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Chapter 22: Problem 63

A spherical water drop \(1.20 \mu \mathrm{m}\) in diameter is suspended in calm air due to a downward-directed atmospheric electric field of magnitude \(E=462 \mathrm{~N} / \mathrm{C}\). (a) What is the magnitude of the gravitational force on the drop? (b) How many excess electrons does it have?

Short Answer

Expert verified
(a) The gravitational force is approximately \(8.88 \times 10^{-15} N\). (b) The drop has about 120 excess electrons.

Step by step solution

01

Calculate the Volume of the Water Drop

The water drop is a sphere, so its volume can be calculated using the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \). The radius of the drop is half the diameter, so \( r = \frac{1.20 \mu m}{2} = 0.60 \mu m = 0.60 \times 10^{-6} m \). Therefore,\[ V = \frac{4}{3} \pi (0.60 \times 10^{-6})^3 \approx 9.05 \times 10^{-19} \text{ m}^3. \]
02

Calculate the Gravitational Force on the Water Drop

To find the force, we first need the mass of the water drop. Using the density of water \( \rho = 1000 \text{ kg/m}^3 \), the mass \( m \) is calculated by \( m = \rho \cdot V \). Substituting, we get \( m = 1000 \times 9.05 \times 10^{-19} = 9.05 \times 10^{-16} \text{ kg} \). The gravitational force \( F_g \) is computed as \( F_g = m \cdot g \), where \( g \approx 9.81 \text{ m/s}^2 \): \[ F_g = 9.05 \times 10^{-16} \times 9.81 \approx 8.88 \times 10^{-15} \text{ N}. \]
03

Calculate the Electric Force on the Water Drop

The water drop is in equilibrium, meaning the electric force \( F_e \) must be equal in magnitude to the gravitational force \( F_g \). Thus, \( F_e = 8.88 \times 10^{-15} \text{ N} \). By definition, \( F_e = |q|E \), where \(|q|\) is the charge on the drop and \(E\) is the electric field strength, 462 N/C. Therefore, \(|q| = \frac{8.88 \times 10^{-15}}{462} \approx 1.92 \times 10^{-17} \text{ C} \).
04

Determine the Number of Excess Electrons

Each electron has a charge of \( 1.60 \times 10^{-19} \text{ C} \). The number of excess electrons \( n \) on the water drop is calculated by dividing the total charge by the charge of a single electron: \[ n = \frac{|q|}{1.60 \times 10^{-19}} = \frac{1.92 \times 10^{-17}}{1.60 \times 10^{-19}} \approx 120 \text{ electrons}. \]

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Key Concepts

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

Gravitational Force and Its Relation to Mass
Gravitational force is a fundamental force that acts between two bodies with mass. In our example, we have a tiny water drop suspended in calm air, and this drop experiences a gravitational pull towards the Earth. The magnitude of this force depends on two main factors: the mass of the object (in this case, the water drop) and the acceleration due to gravity.

To compute this force, we used the equation:
  • \( F_g = m \cdot g \)
Here, \(m\) represents the mass of the water drop, and \(g\) is the acceleration due to gravity, approximately 9.81 m/s虏 on Earth. Since the water drop is spherical, its mass was initially determined from its volume and the known density of water. Once we have its mass, calculating the gravitational force becomes straightforward, revealing the force's pulling power on the small water drop. This gravitational force has to be perfectly balanced by another force to keep the drop suspended.
Understanding Electrical Charge
Electrical charge is a property of particles that cause them to experience a force when placed in an electric field. In our scenario, the water drop has a small charge due to the presence of excess electrons. This charge is essential because it interacts with the Earth's atmospheric electric field to provide the balancing force that holds the water drop in place.

The electric force on the charge is computed using:
  • \( F_e = |q|E \)
where \(|q|\) is the magnitude of the charge, and \(E\) is the electric field strength. Thus, the product of these two gives the electric force that counters the downward gravitational pull. Since each electron carries a charge of \(1.60 \times 10^{-19} \text{ C}\), knowing the total charge \(|q|\) allows us to determine the number of excess electrons, further highlighting how minuscule charges create significant effects.
Equilibrium of Forces: Balancing the Water Drop
Equilibrium of forces is a condition where all forces acting on an object are balanced, resulting in a net force of zero. In our example with the suspended water drop, equilibrium is reached when the downward gravitational force is exactly countered by an upward electric force. This balance is vital to keep the drop from moving either upward or downward.

This balanced state is represented mathematically by:
  • \( F_g = F_e \)
When the gravitational force \( F_g \) and the electric force \( F_e \) are equal in magnitude but opposite in direction, the forces cancel each other out, and the water drop remains stationary. Understanding this equilibrium concept helps us grasp how objects can be held in place by forces that are ever-present and all around us, ensuring that the water drop "floats" in the air with the aid of charged particles.

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Most popular questions from this chapter

Three particles, each with positive charge \(Q,\) form an equilateral triangle, with each side of length \(d\). What is the magnitude of the electric field produced by the particles at the midpoint of any side?

The nucleus of a plutonium- 239 atom contains 94 protons. Assume that the nucleus is a sphere with radius \(6.64 \mathrm{fm}\) and with the charge of the protons uniformly spread through the sphere. At the surface of the nucleus, what are the (a) magnitude and (b) direction (radially inward or outward) of the electric field produced by the protons?

Two charged particles are fixed to an \(x\) axis: Particle 1 of charge \(q_{1}=2.1 \times 10^{-8} \mathrm{C}\) is at position \(x=20 \mathrm{~cm}\) and particle 2 of charge \(q_{2}=-4.00 q_{1}\) is at position \(x=70 \mathrm{~cm} .\) At what coordinate on the axis (other than at infinity) is the net electric field produced by the two particles equal to zero?

A charge of \(20 \mathrm{nC}\) is uniformly distributed along a straight rod of length \(4.0 \mathrm{~m}\) that is bent into a circular arc with a radius of \(2.0 \mathrm{~m} .\) What is the magnitude of the electric field at the center of curvature of the arc?

An electron enters a region of uniform electric field with an initial velocity of \(40 \mathrm{~km} / \mathrm{s}\) in the same direction as the electric field, which has magnitude \(E=50 \mathrm{~N} / \mathrm{C}\). (a) What is the speed of the electron \(1.5 \mathrm{~ns}\) after entering this region? (b) How far does the electron travel during the \(1.5 \mathrm{~ns}\) interval?
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