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Chapter 22: Problem 81

An electric field \(\vec{E}\) with an average magnitude of about \(150 \mathrm{~N} / \mathrm{C}\) points downward in the atmosphere near Earth's surface. We wish to "float" a sulfur sphere weighing \(4.4 \mathrm{~N}\) in this field by charging the sphere. (a) What charge (both sign and magnitude) must be used? (b) Why is the experiment impractical?

Short Answer

Expert verified
(a) +0.0293 C needed. (b) Large charge is impractical to maintain.

Step by step solution

01

Understanding the Force Balance

For the sphere to "float," the electric force acting on the charged sphere must exactly counteract the gravitational force. This means that the upward electric force, which is equal to the charge on the sphere times the electric field strength, must equal the downward gravitational force, which is the weight of the sphere.
02

Apply the Electric Force Formula

The electric force can be calculated using the formula \( F_e = q \cdot E \), where \( F_e \) is the electric force, \( q \) is the charge on the sphere, and \( E \) is the electric field magnitude. Since we want the electric force to counterbalance the gravitational force, we set \( q \cdot E = W \), where \( W = 4.4 \mathrm{~N} \).
03

Solve for the Charge \( q \)

Rearrange the equation \( q \cdot E = W \) to solve for \( q \). We have \[ q = \frac{W}{E} = \frac{4.4 \mathrm{~N}}{150 \mathrm{~N/C}} = 0.0293 \mathrm{~C}. \] Since the electric field points downward, a positive charge must be used to create an upward force.
04

Determine the Impracticality

The experiment is impractical because it requires a very large charge of approximately \(0.0293 \mathrm{~C} \), which is difficult to maintain static on a small object like a sphere without causing discharge or ionization in the surrounding air.

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Key Concepts

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

Electric Field
The electric field is a region around a charged object where other charges experience a force. In this exercise, the electric field has an average magnitude of about 150 N/C and points downward near Earth's surface. This is a common direction for electric fields due to the atmospheric electrical environment.
  • Electric fields are vector fields, meaning they have both magnitude and direction.
  • The direction of the field is the direction of the force a positive test charge would experience.
  • Magnitude is measured in newtons per coulomb (N/C).
Understanding the electric field is vital because it determines how a charged particle, like our sulfur sphere, interacts with its environment. In this case, we use the field to oppose the gravitational force and allow the sphere to "float."
Gravitational Force
The gravitational force is the force acting between two masses due to gravity. On Earth, this force is what gives objects weight, pulling them downward towards the center of the planet. For our sulfur sphere, it weighs 4.4 N, which directly ties into how much electric force is needed.
  • The gravitational force can be determined using the formula: \( F_g = m \cdot g \).
  • Here, \( F_g \) is the gravitational force, \( m \) is the mass of the object, and \( g \approx 9.8 \mathrm{~m/s^2} \) is the acceleration due to Earth's gravity.
  • It's an ever-present force that must be considered when balancing it with electrostatic forces.
Understanding gravitational force helps frame why we need an electric force of equal magnitude for the sphere to counterbalance this natural pull and "float."
Charge Calculation
The charge calculation is the process of determining the amount of charge required to balance another force鈥攊n this case, the gravitational force. We use the balance of forces formula to solve for the charge, \( q \), needed to "float" the sphere against gravity.
  • Electric force is given by \( F_e = q \cdot E \), where \( E \) is the electric field magnitude.
  • To float the sphere, \( F_e \) needs to be equal to its weight, or gravitational force, \( W = 4.4 \mathrm{~N} \).
  • The calculation: \[ q = \frac{W}{E} = \frac{4.4 \mathrm{~N}}{150 \mathrm{~N/C}} = 0.0293 \mathrm{~C}. \]
Understanding this helps us see the direct relationship between the charge required on the sphere and the intensity of the electric field it exists in.
Electrostatics
Electrostatics is the study of electric charges at rest. This includes how charged particles interact, the forces between them, and the fields they produce. In this scenario, electrostatic principles help explain how the sphere can be "floated" by counterbalancing gravitational forces with electric forces.
  • Charged objects exert forces and can influence other charges within their electric field.
  • The strength of electrostatic interactions decreases with distance and increases with the amount of charge involved.
  • While theoretically possible, maintaining a charge of 0.0293 C on a small sphere is difficult practically due to discharge and ionization risks.
Grasping electrostatic concepts is key to understanding how and why charged materials behave as they do in different environments.

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Most popular questions from this chapter

Figure 22-37 shows two charged particles on an \(x\) axis: \(-\bar{q}=\) \(-3.20 \times 10^{-19} \mathrm{C}\) at \(x=-3.00 \mathrm{~m}\) and \(q=3.20 \times 10^{-19} \mathrm{C}\) at \(x=+3.00 \mathrm{~m}\) What are the (a) magnitude and (b) direction (relative to the positive direction of the \(x\) axis) of the net electric field produced at point \(P\) at \(y=4.00 \mathrm{~m} ?\)

The following table gives the charge seen by Millikan at different times on a single drop in his experiment. From the data, calculate the elementary charge \(e\). $$ \begin{array}{lll} \hline 6.563 \times 10^{-19} \mathrm{C} & 13.13 \times 10^{-19} \mathrm{C} & 19.71 \times 10^{-19} \mathrm{C} \\ 8.204 \times 10^{-19} \mathrm{C} & 16.48 \times 10^{-19} \mathrm{C} & 22.89 \times 10^{-19} \mathrm{C} \\ 11.50 \times 10^{-19} \mathrm{C} & 18.08 \times 10^{-19} \mathrm{C} & 26.13 \times 10^{-19} \mathrm{C} \\ \hline \end{array} $$

Find an expression for the oscillation frequency of an electric dipole of dipole moment \(\vec{p}\) and rotational inertia \(I\) for small amplitudes of oscillation about its equilibrium position in a uniform electric field of magnitude \(E\).

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?

An electron with a speed of \(5.00 \times 10^{8} \mathrm{~cm} / \mathrm{s}\) enters an electric field of magnitude \(1.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\), traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed? (c) If the region containing the electric field is \(8.00 \mathrm{~mm}\) long (too short for the electron to stop within it), what fraction of the electron's initial kinetic energy will be lost in that region?
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