/*! 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 ssm At a distance \(r_{1}\) from... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

ssm At a distance \(r_{1}\) from a point charge, the magnitude of the electric field created by the charge is \(248 \mathrm{N} / \mathrm{C}\). At a distance \(r_{2}\) from the charge, the field has a magnitude of \(132 \mathrm{N} / \mathrm{C}\). Find the ratio \(r_{2} / r_{1}\).

Short Answer

Expert verified
The ratio \( \frac{r_2}{r_1} \) is approximately 1.37.

Step by step solution

01

Understanding the Problem

We need to find the ratio \( \frac{r_2}{r_1} \) given the electric field magnitudes at two distances \( r_1 \) and \( r_2 \), with \( E_1 = 248 \text{ N/C} \) and \( E_2 = 132 \text{ N/C} \).
02

Using the Electric Field Formula

The electric field due to a point charge is given by \( E = \frac{kQ}{r^2} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge.
03

Setting Up Equations

We have two equations from the electric fields: \( E_1 = \frac{kQ}{r_1^2} = 248 \) and \( E_2 = \frac{kQ}{r_2^2} = 132 \).
04

Finding the Ratio of Fields

Divide the first equation by the second: \( \frac{E_1}{E_2} = \frac{ \frac{kQ}{r_1^2} }{ \frac{kQ}{r_2^2} } = \frac{r_2^2}{r_1^2} \). This simplifies to \( \frac{E_1}{E_2} = \left(\frac{r_2}{r_1}\right)^2 \).
05

Solving for the Ratio

Substitute the known values: \( \frac{248}{132} = \left(\frac{r_2}{r_1}\right)^2 \). Calculate \( \frac{248}{132} = 1.8788 \).
06

Taking the Square Root

To find \( \frac{r_2}{r_1} \), take the square root: \( \frac{r_2}{r_1} = \sqrt{1.8788} \approx 1.37 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Coulomb's Law
Coulomb's Law is the fundamental principle describing the interaction between charged particles. It states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance separating them. Mathematically, it is represented as:
\[ F = \frac{k \, |q_1 q_2|}{r^2} \]where:
  • \( F \) is the magnitude of the electrostatic force between the charges,
  • \( k \) is Coulomb's constant \( (8.99 \times 10^9 \text{ N m}^2/\text{C}^2) \),
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
  • \( r \) is the distance separating the charges.
Coulomb's Law not only helps in understanding the quantitative aspects of electric force but also plays a key role in deriving other related concepts like the electric field, demonstrating how charges interact within a field.
Point Charge
A point charge refers to an idealized or hypothetical charge located at a single point in space. It is helpful in simplifying the calculations of electric fields and forces, as it assumes the charge is concentrated at a particular location. In reality, charges are distributed over a region, but treating them as point charges allows for easier analysis of their interactions.
When calculating an electric field due to a point charge, the formula used is:
\[ E = \frac{kQ}{r^2} \]where:
  • \( E \) is the electric field strength at a distance \( r \),
  • \( k \) is Coulomb's constant,
  • \( Q \) is the charge causing the field.
It's important to remember that the direction of the electric field depends on the type of charge:
  • A positive point charge creates an outward field direction.
  • A negative point charge creates an inward field direction.
These properties make point charges a pivotal concept in conceptualizing and calculating electric fields.
Distance and Field Relation
The relationship between distance and the electric field is a fundamental concept in electromagnetism. According to Coulomb's Law, the electric field (\( E \)) generated by a point charge decreases with the square of the distance (\( r \)) from the charge. This inverse square law nature means that as you move further away from a charge, the field's strength diminishes rapidly.
In the context of the given exercise, calculating the electric field at two different distances allows us to compare their relative magnitudes. Using the expressions:
  • \( E_1 = \frac{kQ}{r_1^2} \)
  • \( E_2 = \frac{kQ}{r_2^2} \)
We can derive the ratio of these distances by dividing the equations:
\[ \frac{E_1}{E_2} = \left(\frac{r_2}{r_1}\right)^2 \]This tells us the relation of distances is proportional to the square root of the field magnitudes' ratio.
Intuitively, this means if you know the electric field at two distances, you can determine how much further one distance is from the other. This comprehension helps in predicting how fields change as you adjust your position relative to a point charge.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

ssm Two charges are placed on the \(x\) axis. One of the charges \(\left(q_{1}=+8.5 \mu \mathrm{C}\right)\) is at \(x_{1}=+3.0 \mathrm{cm}\) and the other \(\left(q_{2}=-21 \mu \mathrm{C}\right)\) is at \(x_{1}=+9.0 \mathrm{cm} .\) Find the net electric field (magnitude and direction) at (a) \(x=0 \mathrm{cm}\) and (b) \(x=+6.0 \mathrm{cm}\)

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?

In a vacuum, a proton (charge \(=+e,\) mass \(=1.67 \times 10^{-27} \mathrm{kg}\) ) is moving parallel to a uniform electric field that is directed along the \(+x\) axis (see the figure). The proton starts with a velocity of \(+2.5 \times 10^{4} \mathrm{m} / \mathrm{s}\) and accelerates in the same direction as the electric field, which has a value of \(+2.3 \times 10^{3} \mathrm{N} / \mathrm{C} .\) Find the velocity of the proton when its displacement is \(+2.0 \mathrm{mm}\) from the starting point.

Two tiny conducting spheres are identical and carry charges of \(-20.0 \mu \mathrm{C}\) and \(+50.0 \mu \mathrm{C} .\) They are separated by a distance of \(2.50 \mathrm{cm}\) (a) What is the magnitude of the force that each sphere experiences, and is the force attractive or repulsive? (b) The spheres are brought into contact and then separated to a distance of \(2.50 \mathrm{cm} .\) Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.

ssm At three corners of a rectangle (length \(=2 d\), height \(=d\) ), the following charges are located: \(+q_{1}\) (upper left corner), \(+q_{2}\) (lower right corner), and \(-q\) (lower left corner). The net electric field at the (empty) upper right corner is zero. Find the magnitudes of \(q_{1}\) and \(q_{2}\). Express your answers in terms of \(q.\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.