Chapter 18: Problem 49
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is \(\sigma=1.8 \times\) \(10^{-7} \mathrm{C} / \mathrm{m}^{2},\) and the plate separation is \(1.5 \times 10^{-2} \mathrm{m} .\) How fast is the electron moving just before it reaches the positive plate?
Short Answer
Step by step solution
Understand Charge Distribution and Electric Field in a Parallel Plate Capacitor
Calculate the Electric Field between the Plates
Determine the Electron's Potential Energy Change
Calculate the Change in Kinetic Energy
Find the Electron's Final Speed
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Electron Motion
Parallel Plate Capacitor
- Surface Charge Density (\( \sigma \)): This is the amount of charge per unit area on each plate.
- Electric Field (\( E \)): The field is uniform between the plates and its strength can be determined by the formula \( E = \frac{\sigma}{\varepsilon_0} \), where \( \varepsilon_0 \) is the permittivity of free space.
- Plate Separation: The distance between the plates influences the capacitor's capacitance and the electric field strength.
Kinetic Energy
Electric Potential Energy
- The potential energy \( \Delta U \) decreases.
- It converts into kinetic energy, propelling the electron faster as it approaches the positive plate.