Problem 41 - Fundamentals of Physics

Question

An electron is shot directly toward the center of a large metal plate that has surface charge density \(-2.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) If the initial kinetic energy of the electron is \(1.60 \times 10^{-17} \mathrm{~J}\) and if the electron is to stop (due to electrostatic repulsion from the plate) just as it reaches the plate, how far from the plate must the launch point be?

Step-by-Step Solution

Answer

The electron must be launched from a distance of approximately 0.088 m from the plate.

Step 1: Understand the Problem

An electron is moving toward a charged plate and it must stop just before reaching the plate. This means that the kinetic energy of the electron will be converted to electric potential energy as it gets closer to the plate.

Step 2: Identify the Formulas

The electric potential energy due to a charged plate can be calculated using:\[ U = qE \cdot d \]Where:- \( U \) is the electric potential energy,- \( q \) is the charge of the electron \((-1.6 \times 10^{-19} \text{ C})\),- \( E \) is the electric field due to the plate, and- \( d \) is the distance from the plate.The electric field \( E \) due to the plate with uniform charge density \( \sigma \) is given by: \[ E = \frac{\sigma}{2\varepsilon_0} \]where \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot \text{m}^2)\).

Step 3: Relate Kinetic and Potential Energy

The initial kinetic energy \( K \) of the electron will be fully converted into electric potential energy when it stops:\[ K = U \]Given \( K = 1.60 \times 10^{-17} \text{ J} \), we can equate this to the expression for potential energy:\[ K = qE \cdot d \]

Step 4: Calculate the Electric Field

Using the formula for the electric field:\[ E = \frac{\sigma}{2\varepsilon_0} \]Substituting \( \sigma = -2.0 \times 10^{-6} \text{ C/m}^2 \) and \( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot \text{m}^2 \):\[ E = \frac{-2.0 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = -1.13 \times 10^5 \text{ N/C} \]

Step 5: Solve for the Distance

Substituting the known values into the equation \( K = qE \cdot d \):\[ 1.60 \times 10^{-17} = (1.6 \times 10^{-19}) \cdot (-1.13 \times 10^5) \cdot d \]Solving for \( d \):\[ d = \frac{1.60 \times 10^{-17}}{(1.6 \times 10^{-19}) \cdot (-1.13 \times 10^5)} \]\[ d \approx 0.088 \text{ m} \] (The negative sign indicates direction, and distance in physics is a scalar, so we use the absolute value).

Key Concepts

Understanding Electric Fields

Electric fields are a crucial concept in electrostatics. They describe how charged particles interact with each other across space. Imagine an invisible field surrounding every charged object. This field exerts a force on other charged particles that come within its influence. In our scenario with an electron approaching a charge plate, the electric field is created by the charged plate itself.

• The formula for the electric field (\( E \)) created by a large, flat plate with uniform charge density (\( \sigma \)) is given by:

\[E = \frac{\sigma}{2\varepsilon_0}\]where \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).In simpler terms, this formula tells us the strength of the electric field produced by the charged plate. The direction of the electric field depends on the sign of the charge. For a negatively charged plate, like in our problem, the field points towards the plate.

Exploring Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. For any moving object, including an electron, kinetic energy is expressed as:\[K = \frac{1}{2}mv^2\]where \(m\) is the mass of the object and \(v\) is its velocity. However, in electrostatics problems, we often directly deal with the given kinetic energy, especially when forces act over distances.

• In the exercise, the electron starts with a kinetic energy of \(1.60 \times 10^{-17} \text{ J}\).
• As the electron moves closer to the charged plate, it loses kinetic energy because the electrostatic repulsive force does work on it.
• When the electron stops just before reaching the plate, all of its initial kinetic energy has converted into electric potential energy.

This interplay between kinetic and potential energy is a quintessential example of the conservation of energy in a system.

Diving into Electric Potential Energy

Electric potential energy relates to the position of a charged particle within an electric field. When a charged particle moves within this field, its potential energy changes depending on its position relative to other charges.
For a particle with charge \( q \) in an electric field \( E \), the electric potential energy \( U \) is given by:\[U = qE \cdot d\]where \( d \) is the distance moved within the field.

• In our exercise, as the electron approaches the plate, it gains potential energy.
• The electron's initial kinetic energy becomes the maximum potential energy when it stops.
• By setting the initial kinetic energy equal to the final potential energy, we determine how far the electron can travel before stopping.

This relationship, often referred to as work-energy principle, helps us understand how energy is conserved and transformed as charged particles move within an electric field.