Problem 9 - Physics

Question

The potential at location \(A\) is \(452 \mathrm{~V}\). A positively charged particle is released there from rest and arrives at location \(B\) with a speed \(v_{B}\). The potential at location \(C\) is \(791 \mathrm{~V},\) and when released from rest from this spot, the particle arrives at \(B\) with twice the speed it previously had, or \(2 v_{B} .\) Find the potential at \(B\).

Step-by-Step Solution

Answer

The potential at location B is 339 V.

Step 1: Understand the Energy Equation

When a particle is released from rest, the change in its kinetic energy can be equated to the change in its electrical potential energy. The relationship can be written as: \[ \frac{1}{2}mv_B^2 = q(V_A - V_B) \]for case 1, and further, \[ \frac{1}{2}m(2v_B)^2 = q(V_C - V_B) \]for case 2.

Step 2: Use the Relationship Between Kinetic Energy Equations

From the given, in the second case, the speed is twice that of the first case, or \(2v_B\). Plugging it in, the second equation becomes:\[ \frac{1}{2}m(2v_B)^2 = q(V_C - V_B) \] which simplifies to \[ 2mv_B^2 = q(V_C - V_B) \].

Step 3: Set Up Two Linear Equations

Given:1. \( \frac{1}{2}mv_B^2 = q(V_A - V_B) \)2. \( 2mv_B^2 = q(V_C - V_B) \)Divide the second equation by the first to eliminate \(mv_B^2\) and solve for \(V_B\).

Step 4: Solve the Combined Relationship

Combining the equations,\[ \frac{2mv_B^2}{\frac{1}{2}mv_B^2} = \frac{q(V_C - V_B)}{q(V_A - V_B)} \]which simplifies to:\[ 4 = \frac{V_C - V_B}{V_A - V_B} \].This can be rewritten as:\[ 4(V_A - V_B) = V_C - V_B \].

Step 5: Solve for the Potential at B

Substitute given potentials:\[ 4(452 - V_B) = 791 - V_B \].Simplify and solve for \(V_B\).

Step 6: Calculate the Final Expression

Distribute and simplify the equation:\[ 1808 - 4V_B = 791 - V_B \]\[ 1808 - 791 = 4V_B - V_B \]\[ 1017 = 3V_B \]Therefore, \[ V_B = \frac{1017}{3} = 339 \mathrm{~V} \].

Key Concepts

Kinetic Energy

Kinetic energy is the energy of motion. When an object is moving, it possesses kinetic energy. The expression for kinetic energy (\( KE \) ) of an object is given by:

• \( KE = \frac{1}{2} mv^2 \)

where:

• \( m \) is the mass of the object
• \( v \) is the velocity or speed of the object

Understanding kinetic energy is crucial when dealing with moving particles in physics.
In the context of charged particles, when a particle is released and accelerates due to a potential difference, its kinetic energy changes.
This change can be precisely calculated when you know the initial and final speeds of the particle.

Potential Energy

Potential energy refers to the stored energy in a system due to its position or configuration. For charged particles, this energy is related to their position in an electric field.
The electric potential energy (\( U \) ) of a charged particle is defined as:

• \( U = qV \)

where:

• \( q \) is the charge of the particle
• \( V \) is the electric potential at that point

When a charged particle moves through different potentials, its potential energy changes.
This change can transform into kinetic energy, causing the particle to speed up or slow down. The interplay of kinetic and potential energy is a core concept in understanding the behavior of particles in electric fields.

Charged Particles

Charged particles, such as electrons or protons, experience forces when placed in electric fields.
Their behavior under these forces is a fundamental part of electromagnetism. When exploring electric potential, charged particles tend to move from regions of high potential to low potential, spontaneously decreasing potential energy. This movement is harnessed in many applications, from simple electrical circuits to complex devices.
In problems related to potential and kinetic energy, recognizing the role of charge helps predict how a particle will move and at what speeds it will arrive at certain points.

Energy Conservation

Energy conservation is a critical principle in physics stating that energy in a closed system remains constant.
In terms of mechanics, this principle implies that the sum of kinetic and potential energy remains constant if only conservative forces are acting. When a charged particle moves through different potentials, it undergoes transformations between potential and kinetic energy. For instance, when the charged particle in our problem starts at rest and speeds up, its potential energy is converted to kinetic energy.
This implies that:

• The initial potential energy = final kinetic energy at a particular point.

By applying energy conservation, you can derive relationships to solve for unknowns, such as the specific electric potential conditions required for particular kinetic energy values.