significant factor to consider. However, in the case of an electron, due to its very small mass, the effects of gravity may be negligible when compared with the forces from an electric field.

The gravitational force acting on the electron can be determined using the formula: \[ F_g = m \cdot g \] Where:
  • \(m\) represents the mass of the electron
  • \(g\) is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\)
By evaluating this force against the considerably larger electric force \(F = m \cdot a\) due to the electric field, we find that the gravitational force \(F_g\) is much smaller. Thus, its influence on the electron's motion is insignificant, allowing us to justify generally ignoring it in such scenarios.

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Most popular questions from this chapter

Two \(1.20 \mathrm{~m}\) nonconducting rods meet at a right angle. One rod carries \(+2.50 \mu \mathrm{C}\) of charge distributed uniformly along its length, and the other carries \(-2.50 \mu \mathrm{C}\) distributed uniformly along it (Fig. \(\mathbf{P} 2 \mathbf{1 . 8 5}\) ). (a) Find the magnitude and direction of the electric field these rods produce at point \(P,\) which is \(60.0 \mathrm{~cm}\) from each rod. (b) If an electron is released at \(P\), what are the magnitude and direction of the net force that these rods exert on it?

Two point charges are placed on the \(x\) -axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{~m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{~m} .\) What are the magnitude and direction of the total force exerted by these two charges on a negative point charge \(q_{3}=-6.00 \mathrm{nC}\) that is placed at the origin?

A very long, straight wire has charge per unit length \(3.20 \times 10^{-10} \mathrm{C} / \mathrm{m} .\) At what distance from the wire is the electric- field magnitude equal to \(2.50 \mathrm{~N} / \mathrm{C} ?\)

A proton is placed in a uniform electric field of \(2.75 \times 10^{3} \mathrm{~N} / \mathrm{C} .\) Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after \(1.00 \mu \mathrm{s}\) in the field, assuming it starts from rest.

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x=0, y=-1.35 \mathrm{~m}\) (b) at \(x=12.0 \mathrm{~cm}, y=12.0 \mathrm{~cm} ;\) (c) at \(x=-1.10 \mathrm{~m}, y=2.60 \mathrm{~m} ?\) Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\)

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