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Chapter 24: Problem 1

Two small spheres in vacuum are \(1.5 \mathrm{~m}\) apart center-to-center. They carry identical charges. Approximately how large is the charge on each if each sphere experiences a force of \(2 \mathrm{~N}\) ? The diameters of the spheres are small compared to the \(1.5 \mathrm{~m}\) separation. We may therefore approximate them as point charges. Coulomb's Law, \(F_{E}=k_{0} q_{.1} q_{.2} / r^{2}\), leads to $$q_{01} q_{\cdot 2}=q^{2}=\frac{F_{E} r^{2}}{k_{0}}=\frac{(2 \mathrm{~N})(1.5 \mathrm{~m})^{2}}{9 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}}=5 \times 10^{-10} \mathrm{C}^{2}$$ from which \(q=2 \times 10^{-5} \mathrm{C}\).

Short Answer

Expert verified
The charge on each sphere is approximately \(2.24 \times 10^{-5} \mathrm{C}\).

Step by step solution

01

Understand the Problem

We are given two small spheres, each with an identical charge, placed at a distance of 1.5 meters apart. The force experienced by each sphere is 2 Newtons due to their charges. We are to find the charge on each sphere using Coulomb's Law.
02

Apply Coulomb's Law

Coulomb's Law is given by the formula: \( F_E = \frac{k_0 q_1 q_2}{r^2} \), where \( F_E \) is the electrostatic force, \( k_0 \) is Coulomb's constant \((9 \times 10^9 \, \mathrm{N\cdot m^2 / C^2})\), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges. Since the charges are identical, \( q_1 = q_2 = q \), which simplifies the equation to \( F_E = \frac{k_0 q^2}{r^2} \).
03

Rearrange the Formula

To find \( q \), rearrange the formula: \( q^2 = \frac{F_E r^2}{k_0} \). Since \( F_E = 2 \, \mathrm{N} \) and \( r = 1.5 \, \mathrm{m} \), substitute these values into the equation.
04

Substitute and Solve for \( q^2 \)

Substitute the given values into the rearranged formula: \[ q^2 = \frac{(2 \, \mathrm{N})(1.5 \, \mathrm{m})^2}{9 \times 10^9 \, \mathrm{N\cdot m^2 / C^2}} \]. Simplify the expression: \[ q^2 = \frac{(2)(2.25)}{9 \times 10^9} = \frac{4.5}{9 \times 10^9} = 5 \times 10^{-10} \, \mathrm{C^2} \].
05

Calculate \( q \)

To find \( q \), take the square root of \( q^2 \): \( q = \sqrt{5 \times 10^{-10}} \). Therefore, \( q \approx 2.24 \times 10^{-5} \, \mathrm{C} \).
06

Verification

Verify the value of \( q \) by substituting back into Coulomb's Law to check if the force calculated matches the given force. Ensure the calculations are consistent with the original problem statement.

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Key Concepts

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

Electrostatics
Electrostatics is a fascinating field of physics that deals with the study of electric charges at rest. Understanding electrostatics is crucial because it forms the foundation for much of electromagnetism. At its core, electrostatics involves analyzing the effects of stationary charges within various mediums. One key aspect of electrostatics is the force interactions between charges, which are described by Coulomb's Law.
Coulomb's Law helps us determine the magnitude of the force experienced by charges. Electrostatic forces can be attractive or repulsive, depending on the nature of the charges involved. This phenomenon occurs because like charges repel each other, while opposite charges attract. Such interactions are fundamental to exploring more complex concepts in electricity and magnetism.
  • The study of electrostatics is essential for understanding electrical principles.
  • It involves stationary electric charges and their force interactions.
  • Coulomb's Law is a central equation in calculating these forces.
With the use of electrostatics, engineers can design various applications, from simple circuits to advanced electronics.
Point Charges
In electrostatics, the concept of 'point charges' is widely used to simplify problems. By assuming that charges are point-like, we can focus on the essential aspects of their interactions, without the complications introduced by the physical size or shape of the charged bodies.
The point charge model considers a charged body to be concentrated at a single point in space. This assumption makes calculations more straightforward, as it allows us to use Coulomb's Law directly without adjustments for charge distribution.
The scenario described in the exercise is a typical application where we can treat the charged spheres as point charges due to their small diameters in relation to the distance between them.
  • Point charges simplify the analysis of electrostatic interactions.
  • This model ignores the size and shape of the charged objects.
  • The exercise uses point charges to calculate the electrostatic force between two spheres.
Using point charges can provide approximate results that are accurate enough for practical purposes, especially when objects remain at a significant distance from one another compared to their sizes.
Force Calculation
Force calculation in electrostatics primarily involves using Coulomb's Law, a formula that quantifies the electrostatic force between two point charges. The formula is expressed as: \[ F_E = \frac{k_0 q_1 q_2}{r^2} \] where:
  • \( F_E \) is the electrostatic force between the charges.
  • \( k_0 \) is Coulomb's constant, with a value of \( 9 \times 10^9 \, \mathrm{N \cdot m^2 / C^2} \).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the charges.
Coulomb's Law allows us to determine the force without complex equipment, just with knowledge of the charges' magnitudes and their separation distance.
The exercise illustrates this concept by calculating the force between identical charges. By knowing one force magnitude, the distance, and Coulomb鈥檚 constant, we can solve for the unknown charge using: \[ q^2 = \frac{F_E r^2}{k_0} \] This logical approach simplifies the derivation of unknown quantities in similar exercises. Understanding the relationship between force and charge lays the groundwork for more elaborate studies in electromagnetism.

