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An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of \(1.2 \mathrm{cm},\) and the electric field within the capacitor has a magnitude of 2.1 \(\times 10^{6} \mathrm{V} / \mathrm{m}\). What is the kinetic energy of the electron just as it reaches the positive plate?

Short Answer

Expert verified
The kinetic energy of the electron is approximately \(4.032 \times 10^{-15}\) J.

Step by step solution

01

Understanding the Problem

The electron starts from rest at the negative plate of a capacitor and is accelerated by an electric field to the positive plate. We need to find the kinetic energy of the electron at the positive plate.
02

Identifying the Variables

The distance between the plates is given as 1.2 cm (which we convert to meters as 0.012 m), and the electric field magnitude is 2.1 \(\times 10^6\) V/m.
03

Electric Potential Difference (Voltage)

The potential difference \(V\) across the plates can be calculated using \(V = E \cdot d\), where \(E\) is the electric field and \(d\) is the distance between plates. So \(V = 2.1 \times 10^6\, \text{V/m} \times 0.012\, \text{m} = 25200\, \text{V}\).
04

Using Energy Conservation Principle

The work done by the electric field on the electron is equal to the change in kinetic energy. This work done (or electrical potential energy conversion) is given by \(q \cdot V\), where \(q\) is the charge of the electron \((-1.60 \times 10^{-19}\, C)\).
05

Calculating the Kinetic Energy

Since the electron starts from rest, all its potential energy will convert into kinetic energy. Thus, the kinetic energy \(KE\) of the electron is \(KE = q \cdot V = (-1.60 \times 10^{-19}\, C) \times 25200\, V = -4.032 \times 10^{-15}\, J\). The negative sign indicates the direction of force and is usually omitted when considering energy values. Therefore, \(KE = 4.032 \times 10^{-15}\, J\).

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Key Concepts

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

Electron Kinetic Energy
The kinetic energy of an electron is crucial in understanding how electrons move in electric fields. When an electron is released from rest in an electric field, it gets accelerated due to the force exerted by the field. This acceleration causes the electron to gain kinetic energy.
Here's how this process works:- An electron in motion possesses kinetic energy, which can be calculated using the formula: \[ KE = \frac{1}{2} m v^2 \] where \(m\) is the mass and \(v\) is the velocity of the electron.
- If we consider the energy conservation principle, all potential energy that the electron initially had converts into kinetic energy as it moves between two plates of a capacitor.
In the provided problem, the potential energy gained by the electron from the electric field is completely transformed into kinetic energy by the time the electron reaches the positive plate.
Parallel Plate Capacitor
The parallel plate capacitor is a fundamental component in understanding electric fields and potentials. It consists of two conductive plates separated by a distance, creating a uniform electric field between them.
Key features of a parallel plate capacitor include:
  • Uniform Electric Field: The electric field between the plates is uniform, meaning it has a constant magnitude and direction, ideal for calculating potential differences and energy transfers.
  • Distance between Plates: This distance directly affects the potential difference and electric field strength. In this problem, the distance is 1.2 cm, which is crucial for determining the capacitor's characteristics.
  • Capacity to Store Electric Energy: Capacitors store energy in the electric field between the plates, which is later transferred to other components in an electric circuit.
In this exercise, the electric field accelerates the electron from rest, converting its potential energy to kinetic as it travels across the plates.
Potential Difference (Voltage)
Potential difference, often referred to as voltage, is a measure of electric potential energy per charge between two points. In capacitors, it represents the work needed to move a charge from one plate to another.
Important aspects of potential difference in capacitors:- Calculating Voltage: The formula used is \( V = E \cdot d \), where \(E\) is the electric field, and \(d\) is the distance between plates. It helps quantify the energy difference that drives electrons across the capacitor.
- In the given exercise, the voltage across the plates is 25,200 V, calculated by multiplying the electric field magnitude by the separation distance. This substantial voltage is what allows the electron to gain kinetic energy rapidly.
- Voltage's Role in Kinetic Energy: Voltage indicates the potential energy available for conversion into kinetic energy as electrons accelerate across the field.
Electric Potential Energy
Electric potential energy is the energy stored by an object due to its position in an electric field. For an electron in a capacitor, this energy is critical to understanding how it accelerates and gains velocity.
Consider the following about electric potential energy:
  • Charge and Energy Relation: The potential energy is proportional to the charge and the potential difference, calculated as \( U = q \cdot V \), where \( q \) is the electron's charge and \( V \) is the voltage.
  • Energy Conversion: When an electron is in an electric field, its potential energy converts to kinetic energy as it moves. This conversion is complete when the electron reaches the other plate.
  • Practical Implications: Understanding this energy conversion is vital for working with electronic components, where energy transformation dictates their function.
In electric circuits, and particularly in capacitors, this energy transformation underlies the behavior and movement of electrons, making it a core principle in electronics.

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Most popular questions from this chapter

A positive point charge \(\left(q=+7.2 \times 10^{-8} \mathrm{C}\right)\) is surrounded by an equipotential surface \(A,\) which has a radius of \(r_{A}=1.8 \mathrm{m} .\) A positive test charge \(\left(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\right)\) moves from surface \(A\) to another equipotential surface \(B\), which has a radius \(r_{B}\). The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9}\) J. Find \(r_{B}\).

An electric field has a constant value of \(4.0 \times 10^{3} \mathrm{V} / \mathrm{m}\) and is directed downward. The field is the same everywhere. The potential at a point \(P\) within this region is 155 V. Find the potential at the following points: (a) \(6.0 \times 10^{-3} \mathrm{m}\) directly above \(P\). (b) \(3.0 \times 10^{-3} \mathrm{m}\) directly below \(P,\) (c) \(8.0 \times 10^{-3} \mathrm{m}\) directly to the right of \(P\).

A particle is uncharged and is thrown vertically upward from ground level with a speed of \(25.0 \mathrm{m} / \mathrm{s}\). As a result, it attains a maximum height \(h\). The particle is then given a positive charge \(+q\) and reaches the same maximum height \(h\) when thrown vertically upward with a speed of \(30.0 \mathrm{m} / \mathrm{s}\). The electric potential at the height \(h\) exceeds the electric potential at ground level. Finally, the particle is given a negative charge \(-q .\) Ignoring air resistance, determine the speed with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the maximum height h. In all three situations, be sure to include the effect of gravity.

Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 V. A particle has a mass of \(5.00 \times 10^{-2} \mathrm{kg}\) and a charge of \(+4.00 \times 10^{-5} \mathrm{C} .\) The particle has a speed of \(2.00 \mathrm{m} / \mathrm{s}\) on surface \(A .\) A nonconservative outside force is applied to the particle, and it moves to surface \(B\), arriving there with a speed of \(3.00 \mathrm{m} / \mathrm{s}\). How much work is done by the outside force in moving the particle from \(A\) to \(B ?\)

Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.070 V. How much work is done by the electric force when a sodium ion (charge \(=+e\) ) moves from the outside to the inside?

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