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Chapter 18: Problem 61

Interactive LearningWare 18.1 at offers some perspective on this problem. Two tiny spheres have the same mass and carry charges of the same magnitude. The mass of each sphere is \(2.0 \times 10^{-6} \mathrm{~kg} .\) The gravitational force that each sphere exerts on the other is balanced by the electric force. (a) What algebraic signs can the charges have? (b) Determine the charge magnitude.

Short Answer

Expert verified
(a) Charges must be opposite in sign. (b) Charge magnitude is approximately \( 3.34 \times 10^{-16} \text{ C} \).

Step by step solution

01

Understand the Forces Involved

Identify the forces acting on the spheres. The gravitational force (\[ F_g = \frac{G m_1 m_2}{r^2} \] where \( G \) is the gravitational constant, and \( m_1 \) and \( m_2 \) are the masses) and the electric force (\[ F_e = \frac{k q^2}{r^2} \] where \( k \) is Coulomb's constant, and \( q \) is the magnitude of the charge) are in balance.
02

Analyze the Charge Signs

For the gravitational force and the electric force to be balanced, the electric force must be attractive (since gravity is always attractive). Two charges will attract each other only if they are of opposite signs (i.e., one positive and one negative).
03

Set Up the Force Equation

Since the gravitational and electric forces balance each other, set their magnitudes equal: \[ \frac{G m^2}{r^2} = \frac{k q^2}{r^2} \] The \( r^2 \) terms can be canceled from both sides, simplifying the equation.
04

Solve for Charge Magnitude

Rearrange the equation \( G m^2 = k q^2 \) to solve for \( q \):\[ q^2 = \frac{G m^2}{k} \] Then find \( q \):\[ q = \sqrt{\frac{G m^2}{k}} \] where \( m = 2.0 \times 10^{-6} \text{ kg} \), \( G = 6.674 \times 10^{-11} \text{ Nm}^2\text{/kg}^2 \), and \( k = 8.987 \times 10^9 \text{ Nm}^2\text{/C}^2 \).
05

Calculate the Charge Magnitude

Substitute the known values into the equation:\[ q = \sqrt{\frac{(6.674 \times 10^{-11})(2.0 \times 10^{-6})^2}{8.987 \times 10^9}} \]Calculate the numerical result to find:\[ q \approx 3.34 \times 10^{-16} \text{ C} \]

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Key Concepts

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

Gravitational Force
The gravitational force is a fundamental force of nature that acts between any two objects with mass. It always attracts objects towards each other. This force is described mathematically by Newton's Law of Universal Gravitation. The formula for gravitational force is:
  • \[ F_g = \frac{G m_1 m_2}{r^2} \]
where:
  • \( F_g \) is the gravitational force,
  • \( G \) is the gravitational constant (\( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)),
  • \( m_1 \) and \( m_2 \) are the masses of the two objects, and
  • \( r \) is the distance between the centers of the two masses.

This formula tells us that the gravitational force increases with larger masses and decreases with greater distances between them. Unlike other forces, gravitational force is always attractive. In the context of this exercise, the two spheres exert this attractive force on one another due to their masses being significant enough for the force to be measurable.
Coulomb's Law
Coulomb's Law describes the electric force between charged objects. This law plays a crucial role in understanding interactions at an atomic level. It's given by:
  • \[ F_e = \frac{k q_1 q_2}{r^2} \]
where:
  • \( F_e \) is the electric force,
  • \( k \) is Coulomb's constant (\( 8.987 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)),
  • \( q_1 \) and \( q_2 \) are the charges on the two objects, and
  • \( r \) is the distance between the charges.

In this exercise, since the gravitational force is balanced by the electric force, we know that it implies an attractive electric force. This means that the forces are equal in magnitude but opposite in direction to keep the system stable. Therefore, the charges on the two spheres must have opposite signs. Understanding this balance helps to solve for the magnitude of the electric charge.
Charge Magnitude
Charge magnitude refers to the amount or strength of electric charge present on an object. In the exercise, we need to find the charge magnitude such that the electric force balances the gravitational force. To find this magnitude, we utilize the balancing condition:
  • \[ \frac{G m^2}{r^2} = \frac{k q^2}{r^2} \]
By simplifying this equation (since \( r^2 \) is present in both terms), we isolate and solve for \( q \).
  • \[ q^2 = \frac{G m^2}{k} \]
This leads to:
  • \[ q = \sqrt{\frac{G m^2}{k}} \]

Substituting the given values, with mass \( m = 2.0 \times 10^{-6} \, \text{kg} \), we calculate:
  • \[ q = \sqrt{\frac{(6.674 \times 10^{-11})(2.0 \times 10^{-6})^2}{8.987 \times 10^9}} \approx 3.34 \times 10^{-16} \, \text{C} \]
This is the magnitude of the charge needed to balance the gravitational force.

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Most popular questions from this chapter

The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of \(7.1 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} \cdot\) (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion (K \(+\); charge \(=+e\) ) placed inside the membrane.

A vertical wall \((5.9 \mathrm{~m} \times 2.5 \mathrm{~m})\) in a house faces due east. A uniform electric field has a magnitude of \(150 \mathrm{~N} / \mathrm{C}\). This field is parallel to the ground and points \(35^{\circ}\) north of east. What is the electric flux through the wall?

Two spherical objects are separated by a distance of \(1.80 \times 10^{-3} \mathrm{~m}\). The objects are initially electrically neutral and are very small compared to the distance between them. Each object acquires the same negative charge due to the addition of electrons. As a result, each object experiences an electrostatic force that has a magnitude of \(4.55 \times 10^{-21} \mathrm{~N}\). How many electrons did it take to produce the charge on one of the objects?

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of \(8480 \mathrm{~N} / \mathrm{C}\). The mass of the water drop is \(3.50 \times 10^{-9} \mathrm{~kg} .\) (a) Is the excess charge on the water drop positive or negative? Why? (b) How many excess electrons or protons reside on the drop?

A tiny ball (mass \(=0.012 \mathrm{~kg}\) ) carries a charge of \(-18 \mu \mathrm{C}\). What electric field (magnitude and direction) is needed to cause the ball to float above the ground?
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