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Chapter 21: Problem 5

A particle of charge \(+3.00 \times 10^{-6} \mathrm{C}\) is \(12.0 \mathrm{~cm}\) distant from a second particle of charge \(-1.50 \times 10^{-6} \mathrm{C} .\) Calculate the magnitude of the electrostatic force between the particles.

Short Answer

Expert verified
The magnitude of the electrostatic force is approximately 0.281 N.

Step by step solution

01

Identify given values

The charges of the particles are given as \( q_1 = +3.00 \times 10^{-6} \text{ C} \) and \( q_2 = -1.50 \times 10^{-6} \text{ C} \). The distance between the charges \( r \) is given as \( 12.0 \text{ cm} \), which can be converted to meters as \( 0.12 \text{ m} \).
02

Recall Coulomb's Law

Coulomb's Law calculates the electrostatic force \( F \) between two point charges as \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \).
03

Substitute the values into Coulomb's formula

Substitute \( q_1 = +3.00 \times 10^{-6} \text{ C} \), \( q_2 = -1.50 \times 10^{-6} \text{ C} \), and \( r = 0.12 \text{ m} \) into the formula: \[ F = (8.99 \times 10^9) \times \frac{|3.00 \times 10^{-6} \times (-1.50 \times 10^{-6})|}{(0.12)^2} \].
04

Calculate the force

Calculate the values:- The absolute value of the product of charges: \(|3.00 \times 10^{-6} \times -1.50 \times 10^{-6}| = 4.50 \times 10^{-12} \)- The square of the distance: \((0.12)^2 = 0.0144 \text{ m}^2\)Using these values in the equation,\[ F = (8.99 \times 10^9) \times \frac{4.50 \times 10^{-12}}{0.0144} \approx 2.81 \times 10^{-1} \text{ N} \].

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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 one of the fundamental forces between charged particles. When charges interact, this force either pulls them together or pushes them apart, depending on the nature of their charges.
  • Magnitude: It can be calculated using Coulomb's Law, which gives us the formula for determining the strength of the force between two charged objects.
  • Direction: The force is directed along the line connecting the two charges. If the charges are of opposite signs, the force is attractive. If they are of same signs, it is repulsive.
Coulomb's constant, denoted as \( k \), plays a vital role in determining the magnitude of this force. Its value is approximately \( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). This universal constant applies to all charge interactions, no matter their specific magnitudes.
Charge Interaction
Charge interaction refers to how charged particles influence one another through electrostatic forces. The nature of the forces depends on the types of charges interacting:
  • Like charges: Charges with the same sign (positive-positive or negative-negative) repel each other.
  • Unlike charges: Charges with different signs (positive-negative) attract each other.
By understanding these interactions, you can predict the behavior of objects in electric fields. The strength of the interaction depends on the magnitude of the charges and their distance from each other. In the provided exercise example, a positive charge interacts with a negative charge, leading to an attractive force.
Point Charges
Point charges are idealized charged particles or objects that are considered to have no size and hence, concentrate their charge in a single point in space. In the study of electrostatics:
  • Simplification: Treating charges as point charges simplifies calculations, as it allows us to use Coulomb's Law directly.
  • Applications: This model is useful in predicting how charges behave when placed in proximity to one another or in an electric field.
Despite being an idealization, this concept is highly effective in practical scenarios, especially when dealing with small objects or large distances. In the exercise, the charges given ( \(+3.00 \times 10^{-6} \text{ C}\) and \(-1.50 \times 10^{-6} \text{ C}\) ) are treated as point charges for force calculation using Coulomb's Law.

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Most popular questions from this chapter

A particle of charge \(Q\) is fixed at the origin of an \(x y\) coordinate system. At \(t=0\) a particle \((m=0.800 \mathrm{~g}, q=4.00 \mu \mathrm{C})\) is located on the \(x\) axis at \(x=20.0 \mathrm{~cm}\), moving with a speed of \(50.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. For what value of \(Q\) will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of \(0.108 \mathrm{~N}\) when their center-to-center separation is \(50.0 \mathrm{~cm} .\) The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of \(0.0360 \mathrm{~N}\). Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other?

Calculate the number of coulombs of positive charge in 250 \(\mathrm{cm}^{3}\) of (neutral) water. (Hint: A hydrogen atom contains one proton; an oxygen atom contains eight protons.)

We know that the negative charge on the electron and the positive charge on the proton are equal. Suppose, however, that these magnitudes differ from each other by \(0.00010 \%\). With what force would two copper coins, placed \(1.0 \mathrm{~m}\) apart, repel each other? Assume that each coin contains \(3 \times 10^{22}\) copper atoms. (Hint: \(\mathrm{A}\) neutral copper atom contains 29 protons and 29 electrons.) What do you conclude?

What would be the magnitude of the electrostatic force between two \(1.00\) 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?
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