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Chapter 21: Problem 6

(II) Two charged dust particles exert a force of \(3.2 \times 10^{-2} \mathrm{~N}\) on each other. What will be the force if they are moved so they are only one-eighth as far apart?

Short Answer

Expert verified
The new force is 2.048 N.

Step by step solution

01

Understanding the Coulomb's Law Formula

The force between two charges can be calculated using Coulomb's Law, which states that the force is inversely proportional to the square of the distance between the charges. The formula is given by:\[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force between the charges, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.
02

Relating Initial and New Distances

Initially, we have a force of \( F_i = 3.2 \times 10^{-2} \text{ N} \) at a distance \( r_i \). The problem states that the distance becomes one-eighth of the original, so the new distance \( r_f \) is \( \frac{r_i}{8} \).
03

Applying the Distance Change to Coulomb's Law

Since the force is inversely proportional to the square of the distance, we have:\[ F_f = k \frac{|q_1 q_2|}{\left(\frac{r_i}{8}\right)^2} = k \frac{|q_1 q_2|}{\frac{r_i^2}{64}} = 64 \cdot (k \frac{|q_1 q_2|}{r_i^2}) \]This can be simplified to \( F_f = 64F_i \).
04

Calculating the New Force

Plug the initial force into the equation to find the new force:\[ F_f = 64 \times 3.2 \times 10^{-2} \]\[ F_f = 204.8 \times 10^{-2} \]\[ F_f = 2.048 \text{ N} \]
05

Finalizing the Result

We can conclude that the new force when the particles are one-eighth as far apart is \( 2.048 \text{ N} \).

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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 the push or pull experienced between electrically charged objects. Think of electric force as similar to gravity, but instead of mass attracting masses, charges attract or repel other charges. This fundamental interaction is described by the famous Coulomb's Law, which gives us a mathematical way to calculate the amount of force between charged particles. When like charges (both positive or both negative) are near each other, they repel. When opposite charges (one positive, one negative) meet, they attract. This force can be incredibly strong, much stronger than gravity on a small scale. It plays a crucial role in how particles behave and is a cornerstone of electromagnetism.
Inverse Square Law
The inverse square law is a crucial concept in understanding how electric force behaves. According to this law, the force between two charged particles is inversely proportional to the square of the distance between them. This means as distance \( r \) increases, the force \( F \) decreases quite rapidly, specifically following the relationship \[ F \propto \frac{1}{r^2} \]. So, if the distance between particles doubles, the force between them is reduced to a quarter of its original value. This law isn't exclusive to electricity; it also applies to other forces, like gravity and light intensity. The main takeaway is that small changes in distance can have a big effect on the force experienced.
Charged Particles
Charged particles are simply particles that carry an electric charge. They can be either positively charged, like protons, or negatively charged, like electrons. Their behavior is explained by Coulomb's Law, which tells us how these particles interact with each other. Knowing how to calculate the electric force between charged particles is important for understanding many physical phenomena. For example, in atoms, electrons (negatively charged) are bound to nuclei (positively charged) by electric force. This is what keeps the atom from falling apart and making chemistry possible. Understanding charged particles helps us predict outcomes in experiments involving electricity and magnetism.
Distance and Force Relationship
The distance and force relationship is straightforward yet powerful in the realm of electric forces. As per Coulomb's Law, the electric force depends greatly on the distance between two charged particles. The relationship is such that when the distance is reduced, the force increases significantly. If two charged particles are brought closer by a factor of \( \frac{1}{8} \), as seen in the original exercise, the force increases by a factor of 64 because \( (\frac{1}{\frac{1}{8}})^2 = 64 \). This quadratic change underscores the sensitivity of electric forces to distance. This principle is key in designing and understanding electrical systems, as well as in fields like electrostatics and electronics.

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Most popular questions from this chapter

Four equal positive point charges, each of charge \(8.0 \mu \mathrm { C } ,\) are at the corners of a square of side 9.2\(\mathrm { cm } .\) What charge should be placed at the center of the square so that all charges are at equilibrium? Is this a stable or unstable equilibrium in the plane?

(I) A proton is released in a uniform electric field, and it experiences an electric force of \(2.18 \times 10^{-14} \mathrm{~N}\) toward the south. What are the magnitude and direction of the electric field?

(II) Compare the electric force holding the electron in orbit \(\left( r = 0.53 \times 10 ^ { - 10 } \mathrm { m } \right)\) around the proton nucleus of the hydrogen atom, with the gravitational force between the same electron and proton. What is the ratio of these two forces?

(III) A thin glass rod is a semicircle of radius \(R\), Fig. \(21-66\). A charge is nonuniformly distributed along the rod with a linear charge density given by \(\lambda=\lambda_{0} \sin \theta,\) where \(\lambda_{0}\) is a positive constant. Point \(\mathbf{P}\) is at the center of the semicircle. (a) Find the electric field \(\overrightarrow{\mathbf{E}}\) (magnitude and direction) at point P. [Hint: Remember \(\sin (-\theta)=-\sin \theta,\) so the two halves of the rod are oppositely charged.] (b) Determine the acceleration \(\quad\) (magnitude \(\quad\) and direction) of an electron placed at point \(\mathrm{P},\) assuming \(\quad R=1.0 \mathrm{~cm}\) and \(\lambda_{0}=1.0 \mu \mathrm{C} / \mathrm{m}\)

Two small, identical conducting spheres \(A\) and \(B\) are a distance \(R\) apart; each carries the same charge \(Q . ( a )\) What is the force sphere B exerts on sphere A? (b) An identical sphere with zero charge, sphere \(C\) , makes contact with sphere \(B\) and is then moved very far away. What is the net force now acting on sphere A? (c) Sphere \(C\) is brought back and now makes contact with sphere \(A\) and is then moved far away. What is the force on sphere \(A\) in this third case?
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