Question

From collisions with cosmic rays and from the solar wind, the Earth has a net electric charge of approximately \(-6.8 \cdot 10^{5} \mathrm{C} .\) Find the charge that must be given to a \(1.0-\mathrm{g}\) object for it to be electrostatically levitated close to the Earth's surface.

Short answer

Answer: Approximately \(-2.44 \times 10^{-8}\,\mathrm{C}\).

Step-by-step solution

Calculate Gravitational Force

First, we need to calculate the gravitational force acting on the object. We are given the mass of the object as 1.0 g, which we need to convert to kg: \(m = 1.0 \times 10^{-3} \mathrm{kg}\) To find the gravitational force, we use the formula \(F_g = m \cdot g\), where g is the acceleration due to gravity, which is approximately \(9.81\,\mathrm{m/s^2}\). \(F_g = (1.0 \times 10^{-3}\,\mathrm{kg})\cdot (9.81\,\mathrm{m/s^2}) = 9.81 \times 10^{-3}\,\mathrm{N}\)

Set up the equation for electrostatic force

Now, we will set up the equation for the electrostatic force between the Earth and the object. We will use the following formula: \(F_e = k \frac{q_1 \cdot q_2}{r^2}\). Since we want electrostatic levitation, we can set \(F_g = F_e\): \(9.81 \times 10^{-3}\,\mathrm{N} = k \frac{q_1 \cdot q_2}{r^2}\)

Solve for the charge of the object

To solve for the charge of the object, we need to plug in the values that we know. We know that the Earth's charge is approximately \(-6.8 \times 10^{5}\,\mathrm{C}\), and the distance between the object and the Earth's surface can be assumed to be the Earth's radius, which is approximately \(6.37 \times 10^6\,\mathrm{m}\). Also, the Coulomb's constant, k, is \(8.988 \times 10^9\,\mathrm{N\,m^2/C^2}\) Substituting these values into the equation, we get: \(9.81 \times 10^{-3}\,\mathrm{N} = (8.988 \times 10^9\,\mathrm{N\,m^2/C^2})\frac{(-6.8 \times 10^5\,\mathrm{C}) \cdot q_2}{(6.37 \times 10^6\,\mathrm{m})^2}\) Now, we can solve for \(q_2\): \(q_2 = \frac{(9.81 \times 10^{-3}\,\mathrm{N})\cdot(6.37 \times 10^6\,\mathrm{m})^2}{(8.988 \times 10^9\,\mathrm{N\,m^2/C^2})\cdot(-6.8 \times 10^5\,\mathrm{C})} = -2.44 \times 10^{-8}\,\mathrm{C}\) So, the charge that must be given to the 1.0-g object for it to be electrostatically levitated close to the Earth's surface is approximately \(-2.44 \times 10^{-8}\,\mathrm{C}\).

Key concepts

Coulomb's Law

Coulomb's Law is a fundamental principle used to calculate the electrostatic force between two charged objects. It states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
Mathematically, this is expressed as:

• \( F_e = k \frac{q_1 \cdot q_2}{r^2} \)

where:- \( F_e \) is the electrostatic force,- \( k \) is Coulomb's constant \( (8.988 \times 10^9 \, \mathrm{N\,m^2/C^2}) \),- \( q_1 \) and \( q_2 \) are the magnitudes of the charges,- \( r \) is the distance between the charges.

Coulomb's Law helps us understand how charged particles interact and is crucial in determining the force needed for electrostatic levitation. This force must balance gravitational forces to levitate an object, thus enabling it to float above the ground.

Gravitational Force

Gravitational force is the attractive force between two masses, such as the Earth and an object on its surface. It is described by Newton's law of universal gravitation, which states that the force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
For objects near the Earth's surface, this force is calculated with the equation:

• \( F_g = m \cdot g \)

where:- \( F_g \) is the gravitational force,- \( m \) is the object's mass,- \( g \) is the acceleration due to Earth's gravity (approximately \( 9.81\, \mathrm{m/s^2} \)).

This force pulls the object toward the Earth, and must be counteracted by the electrostatic force when levitating an object through electrostatic means.

Electric Charge

Electric charge is a property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of electric charges: positive and negative. Like charges repel each other, whereas opposite charges attract.
Charge is measured in coulombs (\( \mathrm{C} \)), and the net charge can influence electric forces, as described by Coulomb's Law. In the context of electrostatic levitation, the net electric charge of an object or system, like the Earth, determines the magnitude of electrostatic interactions.

When charging an object for levitation, it is given a specific electric charge, which interacts with the Earth's charge to produce an electrostatic force.
For our problem, Earth's charge is given as \(-6.8 \times 10^5\,\mathrm{C}\). The object's charge needs to be calculated carefully to balance the gravitational pull by adjusting its charge to be \(-2.44 \times 10^{-8}\, \mathrm{C}\).
Understanding electric charge is essential for successfully achieving electrostatic levitation.

Electrostatic Levitation

Electrostatic levitation involves balancing electrostatic force against gravitational force, so an object remains suspended in mid-air. This phenomenon requires precise control of charges and forces.

To achieve electrostatic levitation, Coulomb's Law is utilized to design the conditions where the electrostatic force equals the gravitational force acting on the object. The formula used is:

• \( F_g = F_e \)

This equation means the upward electrostatic force must exactly counterbalance the downward gravitational force. In practical terms, careful calculation and manipulation of an object's electric charge are necessary to achieve levitation. The object's mass and distance from Earth are also key factors. This technique opens up a realm of possibilities for precise control without physical support, finding applications in areas like materials testing and scientific experiments. Understanding electrostatic levitation is a fundamental concept in physics, revealing the intricate dance between electric charges and gravitational forces.