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(a) Calculate the potential energy of a system of two small spheres, one carrying a charge of 2.00 \(\mu\)C and the other a charge of \(-\)3.50 \(\mu\)C, with their centers separated by a distance of 0.180 m. Assume that \(U = 0\) when the charges are infinitely separated. (b) Suppose that one sphere is held in place; the other sphere, with mass 1.50 g, is shot away from it. What minimum initial speed would the moving sphere need to escape completely from the attraction of the fixed sphere? (To escape, the moving sphere would have to reach a velocity of zero when it is infinitely far from the fixed sphere.)

Short Answer

Expert verified
(a) The potential energy is approximately -0.349 Joules. (b) The minimum initial speed required is about 21.58 m/s.

Step by step solution

01

Introduction to Potential Energy Calculation

The potential energy of a system of two point charges can be calculated using Coulomb's law. The formula for the potential energy \( U \) of the system is \( U = \frac{k \, q_1 \, q_2}{r} \), where \( k \) is Coulomb’s constant \( 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
02

Substitute Values to Find Potential Energy

Substitute the given values into the potential energy formula. Let \( q_1 = 2.00 \, \mu\text{C} = 2.00 \times 10^{-6} \text{C} \) and \( q_2 = -3.50 \, \mu\text{C} = -3.50 \times 10^{-6} \text{C} \) with \( r = 0.180 \, \text{m} \). Then calculate \( U = \frac{(8.99 \times 10^9) \, (2.00 \times 10^{-6}) \, (-3.50 \times 10^{-6})}{0.180} \).
03

Calculating U

Perform the multiplication and division: \( U = \frac{(8.99 \times 10^9) \, (2.00) \, (-3.50)}{0.180 \times 10^{12}} \). This gives \( U \approx -0.349 \, \text{Joules} \).
04

Introduction to Escape Speed Calculation

To find the required minimum initial speed to escape, use the concept of energy conservation. The kinetic energy of the moving sphere must equal the magnitude of the potential energy at infinity. The kinetic energy \( KE \) is given by \( KE = \frac{1}{2} m v^2 \), where \( m \) is mass and \( v \) is velocity.
05

Find Initial Speed to Escape

Set the kinetic energy equal to the potential energy magnitude: \( \frac{1}{2} m v^2 = |-U| \). Rearrange to solve for \( v \): \( v = \sqrt{\frac{2 |-U|}{m}} \). Substitute \( m = 1.50 \, \text{g} = 1.50 \times 10^{-3} \text{kg} \), and \( |-U| = 0.349 \) Joules.
06

Calculate Escape Velocity

Substitute the values: \( v = \sqrt{\frac{2 \times 0.349}{1.50 \times 10^{-3}}} \). Calculate \( v \), resulting in \( v \approx 21.58 \, \text{m/s} \).

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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 a fundamental principle in physics that describes the force between two charges. It helps us calculate the electric potential energy in systems like the pair of charges in the exercise. This law states that the force \( F \) between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The formula is: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\]where:
  • \( k = 8.99 \times 10^9 \, \text{N\cdot m}^2/\text{C}^2 \) is the electrostatic constant,
  • \( q_1 \) and \( q_2 \) are the charges,
  • \( r \) is the distance between them.
In the context of the exercise, Coulomb's Law is used to compute the potential energy \( U \) for two point charges. The relationship is expressed with a similar formula:\[ U = \frac{k \cdot q_1 \cdot q_2}{r}\]This calculation shows how energy varies with the distance between charges.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It plays a crucial role when determining how fast an object must move to achieve certain effects, as required in the given exercise. The formula for kinetic energy is:\[ KE = \frac{1}{2} m v^2\]where:
  • \( m \) represents the mass of the object,
  • \( v \) stands for its velocity.
In the scenario described, kinetic energy helps determine the initial speed needed by a moving sphere to escape the electric field generated by another charged sphere. By setting the kinetic energy equal to the potential energy magnitude, we ensure that the two forms of energy balance out, allowing the sphere to break free from the influence of the other charge.
Escape Velocity
Escape velocity refers to the minimum speed that an object must have to break free from a gravitational or electric field without further propulsion. For the charged sphere in the exercise, this concept applies when it moves away from the other charged sphere.Using the conservation of energy, we set the initial kinetic energy of the moving sphere equal to the potential energy it needs to overcome:\[ \frac{1}{2} m v^2 = |-U|\]Solving for the initial velocity \( v \), we have:\[ v = \sqrt{\frac{2 |-U|}{m}}\]This equation helps us calculate the speed required to ensure that, as the sphere moves infinitely far away from the other, its velocity eventually becomes zero, effectively describing the escape condition.
Conservation of Energy
The conservation of energy is a key principle in physics where the total energy in a closed system remains constant over time. This principle is pivotal in understanding phenomena like the movement of charged particles, as seen in the exercise.In this case, it relates to balancing the system's potential energy with the kinetic energy of a moving charge. Initially, when a charged sphere is at rest, all energy is stored as potential energy. As the sphere moves, this potential energy converts into kinetic energy until the charge escapes the other’s influence or reaches zero velocity at infinity.The mathematical expression is captured by equating kinetic and potential energies:\[ \frac{1}{2} m v^2 = |-U|\]This formula expresses the idea that the decrease in potential energy corresponds exactly to the increase in kinetic energy, ensuring total energy remains unchanged throughout the process.

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Most popular questions from this chapter

A gold nucleus has a radius of 7.3 \(\times 10^{-15}\) m and a charge of \(+79e\). Through what voltage must an alpha particle, with charge \(+2e\), be accelerated so that it has just enough energy to reach a distance of 2.0 \(\times 10^{-14}\) m from the surface of a gold nucleus? (Assume that the gold nucleus remains stationary and can be treated as a point charge.)

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work is needed to move the alpha particle to the midpoint of one of the sides of the square?

The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. Typical dimensions are about 3.0 cm on a side, with a separation of about 5.0 mm. The potential difference between the plates is 25.0 V. The plates are close enough that we can ignore fringing at the ends. Under these conditions: (a) how much charge is on each plate, and (b) how strong is the electric field between the plates? (c) If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

Three point charges, which initially are infinitely far apart, are placed at the corners of an equilateral triangle with sides \(d\). Two of the point charges are identical and have charge \(q\). If zero net work is required to place the three charges at the corners of the triangle, what must the value of the third charge be?

Two point charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two point charges having opposite signs.

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