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Chapter 21: Problem 4

(I) What is the repulsive electrical force between two protons \(4.0 \times 10 ^ { - 15 } \mathrm { m }\) apart from each other in an atomic nucleus?

Short Answer

Expert verified
The repulsive electrical force is approximately 14.4 N.

Step by step solution

01

Understand the Problem

We need to find the repulsive electrical force between two protons that are a given distance apart. The force is determined using Coulomb's Law, which describes the electrostatic force between two charges.
02

Identify the Formula

The formula for Coulomb's Law is \[ F = k \frac{|q_1 q_2|}{r^2} \]where \( F \) is the force between the charges, \( k \) is Coulomb's constant (\( 8.988 \times 10^9 \, \text{N m}^2/\text{C}^2 \)), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
03

Identify the Values

Since both are protons, each has a charge of \( 1.602 \times 10^{-19} \, \text{C} \). The distance \( r \) is given as \( 4.0 \times 10^{-15} \, \text{m} \).
04

Calculate the Force

Substitute the known values into Coulomb's Law:\[ F = 8.988 \times 10^9 \times \frac{(1.602 \times 10^{-19})^2}{(4.0 \times 10^{-15})^2} \].Calculating the above expression gives\[ F \approx 14.4 \, \text{N} \].
05

Interpret the Result

The repulsive force of approximately 14.4 N indicates a very strong electrostatic interaction, due to the small distance and the nature of charges involved.

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Key Concepts

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

Electrostatic Force
Electrostatic force is a force that arises from the interaction between charged particles. Two types of charges exist: positive and negative. This force is central to Coulomb's Law, which quantifies how strongly these charges interact.
In the context of our problem, we deal with protons, each bearing a positive charge. The electrostatic force between these protons is repulsive, as like charges repel each other.
  • Coulomb's Law: It states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
  • Formula: The formula for electrostatic force is given by \( F = k \frac{|q_1 q_2|}{r^2} \). This formula highlights two key aspects: the charges involved (\( q_1 \) and \( q_2 \)) and the distance (\( r \)) between them.
  • Importance: Understanding this force is crucial because it plays a significant role in the structure of atoms and molecules, contributing to chemical bonding and interactions.
Recognizing how electrostatic forces operate helps us understand atomic stability and the idea of electrical fields.
Proton Charge
Protons are subatomic particles found in the nucleus of an atom, characterized by a positive electrical charge. This charge is exactly equal in magnitude but opposite in sign to that of electrons.
The charge of a proton is a fundamental property in physics and chemistry, contributing to the behavior of atoms and molecules.
  • Value: The charge of a single proton is approximately \( 1.602 \times 10^{-19} \) coulombs.
  • Role in Atoms: Protons define the identity of an element. For instance, hydrogen has one proton, while helium has two. The number of protons in an atom鈥檚 nucleus is referred to as its atomic number.
  • Repulsion and Attraction: The fact that protons repel each other due to their like charge explains why they can only be held together in the nucleus by stronger forces like the strong nuclear force.
Understanding proton charge is essential to comprehending how atoms interact and form stable structures.
Distance Between Charges
The distance between charges is a key factor in determining the strength of the electrostatic force between them. According to Coulomb鈥檚 Law, the force becomes weaker as the distance increases.
This inverse square relationship means small changes in distance can lead to significant differences in force.
  • Formula Dependence: In the formula \( F = k\frac{|q_1 q_2|}{r^2} \), \( r \) represents the distance between the charges, emphasizing how this distance inversely affects the force magnitude by a square factor.
  • Physical Interpretation: In the atomic nucleus, protons are extremely close, at distances often measured in femtometers (\( 10^{-15} \) meters), which is why they experience significant repulsive forces.
  • Relevance to Atomic Structure: Despite the repulsion between protons, the strong nuclear force, which acts over short ranges like the distances between nucleons, overcomes this repulsion to bind the nucleus together.
Recognizing the importance of distance in electrostatics aids in understanding various interactions at the microscopic scale, especially in nuclear physics.

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Most popular questions from this chapter

(III) Suppose a uniformly charged wire starts at point 0 and rises vertically along the positive \(y\) axis to a length \(\ell\) . (a) Determine the components of the electric field \(E _ { x }\) and \(E _ { y }\) at point \(( x , 0 ) .\) That is, calculate \(\vec { \mathbf { E } }\) near one end of a long wire, in the plane perpendicular to the wire. (b) If the wire extends from \(y = 0\) to \(y = \infty ,\) so that \(\ell = \infty ,\) show that \(\vec { \mathbf { E } }\) makes a \(45 ^ { \circ }\) angle to the horizontal for any \(x .\) [Hint: See Example 11 of "Electric Charge and Electric Field" andFig. \(29 . ]\)

(III) \(\operatorname{An} 8.00 \mu \mathrm{C}\) charge is on the \(x\) axis of a coordinate system at \(x=+5.00 \mathrm{~cm} .\) A \(-2.00 \mu \mathrm{C}\) charge is at \(x=-5.00 \mathrm{~cm}\) (a) Plot the \(x\) component of the electric field for points on the \(x\) axis from \(x=-30.0 \mathrm{~cm}\) to \(x=+30.0 \mathrm{~cm} .\) The sign of \(E_{x}\) is positive when \(\overrightarrow{\mathbf{E}}\) points to the right and negative when it points to the left. \((b)\) Make a plot of \(E_{x}\) and \(E_{y}\) for points on the \(y\) axis from \(y=-30.0\) to \(+30.0 \mathrm{~cm}\)

(II) An electron has an initial velocity \(\vec { \mathbf { v } } _ { 0 } = \left( 8.0 \times 10 ^ { 4 } \mathrm { m } / \mathrm { s } \right) \hat { \mathbf { j } }\) . It enters a region where \(\quad \vec { \mathbf { E } } = ( 2.0 \hat { \mathbf { i } } + 8.0 \hat { \mathbf { j } } ) \times 10 ^ { 4 } \mathrm { N } / \mathrm { C }\) (a) Determine the vector acceleration of the electron as a function of time. \(( b )\) At what angle \(\theta\) is it moving (relative to its initial direction) at \(t = 1.0 \mathrm { ns } ?\)

A one-dimensional row of positive ions, each with charge \(+Q\) and separated from its neighbors by a distance \(d\) occupies the right-hand half of the \(x\) axis. That is, there is a \(+Q\) charge at \(x=0, x=+d, x=+2 d, x=+3 d,\) and so on out to \(\infty .(a)\) If an electron is placed at the position \(x=-d,\) determine \(F,\) the magnitude of force that this row of charges exerts on the electron. \((b)\) If the electron is instead placed at \(x=-3 d,\) what is the value of \(F ?\) [Hint: The infinite sum \(\sum_{n=1}^{n=\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6},\) where \(n\) is a positive integer.]

(I) How many electrons make up a charge of \(-38.0 \mu \mathrm{C} ?\)
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