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Chapter 21: Problem 68

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

Expert verified
Answer: Approximately 3.45 × 10^14 electrons need to be extracted from the cat to suspend it in the air.

Step by step solution

01

Calculate the gravitational force acting on the cat

The gravitational force (Fg) acting on the cat can be calculated using the formula: Fg = m * g Where m is the mass of the cat (7.00 kg) and g is the acceleration due to gravity (approximated to 9.81 m/s^2). Fg = 7.00 kg * 9.81 m/s^2 = 68.67 N
02

Determine the necessary electrostatic force to suspend the cat

We need the electrostatic force (Fe) to be equal to the gravitational force to keep the cat suspended in the air. So, Fe = Fg = 68.67 N
03

Calculate the charge on the cat

We can use Coulomb's Law to relate the electrostatic force with the charge on the cat (q). The formula for Coulomb's Law is: Fe = k * (q1 * q2) / r^2 Where k is the electrostatic constant (approximated to 8.99 × 10^9 N m^2/C^2), q1 and q2 are the charges of the cat and the copper plate (in Coulombs), and r is the distance between the cat and the plate (2.00 m). Since the cat and the copper plate have the same charge, we can rewrite the equation as: Fe = k * (q^2) / r^2 Now we can solve for q: q^2 = (Fe * r^2) / k q = sqrt((68.67 N * (2.00 m)^2) / (8.99 × 10^9 N m^2/C^2)) q = sqrt(274.68 / 8.99 × 10^9) q ≈ 5.52 × 10^{-5} C
04

Calculate the number of electrons extracted

The charge q represents the total charge transferred from the cat. Since we know that the charge of one electron is approximately 1.6 × 10^{-19} C, we can find the number of electrons (n) extracted by dividing the total charge by the charge of one electron: n = q / (charge of one electron) n = 5.52 × 10^{-5} C / 1.6 × 10^{-19} C n ≈ 3.45 × 10^{14} So approximately 3.45 × 10^{14} electrons must be extracted from the cat to suspend it in the air.

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Most popular questions from this chapter

Two point charges lie on the \(x\) -axis. If one point charge is \(6.0 \mu C\) and lies at the origin and the other is \(-2.0 \mu C\) and lies at \(20.0 \mathrm{~cm}\), at what position must a third charge be placed to be in equilibrium?

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