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\(\bullet$$\bullet\) 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 -N weight?

Short Answer

Expert verified
They must be approximately 3716 m apart for the electric attraction to equal 650 N.

Step by step solution

01

Understand the Problem

We have two humans each with a 1.0 C charge, one positive and one negative. We need to find out how far apart they need to be so that the electric force between them equals 650 N, which is the gravitational force equivalent to their weight.
02

Use Coulomb's Law

Coulomb's Law is given by the equation \( F = k \frac{|q_1 q_2|}{r^2} \), where \( F \) is the force between two charges, \( k = 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. In this problem, \( q_1 = 1.0 \text{ C} \), \( q_2 = -1.0 \text{ C} \), so \( |q_1 q_2| = 1.0 \text{ C}^2 \). We set \( F = 650 \text{ N} \).
03

Solve for the Distance \( r \)

Rearrange the formula to solve for \( r \): \[ r^2 = k \frac{|q_1 q_2|}{F} \]. Substitute the known values: \[ r^2 = 8.99 \times 10^9 \frac{1.0}{650} \]. Calculate \( r^2 \): \( r^2 = 1.382 \times 10^7 \).
04

Calculate \( r \)

Take the square root of both sides to find \( r \): \( r = \sqrt{1.382 \times 10^7} \). Calculate: \( r \approx 3716 \text{ m} \).

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Key Concepts

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

Electric Force
The electric force is a fundamental concept in physics, depicting the force exerted between charged objects. It's a fascinating interaction, acting both as an attractive and repulsive force, depending on the nature of the charges involved. When you have two objects with opposite charges, the electric force is attractive, pulling them towards each other. Conversely, with like charges, they repel. This force operates over a distance, much like gravity, and its strength diminishes with increased separation between the charges.

In this exercise, we explore the electric force between two humans each carrying a 1.0 Coulomb charge—one positive and one negative. The scenario sets them at a distance where this electric attraction equals the gravitational force exerted by their weight. This unique setup allows us to delve into how electric forces scale with distance and charge magnitude, illustrating their power even when compared to everyday forces like weight.
Gravitational Force
Gravitational force is the natural force of attraction experienced between two masses. On Earth, it's responsible for giving us weight and keeping us grounded. It’s described by Newton’s Law of Universal Gravitation, but in everyday circumstances, it's what makes us perceive our weight as the force with which Earth attracts us. This force is always attractive and acts over long distances, drawing masses toward each other.

In the context of this physics problem, the gravitational force that equals 650 N is the weight of an average human. The problem challenges us to find the distance where the electric force between the charged individuals equals this force. This comparison highlights how electric forces can overpower gravitational ones when charges are significant, despite often being overlooked in daily life.
Coulomb's Constant
Coulomb's constant, symbolized as \( k \), is a crucial figure in electrostatics, appearing in Coulomb's Law. It quantifies the proportionality between electric force, charge magnitudes, and distance between charges. With a value of \( 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2 \), it shows the intensity with which electric forces act at a given distance.

This constant plays a vital role in calculating the electric force between two point charges. In the given exercise, knowing Coulomb's constant allows us to bridge the gap between theoretical charge interactions and practical calculations. It's the backbone of determining how intense the force will be for the specific setup of charges. By incorporating this constant into calculations, we precisely determine the distance at which the electric force matches the gravitational pull on the humans.
Physics Problem Solving
Physics problem-solving often involves understanding underlying principles and applying formulas judiciously. In exercises like this one, step-by-step approaches are vital. Here's how we systematically tackled the presented problem:
  • **Understanding the problem:** Identifying that we need the electric force to equal a specific gravitational force and noting variables, like charge and force magnitude.
  • **Using Coulomb’s Law:** Applying the formula \( F = k \frac{|q_1 q_2|}{r^2} \) helps set up the problem in a solvable equation by substituting known values.
  • **Rearranging and solving:** This involves isolating the variable of interest—in this case, distance \( r \)—from the equation through rearrangement and substitution.
  • **Calculating results:** Finally, performing mathematical calculations to arrive at numerical results, ensuring logical consistency throughout.
By following this logical sequence, physics becomes a methodical process of exploration, where we transform theoretical concepts into practical understanding.

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Most popular questions from this chapter

\(\bullet\) During a violent electrical storm, a car is struck by a falling high-voltage wire that puts an excess charge of \(-850 \mu C\) on the metal car. (a) How much of this charge is on the inner sur- face of the car? (b) How much is on the outer surface?

\(\bullet\) In a rectangular coordinate system, a positive point charge \(q=6.00 \mathrm{nC}\) is placed at the point \(x=+0.150 \mathrm{m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{m}, y=0 .\) Find the \(x\) and \(y\) components and the magnitude and direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{m}, y=0 ;\) (c) \(x=0.150 \mathrm{m}, y=-0.400 \mathrm{m},\) (d) \(x=0, y=0.200 \mathrm{m}\)

\(\bullet$$\bullet\) A total charge of magnitude \(Q\) is distributed uniformly within a thick spherical shell of inner radius \(a\) and outer radius b. (a) Use Gauss's law to find the electric field within the cavity \((r \leq a)\) . (b) Use Gauss's law to prove that the electric field outside the shell \((r \geq b)\) is exactly the same as if all the charge were concentrated as a point charge \(Q\) at the center of the sphere. (c) Explain why the result in part (a) for a thick shell is the same as that found in Example 17.10 for a thin shell. A thick shell can be viewed as infinitely many thin shells.)

\(\bullet\) Signal propagation in neurons. Neurons are components of the nervous system of the body that transmit signals as elec- trical impulses travel along their length. These impulses propa- gate when charge suddenly rushes into and then out of a part of the neutron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e\) enter the axon. How many coulombs of charge enter a 1.5 \(\mathrm{cm}\) length of the axon during this process?

\(\bullet\) \(\bullet\) (a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A\) , the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A\) ? (b) A proton is moving in the uniform electric field of part (a). At point \(A\) , the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s},\) east. What is the speed of the proton at point \(B\) ?

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