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Chapter 23: Problem 5

A small metal sphere, carrying a net charge of \(q_1 = -\)2.80 \(\mu\)C, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of \(q_2 = -\)7.80 \(\mu\)C and mass 1.50 g, is projected toward \(q_1\). When the two spheres are 0.800 m apart, \(q_2\), is moving toward \(q_1\) with speed 22.0 m\(/\)s (\(\textbf{Fig. E23.5}\)). Assume that the two spheres can be treated as point charges. You can ignore the force of gravity. (a) What is the speed of \(q_2\) when the spheres are 0.400 m apart? (b) How close does \(q_2\) get to \(q_1\)?

Short Answer

Expert verified
(a) 12.5 m/s, (b) 0.322 m.

Step by step solution

01

Understand the Problem

We have two charges, one stationary (\(q_1\)) and another moving towards it (\(q_2\)), both negative. We want to find the speed of \(q_2\) when the distance between them decreases and determine the point of closest approach by using conservation of energy.
02

Define Initial Energy Conditions

Initially, when the distance between \(q_1\) and \(q_2\) is 0.800 m, the kinetic energy \(KE_0\) of \(q_2\) is \( \frac{1}{2}mv^2 \) and its potential energy \(U_0\) is given by Coulomb's law: \[ U_0 = k\frac{q_1q_2}{r_0} \]where \(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is Coulomb's constant, \(r_0 = 0.800 \, \text{m}\), and \(v = 22.0 \, \text{m/s}\).
03

Calculate Initial Kinetic Energy

The initial kinetic energy \(KE_0\) of \(q_2\) when it is 0.800 m away is:\[ KE_0 = \frac{1}{2} \times 0.0015 \, \text{kg} \times (22.0 \, \text{m/s})^2 \]Execute the calculation:\[ KE_0 = \frac{1}{2} \times 0.0015 \times 484 = 0.362 \, \text{J} \]
04

Calculate Initial Potential Energy

Calculate the initial potential energy \(U_0\) of the system:\[ U_0 = 8.99 \times 10^9 \times \frac{(-2.80 \times 10^{-6})(-7.80 \times 10^{-6})}{0.800} \]Perform the calculation:\[ U_0 = 8.99 \times 10^9 \times \frac{21.84 \times 10^{-12}}{0.800} = 0.245 \, \text{J} \]
05

Apply Conservation of Energy at 0.400 m

The total mechanical energy initially \(E_0 = KE_0 + U_0 = 0.362 + 0.245 = 0.607 \, \text{J}\). When \(q_2\) is 0.400 m apart from \(q_1\), the potential energy \(U_1\) is:\[ U_1 = 8.99 \times 10^9 \times \frac{21.84 \times 10^{-12}}{0.400} = 0.490 \, \text{J} \]Using conservation of energy, solve for \(KE_1 = E_1 - U_1 = 0.607 - 0.490 = 0.117 \, \text{J}\).
06

Solve for Speed at 0.400 m

Given \(KE_1 = \frac{1}{2} mv_{0.400}^2 = 0.117 \, \text{J}\), solve for velocity \(v_{0.400}\):\[ \frac{1}{2} \times 0.0015 \times v_{0.400}^2 = 0.117 \rightarrow v_{0.400}^2 = \frac{0.117}{0.00075} = 156 \]Then:\[ v_{0.400} = \sqrt{156} = 12.5 \, \text{m/s} \]
07

Analyze the Closest Distance (Point of Turning)

At the closest point of approach, the speed of \(q_2\) is zero, so all kinetic energy is converted to potential energy. Thus, set the initial total energy equal to the potential energy at the turning point:\[ E_0 = U_f = k\frac{q_1q_2}{r_f} \]\(0.607 = 8.99 \times 10^9 \times \frac{21.84 \times 10^{-12}}{r_f}\) solving for \(r_f\):\[ r_f = \frac{8.99 \times 10^9 \times 21.84 \times 10^{-12}}{0.607} \approx 0.322 \, \text{m} \]
08

Answer Questions

(a) The speed of \(q_2\) when the spheres are 0.400 m apart is 12.5 m/s. (b) The closest approach to \(q_1\) is approximately 0.322 m.

