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Equipotential surface \(A\) has a potential of \(5650 \mathrm{V},\) while equipotential surface \(B\) has a potential of 7850 \(\mathrm{V}\) . 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 \(B ?\)

Short Answer

Expert verified
The work done by the outside force is 0.213 J.

Step by step solution

01

Identify the Given Values

We are given the following information: \( V_A = 5650 \mathrm{V} \), \( V_B = 7850 \mathrm{V} \), mass of the particle \( m = 5.00 \times 10^{-2} \mathrm{kg} \), charge \( q = +4.00 \times 10^{-5}\mathrm{C} \), initial speed \( v_A = 2.00 \mathrm{m/s} \), and final speed \( v_B = 3.00 \mathrm{m/s} \).
02

Calculate the Change in Electric Potential Energy

The change in electric potential energy \( \Delta U \) is given by \( \Delta U = q \cdot (V_B - V_A) \). Substitute to calculate: \( \Delta U = 4.00 \times 10^{-5}\mathrm{C} \times (7850 \mathrm{V} - 5650 \mathrm{V}) = 4.00 \times 10^{-5} \mathrm{C} \times 2200 \mathrm{V} = 0.088 \mathrm{J} \).
03

Calculate the Change in Kinetic Energy

The change in kinetic energy \( \Delta KE \) is given by \( \Delta KE = \frac{1}{2}m(v_B^2 - v_A^2) \). Substitute to calculate: \( \Delta KE = \frac{1}{2}(5.00 \times 10^{-2} \mathrm{kg})(3.00^2 - 2.00^2) \mathrm{m}^2/\mathrm{s}^2 \). Simplify the velocity squared: \(\Delta KE = \frac{1}{2}(5.00 \times 10^{-2})(9 - 4) = \frac{1}{2}(5.00 \times 10^{-2}) \times 5 = 0.125 \mathrm{J} \).
04

Use Work-Energy Principle

The work-energy principle states that the work done by nonconservative forces (\( W_{nc} \)) equals the change in kinetic energy plus the change in electric potential energy: \( W_{nc} = \Delta KE + \Delta U \). Substitute the values from previous steps: \( W_{nc} = 0.125 \mathrm{J} + 0.088 \mathrm{J} = 0.213 \mathrm{J} \).

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Key Concepts

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

Equipotential Surface
Equipotential surfaces are imaginary planes where every point is at the same electric potential. This means that a charged particle would not experience any work done by the electric force moving between points on the same equipotential surface.
In the context of the exercise, the equipotential surfaces are labeled as surface A and surface B, with potentials of 5650 V and 7850 V, respectively.
When a particle moves between different equipotential surfaces, there's a change in electric potential energy due to the difference in potential values.
The difference in potentials from A to B ( 7850 ext{ V} - 5650 ext{ V} ) signifies that the particle either gains or loses electric potential energy, depending on its charge and the potential difference. Things to remember about equipotential surfaces:
  • No work is required to move a charge along an equipotential surface.
  • Work is only done when a charge moves between surfaces of different potentials.
  • Equipotential surfaces are always perpendicular to electric field lines.
Work-Energy Principle
The work-energy principle is a cornerstone of physics and explains the relationship between work done, potential energy, and kinetic energy. It states that the work done by all forces acting on an object is equal to the change in the object's kinetic energy.
In terms of electric potential energy, this principle can be expanded to include work done by nonconservative forces: \[ W_{\text{nc}} = \Delta KE + \Delta U \]where \( W_{\text{nc}} \) is the work done by nonconservative forces, \( \Delta KE \) is the change in kinetic energy, and \( \Delta U \) is the change in potential energy.
In the provided exercise, the work-energy principle helps calculate the work done by an external force that moves the particle from one equipotential surface to another, determining how it results in changes to the particle’s energy states (kinetic and potential).
Here's how it influences problem-solving:
  • It allows us to link energy changes to physical movements.
  • Helps us calculate non-obvious forces, like external applied forces, acting on the system.
  • Reassures that energy is conserved in any closed system, even if it converts between forms.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is a fundamental part of the work-energy principle. The kinetic energy ( KE ) of a particle is defined as:\[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the particle and \( v \) is its velocity.
In the context of the exercise, kinetic energy is crucial as it illustrates the change in energy as the particle's speed increases from 2.00 m/s to 3.00 m/s during its movement from equipotential surface A to B.
The change in kinetic energy, usually represented by \( \Delta KE \), is calculated as:\[ \Delta KE = \frac{1}{2}m(v_B^2 - v_A^2) \]where \( v_B \) and \( v_A \) are the final and initial velocities, respectively.
Kinetic energy is always non-negative and increases with an increase in speed. This factor is why it's instrumental to consider changes in kinetic energy while applying the work-energy principle.Key insights about kinetic energy in context:
  • Kinetic energy increases with the square of velocity—faster objects possess more kinetic energy.
  • Mass directly affects kinetic energy; a heavier object at the same speed as a lighter one will have more kinetic energy.
  • A change in kinetic energy is a direct indicator of work being done on or by the object.

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Most popular questions from this chapter

During a particular thunderstorm, the electric potential difference between a cloud and the ground is \(V_{\text { clound }}-V_{\text { ground }}=1.3 \times 10^{8} \mathrm{V},\) with the cloud being at the higher potential. What is the change in an electron's elec- tric potential energy when the electron moves from the ground to the cloud?

A capacitor is constructed of two concentric conducting cylindrical shells. The radius of the inner cylindrical shell is \(2.35 \times 10^{-3} \mathrm{m},\) and that of the outer shell is 2.50 \(\times 10^{-3} \mathrm{m}\) . When the cylinders carry equal and opposite charges of magnitude 1.7 \(\times 10^{-10} \mathrm{C}\) , the electric field between the plates has an average magnitude of \(4.2 \times 10^{4} \mathrm{V} / \mathrm{m}\) and is directed radially outward from the inner shell to the outer shell. Determine (a) the magnitude of the potential difference between the cylindrical shells and \((\text { b) the capacitance of this capacitor. }\)

The same voltage is applied between the plates of two different capacitors. When used with capacitor A, this voltage causes the capacitor to store 11\(\mu \mathrm{C}\) of charge and \(5.0 \times 10^{-5} \mathrm{J}\) of energy. When used with capacitor \(\mathrm{B}\) , which has a capacitance of 6.7\(\mu \mathrm{F}\) , this voltage causes the capacitor to store a charge that has a magnitude of \(q_{\mathrm{B}}\) Determine \(q_{\mathrm{B}}\) .

Multiple-Concept Example 4 to see the concepts that are pertinent here. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 25000 \(\mathrm{V}\) . The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relativity must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

Four identical charges \((+2.0 \mu \mathrm{C} \text { each) are brought from }\) infinity and fixed to a straight line. The charges are located 0.40 \(\mathrm{m}\) apart. Determine the electric potential energy of this group.

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