/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 A point charge \(Q=+87.1 \mu \ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A point charge \(Q=+87.1 \mu \mathrm{C}\) is held fixed at the origin. A second point charge, with mass \(m=0.0576 \mathrm{kg}\) and charge \(q=-2.87 \mu C,\) is placed at the location \((0.323 \mathrm{m}, 0)\) (a) Find the electric potential energy of this system of charges. (b) If the second charge is released from rest, what is its speed when it reaches the point \((0.121 \mathrm{m}, 0) ?\)

Short Answer

Expert verified
(a) -6.914 J at initial; (b) speed is 20.01 m/s.

Step by step solution

01

Define the Formula for Electric Potential Energy

The electric potential energy (U) of a system with two point charges is given by the formula: \( U = \frac{k \cdot Q \cdot q}{r} \) where \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \), \( Q \) and \( q \) are the charges, and \( r \) is the distance between the charges.
02

Calculate Initial Electric Potential Energy

Calculate the electric potential energy when the second charge is positioned at \((0.323 \text{ m}, 0)\). Use the formula: \( U_i = \frac{8.99 \times 10^9 \times 87.1 \times 10^{-6} \times (-2.87 \times 10^{-6})}{0.323} \). Simplify this to \( U_i = -6.914 \text{ J} \).
03

Calculate Final Electric Potential Energy

Calculate the electric potential energy when the charge is at \((0.121 \text{ m}, 0)\). Use the formula: \( U_f = \frac{8.99 \times 10^9 \times 87.1 \times 10^{-6} \times (-2.87 \times 10^{-6})}{0.121} \). Simplify to \( U_f = -18.454 \text{ J} \).
04

Determine Change in Potential Energy and Kinetic Energy Principle

The change in potential energy \( \Delta U = U_f - U_i = -18.454 - (-6.914) = -11.54 \text{ J}.\) By the conservation of energy, this change equals the kinetic energy acquired by the charge: \( \frac{1}{2}mv^2 = -\Delta U \).
05

Solve for the Speed of the Second Charge

Use the equation for kinetic energy: \( \frac{1}{2} \times 0.0576 \times v^2 = 11.54 \). Solving for \( v \), we get \( v^2 = \frac{2 \times 11.54}{0.0576} \rightarrow v = \sqrt{400.694} \approx 20.01 \text{ m/s} \).

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Key Concepts

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

Coulomb's Law
Coulomb's Law describes the electric force between two point charges. These are simply charged particles located at specific points in space. The law states that the force (\( F \)) between two charges is proportional to the absolute product of the charges and inversely proportional to the square of the distance between them.
The mathematical expression for Coulomb's Law is given by:
  • \[F = \frac{k \cdot |Q \cdot q|}{r^2}\]where \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2) \), \( Q \) and \( q \) are the magnitudes of the point charges, and \( r \) is the distance separating the charges.
Coulomb's Law helps us calculate the magnitude of the force, which can be attractive or repulsive: like charges repel and unlike charges attract. This fundamental principle is crucial in understanding how charged particles interact in an electric field.
By applying Coulomb's Law in conjunction with the concept of electric potential energy, we can better analyze how energy transforms as the positions of these charges change.
Point Charges
In physics, a point charge is an idealization where the charge is assumed to be located at a single point in space. This simplifies calculations by ignoring any physical dimensions the charge may have.
Point charges are used in physics to model objects that are charged but are small enough that their size doesn’t significantly affect the calculations of their interactions with other charges.
  • Point charges allow the use of simple equations like those derived from Coulomb's Law to predict behaviors of electric forces and energies.
A real-world application is where we consider tiny charged particles, such as electrons and protons, as point charges due to their relatively minute size compared to the distances over which we calculate forces and energies.
This concept is key in calculations regarding electric potential energy and electric fields because it allows us to assume the charge is acting from one specific location without accounting for any distribution around it.
Conservation of Energy
The principle of conservation of energy states that energy in a closed system remains constant; it can neither be created nor destroyed, but can only change forms. In electric systems involving point charges, this principle is crucial to understand the transition between different types of energy.
In our scenario, as a point charge moves within an electric field, its electric potential energy changes. According to the conservation of energy:
  • The decrease in electric potential energy converts to an increase in kinetic energy, resulting in the motion of the charge.
  • The sum of potential energy and kinetic energy at any point is constant.
Mathematically, this can be expressed as:\[U_i + K_i = U_f + K_f\]where \( U \) represents electric potential energy, and \( K \) is kinetic energy. Thus:
  • \( \Delta U = U_f - U_i \), and the change equals \( \Delta K = K_f - K_i \).
In the step-by-step solution, this principle was employed to find the speed of a moving point charge after it was released from rest, by equating the potential energy lost to the kinetic energy gained.

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Most popular questions from this chapter

A point charge \(Q=+87.1 \mu C\) is held fixed at the origin. A second point charge, with mass \(m=0.0576 \mathrm{kg}\) and charge \(q=-2.87 \mu C,\) is placed at the location \((0.323 \mathrm{m}, 0)\) (a) Find the electric potential energy of this system of charges. (b) If the second charge is released from rest, what is its speed when it reaches the point \((0.121 \mathrm{m}, 0) ?\)

Rutherford's Planetary Model of the Atom In 1911 , Ernest Rutherford developed a planetary model of the atom, in which a small positively charged nucleus is orbited by electrons. The model was motivated by an experiment carried out by Rutherford and his graduate students, Geiger and Marsden. In this experiment, they fired alpha particles with an initial speed of \(1.75 \times 10^{7} \mathrm{m} / \mathrm{s}\) at a thin sheet of gold. (Alpha particles are obtained from certain radioactive decays. They have a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27} \mathrm{kg}\).) How close can the alpha particles get to a gold nucleus (charge \(=+79 e\) ), assuming the nucleus remains stationary? (This calculation sets an upper limit on the size of the gold nucleus. See Chapter 31 for further details.)

A uniform electric field with a magnitude of \(6350 \mathrm{N} / \mathrm{C}\) points in the positive \(x\) direction. Find the change in electric potential energy when a \(+12.5-\mu \mathrm{C}\) charge is moved \(5.50 \mathrm{cm}\) in \((a)\) the positive \(x\) direction, (b) the negative \(x\) direction, and (c) the positive \(y\) direction.

An electronic flash unit for a camera contains a capacitor with a capacitance of \(890 \mu\) F. When the unit is fully charged and ready for operation, the potential difference between the capacitor plates is \(330 \mathrm{V}\). (a) What is the magnitude of the charge on each plate of the fully charged capacitor? (b) Find the energy stored in the "charged-up" flash unit.

A charge of \(20.2 \mu C\) is held fixed at the origin. (a) If a \(-5.25-\mu C\) charge with a mass of \(3.20 \mathrm{g}\) is released from rest at the position \((0.925 \mathrm{m}, 1.17 \mathrm{m}),\) what is its speed when it is halfway to the origin? (b) Suppose the \(-5.25-\mu \mathrm{C}\) charge is released from rest at the point \(x=\frac{1}{2}(0.925 \mathrm{m})\) and \(y=\frac{1}{2}(1.17 \mathrm{m}) .\) When it is halfway to the origin, is its speed greater than, less than, or equal to the speed found in part (a)? Explain. (c) Find the speed of the charge for the situation described in part (b).

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