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Chapter 18: Problem 7

Two charges attract each other with a force of \(1.5 \mathrm{~N}\). What will be the force if the distance between them is reduced to one-ninth of its original value?

Short Answer

Expert verified
The new force is 121.5 N.

Step by step solution

01

Understanding Coulomb's Law

Coulomb's Law states that the force between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. It is given by the formula: \( F = k \frac{q_1 q_2}{r^2} \), where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
02

Initial Force Expression

Let's consider the initial force: \( F_1 = k \frac{q_1 q_2}{r^2} = 1.5 \text{ N} \). This expression gives us the force exerted at the original distance \( r \).
03

New Distance Calculation

The problem states that the distance is reduced to one-ninth of its original value, so the new distance \( r' = \frac{r}{9} \).
04

New Force Expression

Substituting \( r' \) into Coulomb's Law, the new force \( F_2 = k \frac{q_1 q_2}{(r/9)^2} \). Simplifying this gives \( F_2 = k \frac{q_1 q_2}{r^2/81} = 81 \times k \frac{q_1 q_2}{r^2} = 81F_1 \).
05

Calculate the New Force

Given that \( F_1 = 1.5 \text{ N} \), the new force \( F_2 = 81 \times 1.5 \text{ N} = 121.5 \text{ N} \).

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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 electromagnetism, describing the interaction between two charged objects. It arises due to the presence of charges and is either attractive or repulsive, depending on the nature of the charges involved. Coulomb's Law helps us quantify this force. When two charges with the same sign repel each other, the force is pushing them apart. Conversely, when their signs differ, they attract each other, pulling them closer.

The electric force plays a crucial role in chemical bonding, atomic structure, and even in everyday phenomena like static electricity. Its magnitude and direction rely on the respective charges involved and their separation distance.
Point Charges
In physics, a point charge is considered to be a charged object that is small enough such that its size does not affect the analysis of the problem at hand. Essentially, a point charge is an idealized model where all the charge of an object is concentrated at a single point.

Point charges are crucial when using Coulomb's Law to calculate electric forces because they simplify the geometry of the situation. When dealing with point charges, attention is focused purely on the magnitude of the charges and the distance separating them, without worrying about the physical dimensions or shape of the objects.
Inverse Square Law
The inverse square law is a principle stating that a specified quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity. In the context of Coulomb's Law, the electric force between two point charges decreases with the square of the distance between them.

This means if the distance between the charges is doubled, the electric force becomes one fourth as strong. Likewise, reducing the distance brings a significant increase in force. This law illustrates why interactions such as gravitational, acoustic, and electric forces weaken as distance increases.
  • The further apart two charges are, the weaker the force.
  • The closer they are, the stronger the attractive or repulsive force.
Distance and Force Relationship
Understanding the relationship between distance and force is vital in problems involving Coulomb's Law. This relationship is mathematically modeled using the inverse square law, as previously discussed. In applied problems, changing the distance between charges directly affects the magnitude of the electric force.

If the distance between two point charges is reduced to one-ninth of its original value, the force becomes dramatically stronger. Specifically, under these conditions, the force increases by a factor of 81, as the exercise example showcases. This direct relationship helps predict and understand the effects of spatial changes in charge interactions.
  • Reducing distance increases force exponentially.
  • Increasing distance decreases force exponentially.

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Most popular questions from this chapter

A rectangular surface \((0.16 \mathrm{~m} \times 0.38 \mathrm{~m})\) is oriented in a uniform electric field of \(580 \mathrm{~N} / \mathrm{C}\). What is the maximum possible electric flux through the surface?

Concept Questions Two identical metal spheres have charges of \(q_{1}\) and \(q_{2}\). They are brought together so they touch, and then they are separated. (a) How is the net charge on the two spheres before they touch related to the net charge after they touch? (b) After they touch and are separated, is the charge on each sphere the same? Why? Problem Four identical metal spheres have charges of \(q_{\mathrm{A}}=-8.0 \mu \mathrm{C}, q_{\mathrm{B}}=-2.0 \mu \mathrm{C}\) \(q_{\mathrm{C}}=+5.0 \mu \mathrm{C}\), and \(q_{D}=+12.0 \mu \mathrm{C}\). (a) Two of the spheres are brought together so they touch and then they are separated. Which spheres are they, if the final charge on each of the two is \(+5.0 \mu \mathrm{C} ?(\mathrm{~b})\) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is \(+3.0 \mu \mathrm{C}\) (c) How many electrons would have to be added to one of the spheres in part (b) to make it electrically neutral?

Two identical metal spheres have charges of \(q_{1}\) and \(q_{2}\). They are brought together so they touch, and then they are separated. (a) How is the net charge on the two spheres before they touch related to the net charge after they touch? (b) After they touch and are separated, is the charge on each sphere the same? Why? Four identical metal spheres have charges of \(q_{\mathrm{A}}=-8.0 \mu \mathrm{C}, q_{\mathrm{B}}=-2.0 \mu \mathrm{C}\) \(q_{\mathrm{C}}=+5.0 \mu \mathrm{C},\) and \(q_{\mathrm{D}}=+12.0 \mu \mathrm{C} .\) (a) Two of the spheres are brought together so they touch and then they are separated. Which spheres are they, if the final charge on each of the two is \(+5.0 \mu \mathrm{C} ?\) (b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is \(+3.0 \mu \mathrm{C}\) (c) How many electrons would have to be added to one of the spheres in part (b) to make it electrically neutral?

The drawing shows a positive point charge \(+q_{1}\), a second point charge \(q_{2}\) that may be positive or negative, and a spot labeled \(P\), all on the same straight line. The distance \(d\) between the two charges is the same as the distance between \(q_{1}\) and the point \(P\). With \(q_{2}\) present, the magnitude of the net electric field at \(P\) is twice what it is when \(q_{1}\) is present alone. (a) When the second charge is positive, is its magnitude smaller than, equal to, or greater than the magnitude of \(q_{1}\) ? Explain your reasoning. (b) When the second charge is negative, is its magnitude smaller than, equal to, or greater than that in question (a)? Account for your answer.

\(\operatorname{ssm}\) A cube is located with one corner at the origin of an \(x, y, z\) coordinate system. One of the cube's faces lies in the \(x, y\) plane, another in the \(y, z\) plane, and another in the \(x, z\) plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are \(0.20 \mathrm{~m}\) long. A uniform electric field is parallel to the \(x, y\) plane and points in the direction of the \(+y\) axis. The magnitude of the field is \(1500 \mathrm{~N} / \mathrm{C}\). (a) Find the electric flux through each of the six faces of the cube. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss' law predicts, since there is no net charge within the cube
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