Question

Your sister wants to participate in the yearly science fair at her high school and asks you to suggest some exciting project. You suggest that she experiment with your recently created electron extractor to suspend her cat in the air. You tell her to buy a copper plate and bolt it to the ceiling in her room and then use your electron extractor to transfer electrons from the plate to the cat. If the cat weighs \(7.00 \mathrm{~kg}\) and is suspended \(2.00 \mathrm{~m}\) below the ceiling, how many electrons have to be extracted from the cat? Assume that the cat and the metal plate are point charges.

Short answer

Answer: We need to extract approximately \(4.91 * 10^{12}\) electrons from the cat to suspend it in the air, 2 meters below the ceiling.

Step-by-step solution

Calculate the gravitational force on the cat.

To calculate the gravitational force on the cat, we can use the formula \(F_g = mg\), where \(m\) is the mass of the cat and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)). The mass of the cat is \(7.00 \mathrm{~kg}\), so the gravitational force is: \(F_g = 7.00 \mathrm{~kg} * 9.81 \mathrm{~m/s^2} \approx 68.67 \mathrm{~N}\).

Find the electrostatic force between the cat and the metal plate.

The electrostatic force between the cat and the metal plate can be found using Coulomb's law: \(F_e = \cfrac{k \cdot q_1 \cdot q_2}{r^2}\). Here, \(F_e\) is the electrostatic force, \(k\) is the electrostatic constant (\(8.99 * 10^9 \mathrm{~N \cdot m^2 / C^2}\)), \(q_1\) and \(q_2\) are the charges on the cat and the metal plate, and \(r\) is the distance between them (in this case, \(2.00 \mathrm{~m}\)). Since the cat is suspended in the air, we can assume that the gravitational force and the electrostatic force are equal in magnitude: \(F_g = F_e\). We can plug in the values into the equation above: \(68.67 \mathrm{~N} = \cfrac{8.99 * 10^9 \mathrm{~N \cdot m^2 / C^2} \cdot q_1 \cdot q_2}{(2.00 \mathrm{~m})^2}\).

Determine the charge on the cat.

To find the charge on the cat, we need to simplify the equation that we found in the previous step. Since the cat and the metal plate have the same magnitude of charge (though opposite in sign), we can replace \(q_1\) and \(q_2\) with \(q\). The equation becomes: \(68.67 \mathrm{~N} = \cfrac{8.99 * 10^9 \mathrm{~N \cdot m^2 / C^2} \cdot q \cdot q}{(2.00 \mathrm{~m})^2}\). Now, we have to solve for \(q\): \(q = \sqrt{\cfrac{68.67 \mathrm{~N} \cdot (2.00 \mathrm{~m})^2}{8.99 * 10^9 \mathrm{~N \cdot m^2 / C^2}}} \approx 7.85 * 10^{-7} \mathrm{~C}\).

Find the number of electrons extracted from the cat.

To find the number of electrons extracted from the cat, we need to relate the charge to the electrons. Since the charge of one electron is \(1.60 * 10^{-19} \mathrm{~C}\), the number of electrons can be found by dividing the total charge by the charge of one electron: \(n = \cfrac{7.85 * 10^{-7} \mathrm{~C}}{1.60 * 10^{-19} \mathrm{~C}} \approx 4.91 * 10^{12}\) electrons. So, we need to extract approximately \(4.91 * 10^{12}\) electrons from the cat to suspend it in the air, 2 meters below the ceiling.

Key concepts

Coulomb's Law

Imagine two charged particles in space; Coulomb's law gives us a way to calculate the electrostatic force between them. Fundamental to this concept is the understanding that like charges repel each other while opposite charges attract. Coulomb's law states that the magnitude of the electrostatic force (\(F_e\text{, which in our case equals the gravitational force on the cat,}\(F_g\text{) 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 the charges.}

The law is mathematically represented as \[F_e = \frac{k \times q_1 \times q_2}{r^2}\] where \(k\) is the electrostatic 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. For cases like our exercise, this law allows us to predict how many electrons need to be transferred to float the cat by making the gravitational and electrostatic forces equal.

Gravitational Force

Gravitational force is the pull that two bodies exert on each other. It's what keeps the planets in orbit around the sun and what makes objects fall to the ground on Earth. In our scenario, it is what would normally pull down the cat if it weren't for the opposing electrostatic force. The formula to calculate the gravitational force (\(F_g\text{) experienced by an object near the Earth's surface is} \[ F_g = m \times g \] where \(m\) is the object's mass and \(g\) is the acceleration due to gravity. On Earth, \(g\) is approximately \(9.81 m/s^2\). Using this, we calculate that the cat, with a mass of \(7.00 kg\), experiences a gravitational force of \(68.67 N\text{, which is the force needed to be counteracted for electrostatic suspension.}}

Electric Charge

Electric charge is a fundamental property of matter carried by certain subatomic particles, such as protons and electrons. Charges can be positive, as in protons, or negative, as in electrons. Objects can become charged through various methods, including electron transfer.

An important property of charge is that it is conserved; it cannot be created or destroyed, only transferred from one place to another. This conservation plays a pivotal role in our cat's hypothetical levitation. We adjusted the cat's overall charge by extracting electrons, resulting in a net positive charge that interacts with the oppositely charged ceiling plate.

Electron Transfer

Electron transfer is a process where electrons move from one atom, molecule, or material to another. In our exercise, we make use of electron transfer to create an electrostatic suspension. By removing electrons from the cat, we're giving it a net positive charge because it's losing negatively charged electrons.

This transfer of electrons can be done through various methods, including friction, contact, or in our exercise, an 'electron extractor', a device that facilitates the movement of electrons. Finally, to determine the number of electrons we need to remove to suspend the cat, we use the electric charge of a single electron, \(1.60 \times 10^{-19} C\), and the calculated charge needed for suspension. This results in needing to extract approximately \(4.91 \times 10^{12}\) electrons—a prime example of fundamental physics principles applied to a whimsical scenario.