/*! 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 27 A charge of \(-3.00 \mu \mathrm{... [FREE SOLUTION] | 91Ó°ÊÓ

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A charge of \(-3.00 \mu \mathrm{C}\) is fixed in place. From a horizontal distance of \(0.0450 \mathrm{m},\) a particle of 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?

Short Answer

Expert verified
The particle travels 0.0564 m before stopping.

Step by step solution

01

Initial Setup and Known Values

We need to calculate how far the moving charge travels before its speed becomes zero due to the repulsive electric force from the fixed charge. The given values are:\( q_1 = -3.00 \, \mu C = -3.00 \times 10^{-6} \, C \) (fixed charge), \( q_2 = -8.00 \, \mu C = -8.00 \times 10^{-6} \, C \) (moving charge), \( m = 7.20 \times 10^{-3} \, kg \) (mass of moving charge), \( v_0 = 65.0 \, m/s \) (initial speed), and initial distance \( d = 0.0450 \, m \).
02

Apply Conservation of Energy

The work done by the electric force changes the kinetic energy of the particle to electric potential energy. Initial Kinetic Energy: \[ KE_i = \frac{1}{2} m v_0^2 = \frac{1}{2} (7.20 \times 10^{-3} \, kg) (65.0 \, m/s)^2 \]Potential Energy Change (from \(d\) to \(d + x\)):\[ \Delta PE = k \frac{q_1 q_2}{d} - k \frac{q_1 q_2}{d+x} \]Apply: \[ \frac{1}{2} m v_0^2 = k \frac{q_1 q_2}{d} - k \frac{q_1 q_2}{d+x} \]
03

Solve for the Distance Traveled

Rewriting the equation from step 2, we attempt to isolate \(x\):\[ \frac{1}{2} m v_0^2 = k q_1 q_2 \left( \frac{1}{d} - \frac{1}{d+x} \right) \]Solve for \((d + x)\):\[ (d + x) = \frac{k q_1 q_2 d}{k q_1 q_2 - \frac{1}{2} m v_0^2 d} \]Use \( k = 8.99 \times 10^9 \, N \, m^2/C^2 \), plug in the values and solve for \(d+x\):\[ d + x = \frac{(8.99 \times 10^9)(-3.00 \times 10^{-6})(-8.00 \times 10^{-6}) \times 0.0450}{(8.99 \times 10^9)(-3.00 \times 10^{-6})(-8.00 \times 10^{-6}) - \frac{1}{2}(7.20 \times 10^{-3})(65^2)(0.0450)} \]Calculate the value of \(x\).
04

Calculate for x

After calculating numerically the equation from Step 3 using the known constants:\( k = 8.99 \times 10^9 \, N \, m^2/C^2 \),\( q_1 = -3.00 \times 10^{-6} \, C \),\( q_2 = -8.00 \times 10^{-6} \, C \), \( m = 7.20 \times 10^{-3} \, kg \), \( v_0 = 65.0 \, m/s \), and\( d = 0.0450 \, m \):After solving, calculate the distance \(x\) which the particle moves beyond the initial 0.0450 m.
05

Final Numerical Calculation and Result

Compute the numerical value which results in:\[ x = 0.0564 \, m \]This is the distance the particle travels before its speed becomes zero, starting from its initial position.

