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A particle of charge \(+7.5 \mu \mathrm{C}\) is released from rest at the point \(x=60 \mathrm{~cm}\) on an \(x\) axis. The particle begins to move due to the presence of a charge \(Q\) that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved \(40 \mathrm{~cm}\) if (a) \(Q=+20 \mu \mathrm{C}\) and (b) \(Q=-20 \mu \mathrm{C} ?\)

Short Answer

Expert verified
For (a), \(K = \frac{8.99 \times 10^9 \cdot 7.5 \times 10^{-6} \cdot 20 \times 10^{-6}}{0.2} - \frac{8.99 \times 10^9 \cdot 7.5 \times 10^{-6} \cdot 20 \times 10^{-6}}{0.6}\), which simplifies to positive kinetic energy. For (b), switch the sign for \(Q\) and solve, yielding a different value for \(K\) but with opposite sign, indicating attraction and greater speeding up.

Step by step solution

01

Understand the Problem

We need to find the kinetic energy of a charged particle when it moves in an electric field created by another charge. The electric potential energy change will convert into kinetic energy as the particle moves. We'll apply this to two scenarios with different charges at the origin.
02

Determine Initial and Final Positions

The initial position of the particle is at \(x = 60 \text{ cm}\), and it moves until \(x = 20 \text{ cm}\). This means the displacement is \(40 \text{ cm}\), or \(0.4 \text{ m}\).
03

Calculate Initial and Final Electric Potential Energies

The electric potential energy between two charges is given by \(U = \frac{k \, q_1 \, q_2}{r}\), where \(k = 8.99 \times 10^9 \, \text{N}\cdot \text{m}^2/\text{C}^2\). Calculate the initial and final potential energies for both scenarios: \(Q=+20 \mu \text{C}\) and \(Q=-20 \mu \text{C}\).
04

Calculate the Change in Electric Potential Energy

The change in electric potential energy \(\Delta U\) is \(U_{\text{final}} - U_{\text{initial}}\). When \(Q = 20 \mu \text{C}\), this change will indicate conversion from potential to kinetic energy. Calculate \(\Delta U\) for both scenarios.
05

Relate Change in Potential Energy to Kinetic Energy

As the particle starts from rest, the initial kinetic energy is 0. According to conservation of energy, the change in potential energy will be equal to the kinetic energy gained: \(K = \Delta U\). Calculate \(K\) for each scenario.
06

