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An electron enters a region of uniform electric field with an initial velocity of \(30 \mathrm{~km} / \mathrm{s}\) in the same direction as the electric field, which has magnitude \(E=50 \mathrm{~N} / \mathrm{C}\). (a) What is the speed of the electron \(1.5 \mathrm{~ns}\) after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?

Short Answer

Expert verified
(a) The speed is about 16.67 km/s. (b) The distance is about 3.51×10^-5 m.

Step by step solution

01

Calculate the acceleration of the electron

Use the formula for electric force on a charge. The force, \( F \), on the electron is given by \( F = qE \), where \( q \) is the charge of the electron \( (q = -1.6 \times 10^{-19} \, \text{C}) \) and \( E = 50 \, \text{N/C} \) is the electric field.Thus, \( F = (-1.6 \times 10^{-19} \, \text{C}) \times (50 \, \text{N/C}) = -8 \times 10^{-18} \, \text{N} \).Next, use Newton's second law to calculate the acceleration, \( a = \frac{F}{m} \), where \( m = 9.11 \times 10^{-31} \, \text{kg} \) is the mass of the electron:\[ a = \frac{-8 \times 10^{-18} \, \text{N}}{9.11 \times 10^{-31} \, \text{kg}} \approx -8.78 \times 10^{12} \, \text{m/s}^2. \]
02

Determine the speed of the electron after 1.5 ns

Using the initial velocity \( v_0 = 30 \, \text{km/s} = 3 \times 10^4 \, \text{m/s} \), and the acceleration \( a \) calculated in Step 1, apply the kinematic equation \( v = v_0 + at \), where \( t = 1.5 \, \text{ns} = 1.5 \times 10^{-9} \, \text{s} \):\[ v = 3 \times 10^4 \, \text{m/s} + (-8.78 \times 10^{12} \, \text{m/s}^2)(1.5 \times 10^{-9} \, \text{s}) \approx 16.67 \times 10^4 \, \text{m/s} . \]
03

Calculate the distance traveled in 1.5 ns

Use the kinematic equation for distance: \( s = v_0 t + \frac{1}{2}at^2 \):\[ s = (3 \times 10^4 \, \text{m/s})(1.5 \times 10^{-9} \, \text{s}) + \frac{1}{2}(-8.78 \times 10^{12} \, \text{m/s}^2)(1.5 \times 10^{-9} \, \text{s})^2 . \]Calculate each term:- \( v_0 t = (3 \times 10^4)(1.5 \times 10^{-9}) = 4.5 \times 10^{-5} \, \text{m} \).- \( \frac{1}{2} a t^2 = \frac{1}{2} (-8.78 \times 10^{12})(2.25 \times 10^{-18}) \approx -9.88 \times 10^{-6} \, \text{m} \).Thus, \( s = 4.5 \times 10^{-5} - 9.88 \times 10^{-6} \approx 3.51 \times 10^{-5} \, \text{m} \).

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Key Concepts

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

Electron Motion
In the world of physics, understanding electron motion is essential when delving into the study of electric fields. Electrons, being negatively charged particles, are greatly influenced by electric fields. When an electron enters such a field, it experiences a force that causes it to accelerate in a direction depending on the orientation of the field. In our exercise, the electron initially moves at a velocity of 30 km/s in the direction of an electric field with a strength of 50 N/C. Since electrons have a negative charge of \(-1.6 \times 10^{-19} \text{ C}\), they will accelerate in the opposite direction of the field's electric force. This means even though the velocity is initially positive, the acceleration will decelerate the electron initially.
A thorough understanding of electron motion lays the groundwork for more advanced topics in physics and electronics. Key takeaways when considering electron motion include:
  • Electrons are negatively charged particles that respond to electric fields.
  • The force exerted on the electron in an electric field is calculated by out equation \(F = qE\), where \(q\) represents the charge and \(E\) the electric field strength.
Kinematics
Kinematics, the branch of mechanics concerned with the motion of objects, provides crucial insights into our exercise. To determine how an electron behaves in an electric field, kinematic equations are applied under constant acceleration conditions. In this context, the essential equations we use relate to calculating final velocity and distance traveled after a given time.
In essence, the following kinematic equation gives us the final velocity of the electron:
  • \(v = v_0 + at\)
where \(v_0\) is the initial velocity, \(a\) is acceleration, and \(t\) is time.
Additionally, to find out the distance the electron travels within a specific timeframe, we use:
  • \(s = v_0 t + \frac{1}{2}at^2\)
This approach allows us to calculate both the speed and distance covered by the electron in a uniform electric field. When dealing with real-world problems, breaking down the actions and interactions using kinematics is an effective method for understanding motion.
Acceleration
Acceleration is a fundamental concept in understanding how forces influence motion. In our scenario, the electron is subject to a uniform acceleration due to the electric field. This acceleration can be determined using Newton’s second law, formulated as:
  • \(a = \frac{F}{m}\)
where \(F\) is the force on the electron and \(m\) its mass. This formula quantifies how the electric force, calculated with \(F = qE\), affects the motion of the electron. In this exercise:
  • The electric force \(F\) on the electron is \(-8 \times 10^{-18} \text{ N}\).
  • The electron, with a mass of \(9.11 \times 10^{-31} \text{ kg}\), experiences acceleration \(a\) calculated to be approximately \(-8.78 \times 10^{12} \text{ m/s}^2\).
Acceleration is particularly important because it affects how rapidly an object's velocity changes over time. In this context, it slows the electron due to its negative direction, which results from the opposing electron charge and field direction. Understanding acceleration's role within electric fields is crucial for analyzing motion in physics.

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