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Chapter 19: Problem 64

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 \(B ?\)

Short Answer

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

Step by step solution

01

Calculate the Change in Kinetic Energy

The change in kinetic energy, \( \Delta K \), is calculated using the formula: \( \Delta K = \frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2 \) where \( m = 5.00 \times 10^{-2} \) kg is the mass, \( v_f = 3.00 \) m/s is the final speed, and \( v_i = 2.00 \) m/s is the initial speed. Substituting these values, we get: \( \Delta K = \frac{1}{2} (5.00 \times 10^{-2}) (3.00)^2 - \frac{1}{2} (5.00 \times 10^{-2}) (2.00)^2 \) \( \Delta K = 0.5 \times 5.00 \times 10^{-2} \times 9 - 0.5 \times 5.00 \times 10^{-2} \times 4 \) \( \Delta K = 0.225 - 0.1 = 0.125 \) J.
02

Calculate the Change in Electric Potential Energy

The change in electric potential energy, \( \Delta U \), is given by: \( \Delta U = q \Delta V \) where \( q = 4.00 \times 10^{-5} \) C is the charge, and \( \Delta V = V_B - V_A = 7850 - 5650 \) V is the change in potential difference. Substituting these values, we get: \( \Delta U = 4.00 \times 10^{-5} \times 2200 \) \( \Delta U = 0.088 \) J.
03

Calculate the Work Done by the Outside Force

The work done by the outside force, \( W_{external} \), is given by the energy principle: \( W_{external} = \Delta K + \Delta U \) Substituting in the calculated values, we find: \( W_{external} = 0.125 + 0.088 \) \( W_{external} = 0.213 \) J.

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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 an object possesses due to its motion. It's a fundamental concept in physics that helps us understand how objects accelerate and move through space. The formula for kinetic energy is expressed as \( KE = \frac{1}{2} mv^2 \), where \(m\) represents mass and \(v\) is velocity.

In our context, when a particle moves between two equipotential surfaces, its speed changes. Here, the particle starts with a speed of 2.00 m/s on surface A and increases to 3.00 m/s on surface B. Therefore, we must calculate the change in kinetic energy (\( \Delta K \)) to determine how the motion energy has altered between these two positions.

It's crucial to understand that not only does the speed affect kinetic energy, but the mass of the object plays a role too. Even small changes in velocity can lead to significant changes in kinetic energy when mass is involved. Thus, for our particle with mass \( 5.00 \times 10^{-2} \) kg, we calculate \( \Delta K \) to find out how much work needs to be done by an external force to change its speed during its movement from A to B.
Electric Potential Energy
Electric potential energy relates to the position of a charged particle within an electric field. It expresses the potential energy a particle possesses due to its position relative to other charged bodies.

To compute the change in electric potential energy (\( \Delta U \)), we use the formula \( \Delta U = q \Delta V \), where \( q \) is the charge of the particle and \( \Delta V \) is the change in the electric potential between the two surfaces. In our example, the particle moves between surfaces with different potentials, 5650 V to 7850 V, resulting in a \( \Delta V \) of 2200 V.

The particle's charge is \( 4.00 \times 10^{-5} \) C, and when this is applied in the formula above, it results in a \( \Delta U \) of 0.088 J. This tells us how much the electric potential energy of the system has changed due to the particle's displacement in the field. Electric potential energy provides insight into how potential differences can influence motion and behavior of charged objects.
Work Done by Force
The concept of work in physics is quite profound and involves the transfer of energy to or from an object via a force causing displacement. In this specific problem, the work done by an external force helps move the particle across equipotential surfaces.

The key formula here is \( W_{external} = \Delta K + \Delta U \), where the total work done by the outside force is the sum of the changes in both kinetic and electric potential energies. This accounts for the energy needed to change the speed of the particle as well as the energy associated with its position in the electric field.

Calculating these, we find:
  • Change in kinetic energy, \( \Delta K = 0.125 \) J
  • Change in electric potential energy, \( \Delta U = 0.088 \) J
Adding these gives the total work done by the force: \( W_{external} = 0.213 \) J. This computation shows the effort required by the external force to achieve the desired motion and energy state of the particle as it moves from surface A to B.

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Most popular questions from this chapter

A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of \(\operatorname{mass} 7.20 \times 10^{-3} \mathrm{kg}\) and charge \(-8.00 \mu \mathrm{C}\) is fired with an initial speed of \(65.0 \mathrm{m} / \mathrm{s}\) directly toward the fixed charge. How far does the particle travel before its speed is zero?

A particle with a charge of \(-1.5 \mu \mathrm{C}\) and a mass of \(2.5 \times 10^{-6} \mathrm{kg}\) is released from rest at point \(A\) and accelerates toward point \(B\), arriving there with a speed of \(42 \mathrm{m} / \mathrm{s} .\) The only force acting on the particle is the electric force. (a) Which point is at the higher potential? Give your reasoning. (b) What is the potential difference \(V_{\mathrm{B}}-V_{\mathrm{A}}\) between \(\mathrm{A}\) and \(\mathrm{B} ?\)

The electric potential energy stored in the capacitor of a defibrillator is \(73 \mathrm{J},\) and the capacitance is \(120 \mu \mathrm{F}\). What is the potential difference that exists across the capacitor plates?

Two identical point charges \(\left(q=+7.20 \times 10^{-6} \mathrm{C}\right)\) are fixed at diagonally opposite corners of a square with sides of length \(0.480 \mathrm{m}\). A test charge \(\left(q_{0}=-2.40 \times 10^{-8} \mathrm{C}\right),\) with a mass of \(6.60 \times 10^{-8} \mathrm{kg},\) is released from rest at one of the empty corners of the square. Determine the speed of the test charge when it reaches the center of the square.

A positive point charge \(\left(q=+7.2 \times 10^{-8} \mathrm{C}\right)\) is surrounded by an equipotential surface \(A,\) which has a radius of \(r_{A}=1.8 \mathrm{m} .\) A positive test charge \(\left(q_{0}=+4.5 \times 10^{-11} \mathrm{C}\right)\) moves from surface \(A\) to another equipotential surface \(B,\) which has a radius \(r_{B} .\) The work done as the test charge moves from surface \(A\) to surface \(B\) is \(W_{A B}=-8.1 \times 10^{-9}\) J. Find \(r_{B}\).
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