/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 How far must the point charges \... [FREE SOLUTION] | 91Ó°ÊÓ

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How far must the point charges \(q_{1}=+7.22 \mu \mathrm{C}\) and \(q_{2}=-26.1 \mu \mathrm{C}\) be separated for the electric potential energy of the system to be \(-126 \mathrm{J} ?\)

Short Answer

Expert verified
The charges must be separated by 13.38 meters.

Step by step solution

01

Understand the Formula

The electric potential energy \( U \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by the formula \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), where \( k \) is the Coulomb's constant, approximately \( 8.99 \times 10^9 \, \mathrm{N \cdot m^2 / C^2}\).
02

Rearrange the Formula to Solve for r

Given that \( U = -126 \, \mathrm{J} \), we need to solve the equation for \( r \):\[ r = \frac{k \cdot q_1 \cdot q_2}{U} \]
03

Substitute the Known Values into the Equation

Substitute \( q_1 = +7.22 \times 10^{-6} \mathrm{C} \), \( q_2 = -26.1 \times 10^{-6} \mathrm{C} \), and \( U = -126 \mathrm{J} \) into the equation:\[ r = \frac{(8.99 \times 10^9) \cdot (7.22 \times 10^{-6}) \cdot (-26.1 \times 10^{-6})}{-126} \]
04

Calculate the Distance

Calculate the distance \( r \) using the substituted equation:\[ r = \frac{8.99 \times 10^9 \times 7.22 \times 10^{-6} \times 26.1 \times 10^{-6}}{126} \]Simplify and calculate:\[ r \approx 13.38 \, \mathrm{m} \]
05

Conclude the Solution

Based on the calculations, the point charges must be about 13.38 meters apart for the electric potential energy to be \(-126 \mathrm{J} \).

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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 essential when discussing electric forces between charges. It states that the force (F) 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. This can be expressed mathematically as:
  • \( F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \)
In this formula:
  • \( F \) is the force between the charges.
  • \( k \) is the Coulomb's constant, approximately \( 8.99 \times 10^9 \, \mathrm{N \cdot m^2 / C^2} \).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the charges.
The concept of Coulomb's law is integral to understanding how charges interact and influence each other's electric potential energy. It helps to visualize how close or far charges must be to exert a specific force on each other.
Although it primarily deals with force, alterations in distance also affect potential energy, as potential energy stems from the work done by the electric force.
Point Charges
Point charges represent idealized charges located at a single point in space. Despite being theoretical, they help simplify complex problems by assuming all charge resides at one location, often making calculations more feasible.

These charges are critical when calculating interactions within electric fields because they provide a clear point of reference for distance and force measurements. Point charges are conceived as either positive or negative, determining the direction of electric forces:
  • Positive charges repel one another.
  • Negative charges also repel one another.
  • A positive and a negative charge attract each other.
Visualizing how point charges respond to each other’s presence is vital in comprehending more intricate concepts like electric fields and potential energy. They provide a baseline for more detailed studies regarding real-world applications, showing how atoms and molecules comprise charged particles interacting similarly.
Distance Between Charges
The distance between charges is a crucial factor in calculating both electric potential energy and the force as described by Coulomb's law. The distance, represented by \( r \), dictates how strongly charges interact:
  • A shorter distance means stronger interactions, whether attractive or repulsive.
  • A greater distance weakens interactions, allowing charges less influence over each other.
In studying electric potential energy, the distance directly influences energy values. According to the formula \( U = \frac{k \cdot q_1 \cdot q_2}{r} \), adjusting \( r \) changes the potential energy within a system.

When dealing with questions involving potential energy, understanding how varying distances impact the system is fundamental. For instance:
  • Decreasing the distance increases the electric potential energy (if charges have opposite signs, it decreases energy).
  • Increasing the distance decreases energy (or increases it for like-signed charges).
Calculating such parameters involves rearranging the equations to solve for the desired quantity, as seen in exercises measuring the required distance to achieve specific potential energy levels.

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Most popular questions from this chapter

A uniform electric field of magnitude \(4.1 \times 10^{5} \mathrm{N} / \mathrm{C}\) points in the positive \(x\) direction. Find the change in electric potential energy of a \(4.5-\mu C\) charge as it moves from the origin to the points (a) \((0,6.0 \mathrm{m}) ;\) (b) \((6.0 \mathrm{m}, 0) ;\) and (c) \((6.0 \mathrm{m}, 6.0 \mathrm{m})\).

How much work is required to bring three protons, initially infinitely far apart, to a configuration where each proton is \(1.5 \times 10^{-15} \mathrm{m}\) from the other two? (This is a typical separation for protons in a nucleus.)

Consider a parallel-plate capacitor constructed from two circular metal plates of radius \(R\). The plates are separated by a distance of \(1.5 \mathrm{mm}\) (a) What radius must the plates have if the capacitance of this capacitor is to be \(1.0 \mu \mathrm{F} ?\) (b) If the separation between the plates is increased, should the radius of the plates be increased or decreased to maintain a capacitance of \(1.0 \mu \mathrm{F}\) ? Explain. (c) Find the radius of the plates that gives a capacitance of \(1.0 \mu \mathrm{F}\) for a plate separation of \(3.0 \mathrm{mm}\).

Electric Catfish The electric catfish (Malapterurus electricus) is an aggressive fish, \(1.0 \mathrm{m}\) in length, found today in tropical Africa (and depicted in Egyptian hieroglyphics). The catfish is capable of generating jolts of electricity up to \(350 \mathrm{V}\) by producing a positively charged region of muscle near the head and a negatively charged region near the tail. (a) For the same amount of charge, can the catfish generate a higher voltage by separating the charge from one end of its body to the other, as it does, or from one side of the body to the other? Explain. (b) Estimate the charge generated at each end of a catfish as follows:Treat the catfish as a parallel- plate capacitor with plates of area \(1.8 \times 10^{-2} \mathrm{m}^{2},\) separation \(1.0 \mathrm{m},\) and filled with a dielectric with a dielectric constant \(\kappa=95\).

The electrons in a TV picture tube are accelerated from rest through a potential difference of \(25 \mathrm{kV}\). What is the speed of the electrons after they have been accelerated by this potential difference?

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