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Chapter 29: Problem 2

The electric field strength is \(20,000 \mathrm{N} / \mathrm{C}\) inside a parallel plate capacitor with a 1.0 mm spacing. An electron is released from rest at the negative plate. What is the electron's speed when it reaches the positive plate?

Short Answer

Expert verified
The speed of the electron when it reaches the positive plate is obtained as the square root of the term \(2ad\) where \(a\) is the acceleration computed in Step 2 and \(d\) is the distance between the plates.

Step by step solution

01

Understand problem and given information

The electric field strength is \(20,000 \mathrm{N} / \mathrm{C}\), and the distance between the plates (d) is given as 1.0 mm which is \(0.001 m\). An electron is released from rest at the negative plate, hence its initial speed (v) is 0. Electric field (E) force can be used to calculate the acceleration (a) of the electron, and kinematic equation can then determine the final speed of the electron.
02

Calculate the acceleration of the electron

The electric force on the electron (F) is its electric charge (q) times the electric field (E). This can be written as: \(F = qE\). However, force is also mass (m) times acceleration (a), and this is written as \(F = ma\). Combining these two formulas, \(qE = ma\). We can then solve for a: \(a = \frac{qE}{m}\). Substituting the known values: Charge of an electron (q) is \(-1.6x10^{-19} C\), electric field strength (E) is \(20,000 N/C\), and the mass of an electron (m) is \(9.11x10^{-31} kg\).
03

Calculate the final speed of the electron

By using the kinematic equation: \(v^2 = vi^2 + 2ad\). The initial velocity \(vi=0\), the acceleration \(a\) is what we solved for in Step 2, and the distance \((d = 0.001 m)\). Solving this equation gives us the final velocity \(v\) which will be positive and denotes the speed of the electron when it hits the positive plates of the capacitor.

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Key Concepts

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

Parallel Plate Capacitor
A parallel plate capacitor is a simple device that stores electric charge and consists of two conductive plates separated by a certain distance. The space between these plates is typically filled with a non-conductive material called a dielectric. The electric field within the capacitor is uniform and directed from the positive plate to the negative plate.

The strength of this electric field is crucial for understanding how particles like electrons behave inside the capacitor. It is given by the formula: \[ E = \frac{V}{d} \]where \( E \) is the electric field strength, \( V \) is the voltage across the plates, and \( d \) is the distance between the plates. This uniform electric field accelerates the electrons as they move from one plate to the other.
Electron Acceleration
When an electron is released inside a parallel plate capacitor, it begins to accelerate due to the electric field present between the plates. This acceleration occurs because the electric field exerts a force on the electron.

The force acting on the electron can be calculated using:
  • Electric Force \( F = qE \)
  • Where \( q \) is the charge of the electron and \( E \) is the electric field strength.
This force leads to an acceleration given by Newton鈥檚 second law:\[ a = \frac{F}{m} = \frac{qE}{m} \]Here, \( m \) stands for the mass of the electron. As the electron moves from the negative to the positive plate, it picks up speed due to this continuous acceleration.
Kinematic Equations
After determining the acceleration of the electron, we use kinematic equations to find its final speed as it reaches the positive plate. Kinematics allows us to analyze motion under constant acceleration.

The specific equation used is:\[ v^2 = v_i^2 + 2ad \]- \( v \) is the final velocity,- \( v_i \) is the initial velocity (zero in this case),- \( a \) is the acceleration calculated earlier,- \( d \) is the distance between the plates.This formula helps us compute the final speed, making it clear how the initial conditions and the acceleration influence the electron's motion.
Electric Force
The concept of electric force is central to understanding how electrons are moved within a parallel plate capacitor. An electric force is the result of the interaction between charged particles and an electric field.

For an electron in a capacitor, the force is calculated as:\[ F = qE \]This force is what drives the electron from the negative plate towards the positive plate. Since the electron has a negative charge, it experiences a force in the direction opposite to the electric field.
The greater the electric field strength, the stronger the force, and consequently, the greater the acceleration of the electron. Understanding this relationship helps in predicting how different changes in the electric field or distance might impact the motion of electrons.

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Most popular questions from this chapter

Two 10 -cm-diameter electrodes \(0.50 \mathrm{cm}\) apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a \(15 \mathrm{V}\) battery. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes a. While the capacitor is attached to the battery? b. After insulating handles are used to pull the electrodes away from each other until they are \(1.0 \mathrm{cm}\) apart? The electrodes remain connected to the battery during this process. c. After the original electrodes (not the modified electrodes of part b) are expanded until they are \(20 \mathrm{cm}\) in diameter while remaining connected to the battery?

Electrodes of area \(A\) are spaced distance \(d\) apart to form a parallel-plate capacitor. The electrodes are charged to \(\pm q\) a. What is the infinitesimal increase in electric potential energy \(d U\) if an infinitesimal amount of charge \(d q\) is moved from the negative electrode to the positive electrode? b. An uncharged capacitor can be charged to \(\pm Q\) by transferring charge \(d q\) over and over and over. Use your answer to part a to show that the potential energy of a capacitor charged to \(\pm Q\) is \(U_{\mathrm{cap}}=\frac{1}{2} Q \Delta V_{\mathrm{c}}\)

What is the escape speed of an electron launched from the surface of a 1.0 -cm-diameter glass sphere that has been charged to \(10 \mathrm{nC} ?\)

A disk with a hole has inner radius \(R_{\text {in }}\) and outer radius \(R_{\text {ous }}\) the disk is uniformly charged with total charge \(Q\). Find an expression for the on-axis electric potential at distance \(z\) from the center of the disk. Verify that your expression has the correct behavior when \(R_{\text {in }} \rightarrow 0\)

Two 2.0 -cm-diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is \(5.0 \times\) \(10^{5} \mathrm{V} / \mathrm{m}\) a. What is the voltage across the capacitor? b. How much charge is on each disk? c. An electron is launched from the negative plate. It strikes the positive plate at a speed of \(2.0 \times 10^{7} \mathrm{m} / \mathrm{s}\). What was the electron's speed as it left the negative plate?
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