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(I) How much kinetic energy will an electron gain (in joules and eV) if it accelerates through a potential difference of 18,500 V?

Short Answer

Expert verified
The electron gains 2.9637 × 10^-15 Joules and 18,500 eV.

Step by step solution

01

Understanding the Relationship

When a charged particle like an electron accelerates through a potential difference, it gains kinetic energy. This energy is equal to the charge of the electron multiplied by the potential difference it moves through. The formula is given by \[ \text{Kinetic Energy} (E_k) = q \times V \]where \( q \) is the charge of the electron and \( V \) is the potential difference.
02

Electron Charge Value

The charge of an electron is a constant, \( q = -1.602 \times 10^{-19} \) Coulombs. Since we are calculating energy, we will use the magnitude of the charge.
03

Calculate Kinetic Energy in Joules

Using the formula from Step 1:\[ E_k = q \times V \]Substitute the known values:\[ E_k = 1.602 \times 10^{-19} \text{ C} \times 18,500 \text{ V} \]Calculate the result:\[ E_k = 2.9637 \times 10^{-15} \text{ Joules} \]
04

Conversion to Electron Volts

Since 1 electron volt (eV) is defined as \( 1.602 \times 10^{-19} \) Joules, we can calculate the energy in electron volts by dividing the kinetic energy in Joules by this conversion factor:\[ E_k (eV) = \frac{2.9637 \times 10^{-15} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \]Calculate the result:\[ E_k = 18,500 \text{ eV} \]

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Key Concepts

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

Electron Charge
Electrons are one of the fundamental particles in physics, carrying a negative charge. This charge is crucial in determining how electrons behave in electric fields. The charge of an electron is constant and known to be
  • \( q = -1.602 \times 10^{-19} \) Coulombs
When calculating quantities like energy, we focus on the magnitude, disregarding the negative sign. This charge is what makes electrons interact with their surroundings, most notably with electric fields, which is essential for kinetic energy calculations. Understanding electron charge is essential because it defines the electron's ability to gain energy while moving through such fields.
In many physics problems, including calculating kinetic energy, the electron charge's tiny value highlights how we often deal with minuscule energy quantities, especially at atomic and subatomic levels. Despite its small size, this charge plays a massive role in electronic configurations and chemical reactions.
Potential Difference
Potential difference, often referred to as voltage, is a central concept in electricity and electronics. It is the difference in electric potential energy between two points. It is measured in volts (V). When we talk about an electron accelerating through a potential difference, we are referring to it gaining energy as it moves from a point of higher potential to one of lower potential.
This process is analogous to a ball rolling down a hill, where the change in height represents the potential difference. The work done on the electron as it moves through this electric field results in a gain in kinetic energy.
  • The formula to understand this gain in kinetic energy is \( E_k = q \times V \)
  • Where \( E_k \) is the kinetic energy, \( q \) is the charge of the electron, and \( V \) is the potential difference.
A significant voltage, such as 18,500 V, imparts a relatively large amount of kinetic energy to an electron, a critical concept when analyzing the behavior of particles in electric fields like those found in particle accelerators or x-ray tubes.
Energy Conversion
Energy conversion in this context refers to the transformation of electrical potential energy into kinetic energy. When an electron moves through a potential difference, it converts potential energy derived from the electric field into kinetic energy. This kinetic energy is what allows the electron to accelerate.
The conversion is described mathematically as
  • \( E_k (Joules) = q \times V \)
Once we calculate the kinetic energy in joules, we can further convert it to electron volts (eV) for convenience in specific fields like quantum mechanics or high-energy physics.
One electron volt is the energy gained by an electron when it moves through a potential difference of 1 V, equated as
  • \( 1 \, eV = 1.602 \times 10^{-19} \, Joules \)
Thus, by dividing the kinetic energy in joules by this conversion factor, we can express that energy in electron volts. This conversion is valuable because eV is a more fitting unit to express energies at the atomic scale, making it easier for scientists to talk about particle physics and electronic processes.

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Most popular questions from this chapter

A +3.5\(\mu\)C charge is 23 cm to the right of a -7.2\(\mu\)C charge. At the midpoint between the two charges, (\(a\)) determine the potential and (\(b\)) the electric field.

(I) An electric field of 525 V/m is desired between two parallel plates 11.0 mm apart. How large a voltage should be applied?

A parallel-plate capacitor with plate area \(A =\) 2.0 m\(^2\) and plate separation \(d =\) 3.0 mm is connected to a 35-V battery (Fig. 17-51a). (\(a\)) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (\(b\)) With the capacitor still connected to the battery, a slab of plastic with dielectric strength \(K =\) 3.2 is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric (Fig. 17-51b).What are the new values of charge, electric field, capacitance, and the energy stored in the capacitor?

A parallel-plate capacitor with plate area 3.0 cm\(^2\) and airgap separation 0.50 mm is connected to a 12-V battery, and fully charged. The battery is then disconnected. (\(a\)) What is the charge on the capacitor? (\(b\)) The plates are now pulled to a separation of 0.75 mm. What is the charge on the capacitor now? (\(c\)) What is the potential difference between the plates now? (\(d\)) How much work was required to pull the plates to their new separation?

(II) How strong is the electric field between the plates of a 0.80-\(\mu\)F air- gap capacitor if they are 2.0 mm apart and each has a charge of 62\(\mu\)C

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