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Chapter 21: Problem 69

A 10.0 -g mass is suspended \(5.00 \mathrm{~cm}\) above a nonconducting flat plate, directly above an embedded charge of \(q\) (in coulombs). If the mass has the same charge, \(q\), how much must \(q\) be so that the mass levitates (just floats, neither rising nor falling)? If the charge \(q\) is produced by adding electrons to the mass, by how much will the mass be changed?

Short Answer

Expert verified
The change in mass due to the added electrons required for a 10.0 g mass to levitate just above a non-conducting flat plate is approximately 2.96 x 10^-17 kg.

Step by step solution

01

Identifying forces

There are two main forces acting on the mass in this problem: the gravitational force (F_gravity) acting downward, and the electrostatic force (F_electrostatic) acting upward due to the repulsion between the charges. The mass levitates when these forces are balanced or equal.
02

Expressing the gravitational force

The gravitational force (F_gravity) can be expressed using the formula: F_gravity = m * g, where m is the mass (10.0 g) and g is the acceleration due to gravity (9.81 m/s^2). However, the mass given is in grams, so first, we have to convert it to kilograms: m = 10.0 g * (1 kg / 1000 g) = 0.010 kg. Now we can calculate the gravitational force: F_gravity = 0.010 kg * 9.81 m/s^2 = 0.0981 N.
03

Expressing the electrostatic force

The electrostatic force (F_electrostatic) can be expressed using Coulomb's law: F_electrostatic = k * |q1 * q2| / r^2, where k is the Coulomb's constant (8.99 × 10^9 Nm^2/C^2), q1 and q2 are the charges (both equal to q), and r is the distance between the charges (5.00 cm). The distance should be converted to meters: r = 5.00 cm * (1 m / 100 cm) = 0.0500 m.
04

Balancing the forces

To achieve levitation, we must set the electrostatic force equal to the gravitational force: F_electrostatic = F_gravity. Substitute the expressions obtained earlier: k * |q1 * q2| / r^2 = m * g. Knowing that q1=q2=q, we have k * q^2 / r^2 = m * g.
05

Solving for q

To find the charge q, we need to rearrange the equation and solve for q: q^2 = (m * g * r^2) / k. Plugging in the known values: q^2 = (0.010 kg * 9.81 m/s^2 * (0.0500 m)^2) / (8.99 × 10^9 Nm^2/C^2) ≈ 27.11 × 10^{-12} C^2. Taking the square root: q ≈ 5.21 × 10^{-6} C.
06

Calculating the number of electrons and mass change

To find the number of electrons added to produce the charge q, we need to use the elementary charge (e = 1.602 × 10^{-19} C): n = q / e ≈ (5.21 × 10^{-6} C) / (1.602 × 10^{-19} C) ≈ 3.25 × 10^{13} electrons. To calculate the change in mass due to these electrons, we need to use the mass of an electron (m_e = 9.11 × 10^{-31} kg): mass_change = n * m_e ≈ (3.25 × 10^{13}) * (9.11 × 10^{-31} kg) ≈ 2.96 × 10^{-17} kg.

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Most popular questions from this chapter

A small ball with a mass of \(30.0 \mathrm{~g}\) and a charge of \(-0.200 \mu \mathrm{C}\) is suspended from the ceiling by a string. The ball hangs at a distance of \(5.00 \mathrm{~cm}\) above an insulating floor. If a second small ball with a mass of \(50.0 \mathrm{~g}\) and a charge of \(0.400 \mu \mathrm{C}\) is rolled directly beneath the first ball, will the second ball leave the floor? What is the tension in the string when the second ball is directly beneath the first ball?

You bring a negatively charged rubber rod close to a grounded conductor without touching it. Then you disconnect the ground. What is the sign of the charge on the conductor after you remove the charged rod? a) negative d) cannot be determined from b) positive the given information c) no charge

How many electrons are required to yield a total charge of 1.00 C?

Which of the following situations produces the largest net force on the charge \(Q\) ? a) Charge \(Q=1 \mathrm{C}\) is \(1 \mathrm{~m}\) from a charge of \(-2 \mathrm{C}\). b) Charge \(Q=1 \mathrm{C}\) is \(0.5 \mathrm{~m}\) from a charge of \(-1 \mathrm{C}\) c) Charge \(Q=1 \mathrm{C}\) is halfway between a charge of \(-1 \mathrm{C}\) and a charge of \(1 \mathrm{C}\) that are \(2 \mathrm{~m}\) apart. d) Charge \(Q=1 \mathrm{C}\) is halfway between two charges of \(-2 \mathrm{C}\) that are \(2 \mathrm{~m}\) apart. e) Charge \(Q=1 \mathrm{C}\) is a distance of \(2 \mathrm{~m}\) from a charge of \(-4 \mathrm{C}\).

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