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Chapter 21: Problem 53

What would be the magnitude of the electrostatic force between two \(1.00 \mathrm{C}\) point charges separated by a distance of (a) \(1.00 \mathrm{~m}\) and (b) \(1.00 \mathrm{~km}\) if such point charges existed (they do not) and this configuration could be set up?

Short Answer

Expert verified
(a) The force is \(8.99 \times 10^9 \mathrm{~N}\). (b) The force is \(8.99 \times 10^3 \mathrm{~N}\).

Step by step solution

01

Identify the relevant formula

The magnitude of the electrostatic force between two point charges is given by Coulomb's Law, which is: \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \). Here, \( k \) is Coulomb's constant \( 8.99 \times 10^9 \mathrm{~N~m^2/C^2} \), \( q_1 \) and \( q_2 \) are the charges, each \( 1.00 \mathrm{~C} \), and \( r \) is the separation distance.
02

Calculate the force for distance 1 meter

Using the formula from Step 1 with \( r = 1.00 \mathrm{~m} \), substitute in the given values: \( F = \frac{8.99 \times 10^9 \cdot |1.00 \cdot 1.00|}{(1.00)^2} = 8.99 \times 10^9 \mathrm{~N} \).
03

Calculate the force for distance 1 kilometer

Convert the distance to meters: \( 1.00 \mathrm{~km} = 1000 \mathrm{~m} \). Use the formula with \( r = 1000 \mathrm{~m} \): \( F = \frac{8.99 \times 10^9 \cdot |1.00 \cdot 1.00|}{(1000)^2} = \frac{8.99 \times 10^9}{10^6} = 8.99 \times 10^3 \mathrm{~N} \).

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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 at the core of understanding the electrostatic force between two point charges. It helps us quantify the attractive or repulsive force due to electric charges. The law is expressed mathematically as:\[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:
  • \( F \) is the magnitude of the force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \mathrm{~N~m^2/C^2} \).
  • \( q_1 \) and \( q_2 \) are the values of the two charges in Coulombs.
  • \( r \) is the distance between the centers of the two charges in meters.
This formula shows that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Thus, any change in the magnitudes of the charges or the distance between them greatly affects the force experienced.
Point Charges
Point charges are idealized models used in physics to simplify complex electric interactions. When dealing with real-world problems, point charges help focus on the basic principles of electrostatics without the complexity of considering the physical size or shape of the objects. These charges are treated as if all their charge is concentrated at a single point in space, making calculations for electrostatic forces more straightforward. The use of point charges assumes no spatial extent, streamlining the application of Coulomb's Law. In actuality, such high-magnitude or isolated charges do not exist due to electrostatic limits, but the concept serves as a crucial stepping stone in mastering electric force computations and understanding ion interaction at a more abstract level.
Distance and Force Relationship
The distance between two point charges is critical in determining the magnitude of the electrostatic force between them. In Coulomb's Law, the force \( F \) is inversely proportional to the square of the distance \( r \) between the charges. This relationship is expressed in the formula as:\[ F \propto \frac{1}{r^2} \]This means that as the distance \( r \) increases, the force \( F \) decreases rapidly. If the distance is doubled, the force becomes one-fourth of its original value. Conversely, if the distance is halved, the force increases by four times.Understanding this relationship is essential for grasping the effects of distance on electrostatic interactions, highlighting how electrical forces diminish with increasing separation. The exercise exemplifies this when comparing the forces at distances of 1 meter and 1 kilometer, illustrating how dramatically the force decreases over larger distances.

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Most popular questions from this chapter

An electron is in a vacuum near Earth's surface and located at \(y=0\) on a vertical \(y\) axis. At what value of \(y\) should a second electron be placed such that its electrostatic force on the first electron balances the gravitational force on the first electron?

Electrons and positrons are produced by the nuclear transformations of protons and neutrons known as beta decay. (a) If a proton transforms into a neutron, is an electron or a positron produced? (b) If a neutron transforms into a proton, is an electron or a positron produced?

If a cat repeatedly rubs against your cotton slacks on a dry day, the charge transfer between the cat hair and the cotton can leave you with an excess charge of \(-2.00 \mu \mathrm{C}\). (a) How many electrons are transferred between you and the cat? You will gradually discharge via the floor, but if instead of waiting, you immediately reach toward a faucet, a painful spark can suddenly appear as your fingers near the faucet. (b) In that spark, do electrons flow from you to the faucet or vice versa? (c) Just before the spark appears, do you induce positive or negative charge in the faucet? (d) If, instead, the cat reaches a paw toward the faucet, which way do electrons flow in the resulting spark? (e) If you stroke a cat with a bare hand on a dry day, you should take care not to bring your fingers near the cat's nose or you will hurt it with a spark. Considering that cat hair is an insulator, explain how the spark can appear.

What must be the distance between point charge \(q_{1}=26.0 \mu \mathrm{C}\) and point charge \(q_{2}=-47.0 \mu \mathrm{C}\) for the electrostatic force between them to have a magnitude of \(5.70 \mathrm{~N} ?\)

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters (Fig. \(21-22 a\) ). The electrostatic force acting on sphere 2 due to sphere 1 is \(\vec{F}\). Suppose now that a third identical sphere 3 , having an insulating handle and initially neutral, is touched first to sphere 1 (Fig. \(21-22 b\) ), then to sphere 2 (Fig. \(21-22 c\) ), and finally removed (Fig. \(21-22 d\) ). The electrostatic force that now acts on sphere 2 has magnitude \(F^{\prime} .\) What is the ratio \(F^{\prime} / F ?\)
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