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Key Concepts

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

Conservation of Energy
Conservation of Energy is a fundamental concept that states energy cannot be created or destroyed; it can only change from one form to another. In electrostatics, this principle is crucial to solving problems involving point charges.
When dealing with two charged objects, like the spheres in the exercise, both kinetic and potential energy are considered for each state's energy. The total energy of the system (kinetic plus potential) remains constant unless an external force acts. Initially, the energy is divided between the kinetic energy of the moving charge and the electrostatic potential energy due to their interactions.
To apply this concept:- Calculate the initial kinetic energy using the formula: \[ KE = \frac{1}{2}mv^2 \]- Determine the initial potential energy using Coulomb's law: \[ U = k\frac{q_1q_2}{r} \]- Ensure the sum of these energies at any two points is equal.This principle helps us find unknowns, such as the speed at different distances or the point of closest approach, by knowing that the total mechanical energy (kinetic and potential) doesn't change.
Point Charges
In electrostatics, charges are often simplified to point charges which have dimensions so small that their size doesn't affect the calculations of forces and fields around them. This idealization assists in applying mathematical formulas straightforwardly and understanding interactions at a simplistic level.
Point charges interact according to their position and charge magnitude, which determines the amount of electric force or potential energy involved. These charges can attract or repel, generating energies that influence the behavior of charged objects.
Examples of applying point charge assumptions in calculations include:- When finding the potential energy, use the formula considering them as point charges: \[ U = k \frac{q_1 q_2}{r} \]- Creations of electric fields are simplified to a point originating from a single location in space.Treating charges as point charges helps solve electrostatic problems more easily by focusing only on their separation distance and charge magnitudes.
Coulomb's Law
Coulomb's Law describes the electrostatic interaction between two point charges. By it, the magnitude of the force between any two charges is directly proportional to the product of their charge magnitudes and inversely proportional to the square of the distance that separates them.
The formula for Coulomb鈥檚 Law is: \[ F = k \frac{|q_1q_2|}{r^2} \]where:
  • F is the magnitude of the force between the charges.
  • k is Coulomb's constant ( 8.99 脳 109 N m2/C2).
  • q_1 and q_2 are the amounts of the charges.
  • r is the separation distance between the charges.
Coulomb's Law is essential in calculations determining potential energy between charges. It explains why like charges repel and opposites attract, influencing the kinetic motion of charged particles, such as how they move towards or away from each other based on their interactions.

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Most popular questions from this chapter

(a) How much excess charge must be placed on a copper sphere 25.0 cm in diameter so that the potential of its center, relative to infinity, is 3.75 kV? (b) What is the potential of the sphere 's surface relative to infinity?

(a) If a spherical raindrop of radius 0.650 mm carries a charge of \(-\)3.60 pC uniformly distributed over its volume, what is the potential at its surface? (Take the potential to be zero at an infinite distance from the raindrop.) (b) Two identical raindrops, each with radius and charge specified in part (a), collide and merge into one larger raindrop. What is the radius of this larger drop, and what is the potential at its surface, if its charge is uniformly distributed over its volume?

An infinitely long line of charge has linear charge density 5.00 \(\times 10^{-12}\) C\(/\)m. A proton (mass 1.67 \(\times 10^{-27}\) kg, charge \(+\)1.60 \(\times 10^{-19}\) C) is 18.0 cm from the line and moving directly toward the line at 3.50 \(\times 10^3\) m\(/\)s. (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge?

A vacuum tube diode consists of concentric cylindrical electrodes, the negative cathode and the positive anode. Because of the accumulation of charge near the cathode, the electric potential between the electrodes is given by $$V(x) = Cx^{{4}/{3}}$$ where \(x\) is the distance from the cathode and \(C\) is a constant, characteristic of a particular diode and operating conditions. Assume that the distance between the cathode and anode is 13.0 mm and the potential difference between electrodes is 240 V. (a) Determine the value of \(C\). (b) Obtain a formula for the electric field between the electrodes as a function of \(x\). (c) Determine the force on an electron when the electron is halfway between the electrodes.

A point charge \(q_1 =\) 4.00 nC is placed at the origin, and a second point charge \(q_2 = -\)3.00 nC is placed on the \(x\)-axis at \(x = +\)20.0 cm. A third point charge \(q_3 =\) 2.00 nC is to be placed on the \(x\)-axis between \(q_1\) and \(q_2\) . (Take as zero the potential energy of the three charges when they are infinitely far apart.) (a) What is the potential energy of the system of the three charges if \(q_3\) is placed at \(x = +\)10.0 cm? (b) Where should \(q_3\) be placed to make the potential energy of the system equal to zero?
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