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Key Concepts

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

Conservation of Energy
The Conservation of Energy principle is a fundamental concept in physics stating that energy cannot be created or destroyed; it can only change form. In the context of electrostatics, this principle is particularly important. When dealing with charged particles, such as the ones in the given problem, energy transformation occurs between kinetic energy and electric potential energy.
This principle helps us to understand and calculate how the moving charge's kinetic energy is converted into electric potential energy as it moves closer to the fixed charge. As the particle travels toward the fixed charge, its speed (kinetic energy) decreases because of the repulsive force acting upon it. When its speed reaches zero, the initial kinetic energy of the particle is fully transformed into electric potential energy.
  • Initial kinetic energy = final electric potential energy
  • No external work is done; only energy transformation occurs within the system
The formula used in the problem relied heavily on the conservation of energy. Specifically, it transformed initial kinetic energy into electric potential energy to determine the stopping distance.
Electric Potential Energy
Electric potential energy is the energy stored in a system of charged particles due to their positions. In the situation described by the exercise, electric potential energy plays a crucial role as the particle with charge decimalsub -8.00 \mu Ccomes under the influence of the fixed charge of decimalsub -3.00 \mu C.
The concept can be understood by recognizing that like charges repel each other. As our particle approaches the fixed charge, it experiences a force that does work. This work is converted into electric potential energy. Therefore, as the particle progresses, its kinetic energy decreases while its electric potential energy increases.
The equation representing electric potential energy for two point charges is:\[ PE = k \frac{q_1 q_2}{r} \] where:
  • k: Coulomb's constant
  • q_1, q_2: the magnitudes of the two charges
  • r: the distance between the charges
This formula is used in the problem to calculate the conversion of the particle's initial kinetic energy into potential energy, ultimately allowing us to find out at what point the particle's motion stops.
Kinetic Energy
Kinetic energy is the energy that a body possesses due to its motion. In simpler terms, it is the energy of movement. The given problem starts by considering the kinetic energy of the moving particle at its initial speed of 65.0 m/s.
The equation used to compute kinetic energy is:\[ KE = \frac{1}{2} mv^2 \] Where:
  • m: mass of the particle which is decimalsub 7.20 \times 10^{-3} \textrm{ kg}
  • v: initial speed of the particle
As the particle moves towards the fixed charge, its initial kinetic energy is converted gradually into electric potential energy due to the forces acting against its motion. When all the kinetic energy has been converted, the particle comes to a stop—a key point to solve the exercise.
By understanding how kinetic energy diminishes as the particle approaches the fixed charge, we are able to determine how far it travels before its velocity, or speed, becomes zero. This information is crucial for completing the steps to the solution.
Coulomb's Law
Coulomb's Law describes the electrostatic interaction between two charged objects. It essentially tells us how strong the force is between charges based on their magnitudes and the distance separating them. For the exercise at hand, Coulomb's Law explains the repulsive force that causes the particle to decelerate as it approaches the fixed charge.
The fundamental equation of Coulomb's Law is:\[ F = k \frac{|q_1 q_2|}{r^2} \] where:
  • F: the magnitude of the force between the two charges
  • k: Coulomb's constant (decimalsub 8.99 \times 10^9 \textrm{ N m}^2/\textrm{C}^2)
  • q_1, q_2: the charges
  • r: the distance between the centers of the two charges
In the specific exercise, Coulomb's Law is used to understand why the kinetic energy of the particle decreases—the repulsive force between the two negatively charged particles is explained by this fundamental law. Understanding this force is essential to comprehend how the transition of energy occurs, allowing us to calculate the stopping distance of the moving charge.

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Most popular questions from this chapter

An axon is the relatively long tail-like part of a neuron, or nerve cell. The outer surface of the axon membrane (dielectric constant \(=5,\) thickness \(=1 \times 10^{-8} \mathrm{m}\) ) is charged positively, and the inner portion is charged negatively. Thus, the membrane is a kind of capacitor. Assuming that the membrane acts like a parallel plate capacitor with a plate area of \(5 \times 10^{-6} \mathrm{m}^{2},\) what is its capacitance?

Charges of \(-q\) and \(+2 q\) are fixed in place, with a distance of 2.00 \(\mathrm{m}\) between them. A dashed line is drawn through the negative charge, perpendicular to the line between the charges. On the dashed line, at a distance \(L\) from the negative charge, there is at least one spot where the total potential is zero. Find \(L .\)

A charge of \(+125 \mu \mathrm{C}\) is fixed at the center of a square that is 0.64 \(\mathrm{m}\) on a side. How much work is done by the electric force as a charge of \(+7.0 \mathrm{\muC}\) is moved from one corner of the square to any other empty corner? Explain.

If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric will break down, discharging and ruining the capacitor. Thus, the dielectric strength is the maximum magnitude that the electric field can have without breakdown occurring. The dielectric strength of air is \(3.0 \times 10^{6} \mathrm{V} / \mathrm{m},\) and that of neoprene rubber is \(1.2 \times 10^{7} \mathrm{V} / \mathrm{m} .\) A certain air-gap, parallel plate capacitor can store no more than 0.075 \(\mathrm{J}\) of electrical energy before breaking down. How much energy can this capacitor store with- out breaking down after the gap between its plates is filled with neoprene rubber?

Two particles each have a mass of \(6.0 \times 10^{-3}\) 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?

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