/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 The earth has a net electric cha... [FREE SOLUTION] | 91Ó°ÊÓ

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The earth has a net electric charge that causes a field at points near its surface equal to \(150 \mathrm{~N} / \mathrm{C}\) and directed in toward the center of the earth. (a) What magnitude and sign of charge would a \(60-\mathrm{kg}\) human have to acquire to overcome his or her weight by the force exerted by the earth's electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of \(100 \mathrm{~m}\) ? Is use of the earth's electric field a feasible means of flight? Why or why not?

Short Answer

Expert verified
(a) The charge needed is \(+3.92 \mathrm{~C}\). (b) The repulsive force is \(13.9 \times 10^6 \mathrm{~N}\). Flight via Earth's field is not feasible due to extreme charge.

Step by step solution

01

Understand the Force Due to Electric Field

Given that the electric field (E) from Earth is \(150 \mathrm{~N/C}\), we need to find the charge (q) required by a person to have an upward force equal to their weight. The formula for electric force is \( F_e = qE \), where \( F_e \) is the force due to the electric field.
02

Calculate the Weight

The weight of the person is the gravitational force, \( F_g \), given by \( F_g = mg \). Here, \( m = 60 \mathrm{~kg} \) and \( g = 9.8 \mathrm{~m/s^2} \), so \( F_g = 60 \times 9.8 = 588 \mathrm{~N} \).
03

Equate Forces to Find Charge

To overcome the person's weight using the electric field, set \( F_e = F_g \), so \( qE = mg \). Solving for \( q \), we have \( q = \frac{mg}{E} = \frac{588}{150} \). Thus, \( q \approx 3.92 \mathrm{~C} \).
04

Determine the Sign of the Charge

Since the electric field is directed inward, the charge must be positive to exert an upward force. Thus, the charge is \(+3.92 \mathrm{~C}\).
05

Calculate Force of Repulsion

Using Coulomb's Law, the force between two charges is \( F = \frac{k|q_1 q_2|}{r^2} \), where \( k = 8.99 \times 10^9 \mathrm{~N\cdot m^2/C^2} \). Setting \( q_1 = q_2 = 3.92 \mathrm{~C} \) and \( r = 100 \mathrm{~m} \), we get \[ F = \frac{8.99 \times 10^9 \times (3.92)^2}{100^2} \approx 13.9 \times 10^6 \mathrm{~N} \].
06

Evaluate Feasibility for Flight

The required charge of \(+3.92 \mathrm{~C}\) is extremely high and impractical to acquire due to safety and technological limitations, making this means of flight infeasible.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is fundamental for understanding the interactions between electric charges. It describes the force between two point charges. The formula is:
  • \( F = \frac{k|q_1 q_2|}{r^2} \)
This means the force \( F \) is proportional to the product of the magnitudes of the charges, \( q_1 \) and \( q_2 \), inversely proportional to the square of the distance \( r \) between them, and dependent on the constant \( k \), known as Coulomb's constant (\(8.99 \times 10^9 \text{ N }\cdot\text{m}^2/\text{C}^2\)).
The force can be attractive or repulsive, depending on the signs of the charges. Like charges repel, while opposite charges attract. Understanding this law is crucial for solving problems related to electric forces, such as how two people would repel each other if they both had the same charge.
Gravitational Force
Gravitational force is the force of attraction between two masses. On Earth, this force gives us our weight. The weight of an object is calculated by multiplying its mass by the acceleration due to gravity, usually denoted \(g\).
For a person weighing 60 kg on Earth, the gravitational force is:
  • \( F_g = mg = 60 \times 9.8 = 588 \text{ N} \)
This formula highlights the direct relationship between mass and gravitational force.
While solving physics problems, understanding gravitational force is essential, especially when considering other forces like electric force that might oppose or counteract gravitational pull.
Electric Force
Electric force is the force exerted by an electric field on a charge. The magnitude of this force is given by the formula:
  • \( F_e = qE \)
Where \( F_e \) is the electric force, \( q \) is the charge, and \( E \) is the electric field.
In this exercise, the electric field from Earth is \(150 \text{ N/C}\). For a person to counteract their weight using this electric field, the electric force must equal their gravitational force.
Hence:
  • \( qE = mg \)
This balance of forces allows the calculation of the charge required to overcome gravitational force, highlighting the interplay between different physical forces.
Charge Calculation
Calculating charge is crucial for determining how much charge is needed to achieve a certain electric force. From the given exercise, to overcome the gravitational force with an electric force, we compute:
  • \( q = \frac{mg}{E} = \frac{588}{150} \approx 3.92 \text{ C} \)
This result shows the amount of charge required for a 60 kg person to levitate in Earth's electric field.
It's also important to consider the sign of the charge. Because the Earth's electric field is directed inward, a positive charge would produce an upward force, essential for levitation.
However, this calculated charge is unrealistically high for everyday use, demonstrating that using such a method for flight is not feasible with current technology.

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Most popular questions from this chapter

A \(+2.00-n C\) point charge is at the origin, and a second \(-5.00-n C\) point charge is on the \(x\)-axis at \(x=0.800 \mathrm{~m} .\) (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\)-axis: (i) \(x=0.200 \mathrm{~m}\); (ii) \(x=-0.200 \mathrm{~m}\). (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

A thin non-conducting ring of radius \(R\) has a linear charge density \(\lambda=\lambda_{0} \cos \theta\), where \(\lambda_{0}\) is the value of \(\lambda\) at \(\theta=0\). Find net electric dipole moment for this charge distribution.

Two very large parallel sheets are \(5.00 \mathrm{~cm}\) apart. Sheet A carries a uniform surface charge density of \(-9.50 \mu \mathrm{C} / \mathrm{m}^{2}\), and sheet \(B\), which is to the right of \(A\), carries a uniform charge density of \(-11.6 \mu \mathrm{C} / \mathrm{m}^{2}, .\) Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) \(4.00 \mathrm{~cm}\) to the right of sheet \(A ;\) (b) \(4.00 \mathrm{~cm}\) to the left of sheet \(A ;\) (c) \(4.00 \mathrm{~cm}\) to the right of sheet \(B\).of the net force that these wires exert on it?

A negative charge of \(-0.550 \mu \mathrm{C}\) exerts an upward \(0.200-\mathrm{N}\) force on an unknown charge \(0.300 \mathrm{~m}\) directly below it. (a) What is the unknown charge (magnitude and sign)? (b) What are the magnitude and direction of the force that the unknown charge exerts on the \(-0.550-\mu \mathrm{C}\) charge?

Three point charges are arranged on a line. Charge \(q_{3}=+5.00 \mathrm{nC}\) and is at the origin. Charge \(q_{2}=-3.00 \mathrm{nC}\) and is at \(x=+4.00 \mathrm{~cm}\). Charge \(q_{1}\) is at \(x=+2.00 \mathrm{~cm}\). What is \(q_{1}\) (magnitude and sign) if the net force on \(q_{3}\) is zero?

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