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The potential at location \(A\) is 452 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 \(B\).

Short Answer

Expert verified
The potential at B is 113 V.

Step by step solution

01

Understanding the Problem

We have a positively charged particle moving between points A, B, and C with given potentials at A and C. We know the speed at B from both A and C scenarios, using potential difference and conservation of energy. We need to find the potential at B.
02

Apply Energy Conservation from A to B

Use the conservation of energy principle, where the initial energy at A equals the final energy at B. The change in potential energy is equal to the change in kinetic energy. The equation is \( q(V_A - V_B) = \frac{1}{2}mv_B^2 \), where \( q \) is the charge and \( m \) is the mass of the particle.
03

Express Speed from A to B in Terms of Potential Difference

We rearrange the equation from the previous step to express speed: \( v_B = \sqrt{\frac{2q(V_A - V_B)}{m}} \). Use this equation for the given potential at A.
04

Apply Energy Conservation from C to B with Modified Speed

Similarly, use the conservation of energy principle from C to B, with speed doubled: \( q(V_C - V_B) = \frac{1}{2}m(2v_B)^2 \). Simplify to \( q(V_C - V_B) = 2mv_B^2 \).
05

Express Speed from C to B in Terms of Potential Difference

Rearrange the equation for speed in terms of potential difference when released at C: \( 2v_B^2 = \frac{2q(V_C - V_B)}{m} \). Then, equate the two expressions for \( v_B^2 \).
06

Solve for the Potential at B

Set the expressions for \( v_B^2 \) equal from both scenarios and solve:\[ \frac{q(V_A - V_B)}{m} = \frac{q(V_C - V_B)}{2m} \].This simplifies to:\[ V_A - V_B = \frac{V_C - V_B}{2} \]Rearrange to solve for \( V_B \).
07

Substitute Given Values and Calculate

Plug in the values given: \( V_A = 452 \) V and \( V_C = 791 \) V into the equation:\[ 452 - V_B = \frac{791 - V_B}{2} \]Multiply every term by 2 and solve for \( V_B \):\[ 2(452 - V_B) = 791 - V_B \]\[ 904 - 2V_B = 791 - V_B \].Rearrange and solve for \( V_B \).
08

Final Calculation

Subtract 791 from both sides:\[ 904 - 791 = 2V_B - V_B \]\[ 113 = V_B \].Thus, the potential at B is 113 V.

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Key Concepts

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

Energy Conservation
In physics, energy conservation is a fundamental principle stating that energy cannot be created or destroyed; it can only change forms. When dealing with a charged particle moving in an electric field, energy conservation can be expressed through the transformation between potential energy and kinetic energy.
This principle is critical in solving problems that involve particle motion between points with different electric potentials. It assures us that the total energy of the system remains constant.
  • Initial potential energy at a starting point (due to electric potential) converts to kinetic energy as the particle moves.
  • In our exercise, energy conservation allows us to equate the initial potential energy difference between points A and B (or C and B) with the particle's final kinetic energy at point B.
This is summarized in equations that connect the electric potentials at different points to the speed of the particle, revealing the relationship between electric potential and particle motion.
Potential Difference
Potential difference, often referred to as voltage, is the difference in electric potential between two points in a field. It plays a significant role in determining how much work is done when a charged particle moves between these points.
In the problem, the potential differences from A to B and C to B influence the speed of a charged particle. This difference dictates how much energy is available to convert into kinetic energy, as the particle moves through the electric field.
  • The potential difference between points A and B is used to calculate the kinetic energy gained by the particle as it moves to B.
  • Similarly, the potential difference from C to B influences the energy transformation, resulting in different speeds.
Understanding potential difference helps explain why the particle manages differing speeds when released from rest at different potentials.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. For a charged particle, kinetic energy is gained as it accelerates between points of differing electric potentials.
The concept here is that the potential energy difference is directly converted into kinetic energy, allowing us to use energy conservation equations. In our exercise, the kinetic energy at point B can be determined from the particle's speed and mass upon arriving there from both A or C.
  • Kinetic energy is mathematically expressed as \( KE = \frac{1}{2} mv^2 \). Here, \( m \) is the mass and \( v \) is the speed of the particle.
  • In the problem, as the particle moves from A to B or C to B, the change in speed can be expressed in terms of the potential energy differences.
The variance in speed is a direct manifestation of how potential difference modifies kinetic energy as per energy conservation principles.
Charged Particle Motion
The motion of a charged particle in an electric field is determined by several factors, including charge, mass, and the electric potentials of the points it moves between.
In the given problem, understanding the fundamentals of charged particle motion allows us to predict how the particle will behave as it moves from A to B or C to B. This motion is typically influenced by the electric forces acting upon the particle, governed by the potential differences provided.
  • The charge of the particle interacts with the electric field, causing acceleration towards regions of lower potential energy.
  • As the particle moves and the potential energy changes, it gains or loses kinetic energy, reflecting changes in speed.
By understanding these principles, we can apply equations to model and calculate the potential and kinetic states of the charged particle as it travels, using the information about initial and final speeds as well as known potentials.

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Most popular questions from this chapter

The membrane that surrounds a certain type of living cell has a surface area of \(5.0 \times 10^{-9} \mathrm{m}^{2}\) and a thickness of \(1.0 \times 10^{-8} \mathrm{m} .\) Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of \(5.0 .\) (a) The potential on the outer surface of the membrane is \(+60.0 \mathrm{mV}\) greater than that on the inside surface. How much charge resides on the outer surface? (b) If the charge in part (a) is due to positive ions (charge \(+e\) ), how many such ions are present on the outer surface?

The electronic flash attachment for a camera contains a capacitor for storing the energy used to produce the flash. In one such unit, the potential difference between the plates of an \(850-\mu \mathrm{F}\) capacitor is \(280 \mathrm{V}\). (a) Determine the energy that is used to produce the flash in this unit. (b) Assuming that the flash lasts for \(3.9 \times 10^{-3}\) s, find the effective power or "wattage" of the flash.

Identical \(+1.8 \mu \mathrm{C}\) charges are fixed to adjacent corners of a square. What charge (magnitude and algebraic sign) should be fixed to one of the empty corners, so that the total electric potential at the remaining empty corner is \(0 \mathrm{V} ?\)

Two particles each have a mass of \(6.0 \times 10^{-3} \mathrm{kg}\). One has a charge of \(+5.0 \times 10^{-6} \mathrm{C},\) and the other has a charge of \(-5.0 \times 10^{-6} \mathrm{C} .\) They are initially held at rest at a distance of \(0.80 \mathrm{m}\) apart. Both are then released and accelerate toward each other. How fast is each particle moving when the separation between them is one-third its initial value?

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 \(B ?\)

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