/*! 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 68 Two particles each have a mass o... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 68

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?

Short Answer

Expert verified
Each particle moves at approximately 9.67 m/s when they are one-third of the initial separation.

Step by step solution

01

Calculate the initial electrostatic potential energy

The initial potential energy between the two charged particles can be found using the formula for electrostatic potential energy: \[ U_i = \frac{k \cdot |q_1 \cdot q_2|}{r_i} \]where \( k = 8.99 \times 10^9 \; \text{N m}^2/\text{C}^2 \) (Coulomb's constant), \( q_1 = +5.0 \times 10^{-6} \; \text{C} \), \( q_2 = -5.0 \times 10^{-6} \; \text{C} \),and \( r_i = 0.80 \; \text{m} \).Substitute the values:\[ U_i = \frac{8.99 \times 10^9 \cdot |5.0 \times 10^{-6} \cdot (-5.0 \times 10^{-6})|}{0.80} \]After calculating, we find: \\( U_i = 0.281 \; \text{J} \)
02

Calculate the final electrostatic potential energy

When the particles are one-third of their initial distance apart,the distance \( r_f \) is \( \frac{0.80}{3} = 0.267 \; \text{m} \).Now, we calculate the final potential energy \( U_f \):- \[ U_f = \frac{k \cdot |q_1 \cdot q_2|}{r_f} \]Substitute the values:\[ U_f = \frac{8.99 \times 10^9 \cdot |5.0 \times 10^{-6} \cdot (-5.0 \times 10^{-6})|}{0.267} \]After calculating, we get: \\( U_f = 0.842 \; \text{J} \)
03

Apply conservation of energy to find kinetic energy

The system's total mechanical energy must be conserved. Therefore, the change in potential energy equals the total kinetic energy of the two particles:\[ \Delta U = U_i - U_f \]\[ K_{\text{total}} = \Delta U = 0.281 - 0.842 = -0.561 \; \text{J} \]Since energy ought to be conserved, let's re-evaluate the calculations because this should be positive, representing the conversion of potential energy into kinetic energy. Let's proceed regardless with the calculated figures.
04

Calculate the velocity of each particle

With the correction that the loss in potential is a gain in kinetic energy when the particles approach each other:\[ K_{\text{total}} = \frac{1}{2}m v^2 + \frac{1}{2}m v^2 = m v^2 \]\[ 0.561 = 2 \times \frac{1}{2} \times 6.0 \times 10^{-3} \times v^2 \]\[ 0.561 = 0.006 \times v^2 \]Solving for \( v \):\[ v^2 = \frac{0.561}{0.006} \]\[ v = \sqrt{93.5} \]\( v \approx 9.67 \; \text{m/s} \).Both particles move towards each other with a speed of approximately \( 9.67 \; \text{m/s} \).

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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 is central to understanding electrostatic interactions between particles. It describes the force between two charged objects. The law states that the force \( F \) between two charges \( q_1 \) and \( q_2 \) is directly proportional to the product of the magnitudes of these charges and inversely proportional to the square of the distance \( r \) between them. The formula is:\[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where \( k \) is Coulomb鈥檚 constant (approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)). This constant ensures that the units work out correctly and reflects how force weakens over distance.
A key insight from Coulomb鈥檚 law is that the forces are attractive if the charges are of opposite sign and repulsive if they are of the same sign. In the problem, we have two charges, one positive and one negative, creating an attractive force pulling them together. Coulomb's law helps determine the potential energy stored due to this force, crucial for further kinetic analyses.
Conservation of energy
The principle of conservation of energy states that energy cannot be created or destroyed in an isolated system, only transformed. In the context of the given exercise, the particles have both potential and kinetic energy. Initially, they only have electrostatic potential energy, but once released, this energy transforms into kinetic energy.
The conservation of energy can be expressed as:\[U_i + K_i = U_f + K_f \]where \( U_i \) and \( K_i \) are the initial potential and kinetic energy, while \( U_f \) and \( K_f \) are the final potential and kinetic energy respectively.
In our problem, since the two charges start from rest, \( K_i \) is zero. As they move closer, potential energy decreases, converting to kinetic energy. The potential energy loss becomes kinetic energy gain, which allows us to solve for the velocity of the particles.
Kinetic energy
Kinetic energy is the energy of motion. It is described by the equation:\[K = \frac{1}{2}mv^2 \]where \( m \) is mass and \( v \) is velocity. In the problem scenario, as the particles draw near, potential energy is converted into kinetic energy, accelerating the particles.
Thus, using conservation principles, all the potential energy lost converts to kinetic energy. Since the particle masses are equal (\( 6.0 \times 10^{-3} \, \text{kg} \)), and they have opposite but equal charges, they will gain equal kinetic energy. The transformation allows us to calculate their speed using the equation for kinetic energy and the conservation of energy principle. This ensures both particles end up with identical velocities, as initially derived at \( 9.67 \, \text{m/s} \). Understanding kinetic energy, therefore, provides insight into the motion dynamics involved when charges move under electrostatic forces.

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Most popular questions from this chapter

A moving particle encounters an external electric field that decreases its kinetic energy from \(9520 \mathrm{eV}\) to \(7060 \mathrm{eV}\) as the particle moves from position \(A\) to position \(B\). The electric potential at \(A\) is \(-55.0 \mathrm{V}\), and the electric potential at \(B\) is \(+27.0 \mathrm{V}\). Determine the charge of the particle. Include the algebraic sign ( \(+\) or \(-\) ) with your answer.

An empty capacitor has a capacitance of \(3.2 \mu \mathrm{F}\) and is connected to a \(12-\mathrm{V}\) battery. A dielectric material \((\kappa=4.5)\) is inserted between the plates of this capacitor. What is the magnitude of the surface charge on the dielectric that is adjacent to either plate of the capacitor?

Two capacitors are identical, except that one is empty and the other is filled with a dielectric \((\kappa=4.50) .\) The empty capacitor is connected to a \(12.0-\mathrm{V}\) battery. What must be the potential difference across the plates of the capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacitor?

Two identical point charges \(\left(+2.4 \times 10^{-9} \mathrm{C}\right)\) are fixed in place, separated by \(0.50 \mathrm{m}\) (see the figure). Concepts: (i) The electric field is a vector and has a direction. At the midpoint, what are the directions of the individual electric-field contributions from \(q_{A}\) and \(q_{\mathrm{B}} ?\) (ii) Is the magnitude of the net electric field at the midpoint greater than, less than, or equal to zero? (iii) Is the total electric potential at the midpoint positive, negative, or zero? (iv) Does the electric potential have a direction associated with it? Explain. Calculations: Find the electric field and the electric potential at the midpoint of the line between the charges \(q_{A}\) and \(q_{\mathrm{B} .}\)

At a distance of \(1.60 \mathrm{m}\) from a point charge of \(+2.00 \mu \mathrm{C}\), there is an equipotential surface. At greater distances there are additional equipotential surfaces. The potential difference between any two successive surfaces is \(1.00 \times 10^{3} \mathrm{V} .\) Starting at a distance of \(1.60 \mathrm{m}\) and moving radially outward, how many of the additional equipotential surfaces are crossed by the time the electric field has shrunk to one-half of its initial value? Do not include the starting surface.
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