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Chapter 1: Problem 5

An average human weighs about \(650 \mathrm{~N}\). If two such generic humans each carried \(1.0\) coulomb of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their \(650-\mathrm{N}\) weight?

Short Answer

Expert verified
The humans must be about 3718 meters apart.

Step by step solution

01

Understanding the Forces Involved

The problem is asking for the distance at which the electric attraction between two charged humans equals their weight. The electric force between two charges is given by Coulomb's Law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where \( F \) is the force, \( k \) is Coulomb's constant \( (8.99 \times 10^9\, \mathrm{Nm^2/C^2}) \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
02

Set Up the Equation

We know the electric force \( F \) should equal the weight of the human, which is \( 650 \text{ N} \). Since each human is carrying \( 1.0 \text{ C} \) of charge, we have \( q_1 = 1.0 \text{ C} \) and \( q_2 = -1.0 \text{ C} \). Therefore, the equation becomes:\[ 650 = \frac{8.99 \times 10^9 \cdot |1.0 \cdot (-1.0)|}{r^2} \]
03

Solve for Distance \( r \)

Rearrange the equation to solve for \( r \):\[ r^2 = \frac{8.99 \times 10^9}{650} \]Solve this equation step by step to find \( r \).
04

Calculate \( r^2 \) Value

First, calculate the right-hand side of the equation:\[ r^2 = \frac{8.99 \times 10^9}{650} \approx 1.383 \times 10^7 \text{ m}^2 \]
05

Find \( r \) by Taking the Square Root

To find \( r \), take the square root of \( 1.383 \times 10^7 \text{ m}^2 \): \[ r = \sqrt{1.383 \times 10^7} \approx 3718 \text{ m} \]

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Key Concepts

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

Electric Force
Electric force is a fundamental concept in physics that describes the interaction between charged particles. It's the "push" or "pull" you experience due to electric charges. Imagine walking down a street. If you had a positive electric charge, you might feel pulled towards a negatively charged person. This attraction is called the electric force.
  • The strength of this force depends on the amount of charge each object has and the distance between them.
  • If charges are opposite, they attract; if they are the same, they repel.
  • The size of the force can be calculated using Coulomb's Law.

Electric forces can be quite strong. In the given problem, the electric attraction between two humans carrying excess charges can be as strong as their weight on Earth. This illustrates how powerful electric forces can be when charges are significant.
Coulomb
The term "Coulomb" refers to the SI unit of electric charge. It is named after Charles-Augustin de Coulomb, who did extensive research on electric force. One coulomb is a lot of charge - and it might sound surprising, but everyday objects usually carry much less charge.
  • A single coulomb represents a tremendous amount of charge, akin to the charge transferred by a current of one ampere flowing for one second.
  • Charges are typically measured in smaller units like milliCoulombs (mC) or microCoulombs (碌C) in everyday contexts.
  • Since both humans in our scenario carry 1.0 coulomb each with opposite signs, they attract each other strongly.

The study of Coulombs helps us understand the magnitude of charges involved in electric interactions and how they can result in significant forces even over relatively large distances.
Distance Calculation
Calculating the distance between charged objects where the electric force equals a known force, like weight, involves using Coulomb's Law. Here, we need to arrange the formula to solve for distance:
  • The key formula is: \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
  • In the given exercise, we know the force \( F \) is the weight, 650 N.
  • We solve the equation for \( r \), which simplifies to finding \( r = \sqrt{\frac{k \cdot |q_1 \cdot q_2|}{F}} \).

In the example with the two humans, this calculated distance was about 3718 meters. This demonstrates how dramatically electric forces diminish with increasing distance, showing the significance of having a squared distance in the denominator.

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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).

Two small spheres spaced \(20.0 \mathrm{~cm}\) apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is \(4.57 \times 10^{-21} \mathrm{~N} ?\)

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?

The ammonia molecule \(\left(\mathrm{NH}_{3}\right)\) has a dipole moment of \(5.0 \times 10^{-30} \mathrm{C} \cdot \mathrm{m}\). Ammonia molecules in the gas phase are placed in a uniform electric field \(\overrightarrow{\boldsymbol{E}}\) with magnitude \(1.6 \times\) \(10^{6} \mathrm{~N} / \mathrm{C}\). What is the change in electric potential energy when the dipole moment of a molecule changes its orientation with respect to \(\vec{E}\) from parallel to perpendicular?

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?
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