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Most popular questions from this chapter

Three point charges are placed at the following locations on the \(x\) -axis: \(+2.0 \mu \mathrm{C}\) at \(x=0,-3.0 \mu \mathrm{C}\) at \(x=40 \mathrm{~cm},-5.0 \mu \mathrm{C}\) at \(x=120 \mathrm{~cm} .\) Find the force \((a)\) on the \(-3.0 \mu \mathrm{C}\) charge,\((b)\) on the \(-5.0 \mu \mathrm{C}\) charge.

Determine the acceleration of a proton \(\left(q=+e, m=1.67 \times 10^{-27} \mathrm{~kg}\right)\) immersed in an electric field of strength \(0.50 \mathrm{kN} / \mathrm{C}\) in vacuum. How many times is this acceleration greater than that due to gravity?

Compute \((a)\) the electric field \(E\) in air at a distance of \(30 \mathrm{~cm}\) from a point charge \(q_{* 1}=5.0 \times 10^{-9} \mathrm{C}\), (b) the force on a charge \(q_{.2}=4.0 \times 10^{-10} \mathrm{C}\) placed \(30 \mathrm{~cm}\) from \(q_{.1}\), and \((c)\) the force on a charge \(q_{.3}=-4.0 \times 10^{-10} \mathrm{C}\) placed \(30 \mathrm{~cm}\) from \(q_{.1}\) (in the absence of \(\left.q_{.2}\right)\) (a) \(E=k_{0} \frac{q_{01}}{r^{2}}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{5.0 \times 10^{-9} \mathrm{C}}{(0.30 \mathrm{~m})^{2}}=0.50 \mathrm{kN} / \mathrm{C}\) directed away from \(q_{* 1}\) (b) \(F_{E}=E q_{.2}=(500 \mathrm{~N} / \mathrm{C})\left(-4.0 \times 10^{-10} \mathrm{C}\right)=2.0 \times 10^{-7} \mathrm{~N}=0.20 \mu \mathrm{N}\) directed away from \(q_{* 1}\). (c) \(\mathrm{F}_{E}=E q_{.3}=(500 \mathrm{~N} / \mathrm{C})\left(-4.0 \times 10^{-10} \mathrm{C}\right)=-0.20 \mu \mathrm{N}\) This force is directed toward \(q_{.1}\)

Charges of \(+2.0,+3.0\), and \(-8.0 \mu \mathrm{C}\) are placed in air at the vertices of an equilateral triangle of side \(10 \mathrm{~cm}\). Calculate the magnitude of the force acting on the \(-8.0 \mu \mathrm{C}\) charge due to the other two charges.

A small, \(0.60-\mathrm{g}\) ball in air carries a charge of magnitude \(8.0 \mu \mathrm{C}\). It is suspended by a vertical thread in a downward \(300 \mathrm{~N} / \mathrm{C}\) electric field. What is the tension in the thread if the charge on the ball is \((a)\) positive, (b) negative?
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