Substituting Values and Finding Kinetic Energy

For the case \(Q = +20 \mu \text{C}\), substitute into the formula: \(\Delta U = \frac{8.99 \times 10^9 \cdot 7.5 \times 10^{-6} \cdot 20 \times 10^{-6}}{0.2} - \frac{8.99 \times 10^9 \cdot 7.5 \times 10^{-6} \cdot 20 \times 10^{-6}}{0.6}\). For \(Q = -20 \mu \text{C}\), switch the sign of \(Q\) and substitute. Simplify to find \(K\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that a particle possesses due to its motion. When a particle starts from rest, its initial kinetic energy is zero. As it begins to move under the influence of forces, such as electric fields, the kinetic energy increases. This increase in kinetic energy is due to the work done on the particle or the conversion of potential energy into kinetic energy. In the exercise, the particle moves due to the electric field created by another charge, and we calculate this kinetic energy to understand how fast the particle is moving.
  • Formula: The kinetic energy (\( K \)) is calculated as \( K = \frac{1}{2} m v^2 \), where \( m \) is the mass of the particle and \( v \) is its velocity.
  • The relationship with potential energy: As the particle moves in the electric field, its electric potential energy decreases, leading to an increase in kinetic energy.
Remember, a particle in motion has energy that is directly proportional to the square of its velocity, which means a small change in speed results in a large change in kinetic energy.
Charge Interaction
Charge interaction is a fundamental concept in electricity where charges exert forces on each other. These interactions are described by Coulomb's Law, which states that the force between two charges is proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This exercise deals with such interactions, where a particle with charge \(+7.5 \mu C\) interacts with a fixed charge \(Q\) located at the origin.
  • Like charges repel: In scenario (a) with \(Q = +20 \mu C\), the two like charges will repel each other.
  • Opposite charges attract: In scenario (b) with \(Q = -20 \mu C\), the charges are opposite and will attract each other.
  • The strength of interaction decreases with distance: The further apart the charges are, the weaker the force and vice versa.
Understanding charge interaction helps us predict the motion and behavior of charges under electric forces, which is crucial in calculating potential and kinetic energies.
Electric Field
An electric field is a region around a charged particle where it exerts a force on other charged particles. The concept of an electric field is critical in explaining how charges interact without direct contact. In this exercise, the electric field generated by the charge \(Q\) influences the motion of the \(+7.5 \mu C\) charge when released.
  • Field direction: The direction of the electric field depends on the sign of the charge creating it. Positive charges create fields that radiate outwards, while negative charges produce fields directed inward.
  • Field strength: The strength of the electric field \(E\) at a point is calculated as \(E = \frac{k \, |Q|}{r^2}\), where \(k\) is Coulomb's constant and \(r\) is the distance from the charge.
  • Impact on motion: The electric field influences the velocity and acceleration of the moving charge, thus affecting its kinetic energy.
Knowing how electric fields operate gives us the ability to predict how a charge will behave when entering a field and how its energy will change.
Conservation of Energy
The conservation of energy principle states that the total energy in an isolated system remains constant over time. This principle applies to the exercise, as the particle moves in an electric field. The system's total energy is shared between kinetic energy and electric potential energy.
  • Initial condition: The particle starts with potential energy and zero kinetic energy since it is released from rest.
  • Energy transformation: As the particle moves, its potential energy decreases while its kinetic energy increases, maintaining a constant total energy in the system.
  • Final condition: The change in potential energy equals the gain in kinetic energy, as energy is conserved. Hence, \(\Delta U = K\).
Understanding the conservation of energy helps us solve problems involving moving charges and predict the outcomes of various interactions within electric fields.

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Most popular questions from this chapter

Identical \(50 \mu \mathrm{C}\) charges are fixed on an \(x\) axis at \(x=\pm 3.0 \mathrm{~m} .\) A particle of charge \(q=-15 \mu \mathrm{C}\) is then released from rest at a point on the positive part of the \(y\) axis. Due to the symmetry of the situation, the particle moves along the \(y\) axis and has kinetic energy \(1.2 \mathrm{~J}\) as it passes through the point \(x=0, y=4.0 \mathrm{~m}\) (a) What is the kinetic energy of the particle as it passes through the origin? (b) At what negative value of \(y\) will the particle momentarily stop?

An infinite nonconducting sheet has a surface charge density \(\sigma=+5.80 \mathrm{pC} / \mathrm{m}^{2}\). (a) How much work is done by the electric field due to the sheet if a particle of charge \(q=+1.60 \times 10^{-19} \mathrm{C}\) is moved from the sheet to a point \(P\) at distance \(d=3.56 \mathrm{~cm}\) from the sheet? (b) If the electric potential \(V\) is defined to be zero on the sheet, what is \(V\) at \(\overline{P ?}\)

A solid conducting sphere of radius \(3.0 \mathrm{~cm}\) has a charge of 30 nC distributed uniformly over its surface. Let \(A\) be a point \(1.0 \mathrm{~cm}\) from the center of the sphere, \(S\) be a point on the surface of the sphere, and \(B\) be a point \(5.0 \mathrm{~cm}\) from the center of the sphere. What are the electric potential differences (a) \(V_{S}-V_{B}\) and (b) \(V_{A}-V_{B} ?\)

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by \(-1.0 \mathrm{~V}\) during one revolution. Assuming the shuttle is a sphere of radius \(10 \mathrm{~m}\), estimate the amount of charge it collects.

Here are the charges and coordinates of two charged particles located in an \(x y\) plane: \(q_{1}=+3.00 \times 10^{-6} \mathrm{C}, x=+3.50 \mathrm{~cm}\) \(y=+0.500 \mathrm{~cm}\) and \(q_{2}=-4.00 \times 10^{-6} \mathrm{C}, x=-2.00 \mathrm{~cm}, y=+1.50 \mathrm{~cm}\) How much work must be done to locate these charges at their given positions, starting from infinite separation?

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