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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 \(132 \mathrm{~N} /\) C. Find the ratio \(r_{2} / r_{1}\)

Short Answer

Expert verified
The ratio \(r_2 / r_1\) is approximately 0.731.

Step by step solution

01

Understand the Formula for Electric Field Due to a Point Charge

The electric field due to a point charge is given by the equation:\[ E = k \frac{|q|}{r^2} \]where \(E\) is the electric field, \(k\) is the Coulomb's constant, \(q\) is the charge, and \(r\) is the distance from the charge. Given the problem, the electric fields at two different distances \(r_1\) and \(r_2\) are known.
02

Set Up the Ratio of Electric Fields

The electric field at \(r_1\) is given by:\[ E_1 = k \frac{|q|}{r_1^2} \]The electric field at \(r_2\) is given by:\[ E_2 = k \frac{|q|}{r_2^2} \]We need to form a ratio \(\frac{E_2}{E_1}\).
03

Substitute Given Values and Solve for the Ratio of Distances

Substitute the given electric field magnitudes into the ratio:\[ \frac{E_2}{E_1} = \frac{k \frac{|q|}{r_2^2}}{k \frac{|q|}{r_1^2}} = \frac{132}{248} \]Simplifying this gives:\[ \frac{r_1^2}{r_2^2} = \frac{248}{132} \]Take the square root of both sides to find the ratio \( \frac{r_2}{r_1} \):\[ \frac{r_2}{r_1} = \sqrt{\frac{132}{248}} \]
04

Perform the Calculation

Calculate \( \sqrt{\frac{132}{248}} \).First, simplify the fraction:\[ \frac{132}{248} = \frac{33}{62} \]Taking the square root gives:\[ \frac{r_2}{r_1} = \sqrt{\frac{33}{62}} \]
05

Final Computation

Compute the square root of \( \frac{33}{62} \):\[ \frac{r_2}{r_1} \approx 0.731 \]

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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 helps us understand the relationship between charges and forces. It states that the electric force between two charged objects is directly proportional to the product of their charges. Additionally, it's inversely proportional to the square of the distance between them. This force can be attractive or repulsive depending upon the nature of the charges.
Coulomb's law is fundamental in calculating electric fields created by point charges. The mathematical expression for Coulomb’s law is:
  • \[ F = k \frac{|q_1 q_2|}{r^2} \]
where:
  • \( F \) is the magnitude of the force between the charges,
  • \( k \) is the Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \),
  • \( q_1 \) and \( q_2 \) are the amounts of the charges, and
  • \( r \) is the distance between the charges.
The understanding of this law is essential for calculating other properties such as electric fields and related problems.
point charge
A point charge is a hypothetical charge that is considered to exist at a single point in space. In real-world scenarios, charges are distributed over a volume; however, to simplify calculations, we often assume they are concentrated at a point.
Point charges are typically used in physics to simplify problems involving electric fields and forces. They allow for straightforward calculations using Coulomb's law. Because we treat point charges as concentrated at a single location, calculating the electric field or force becomes less complicated.
When dealing with point charges, remember:
  • Charges can be positive or negative.
  • They interact through electric fields.
  • Their effects can be directly calculated using formulas like those derived from Coulomb's law.
Understanding the concept of point charge is crucial in electric field problems.
distance ratio
The ratio of distances in electric field problems is used to compare the effects of different separations on field strength. Given two distances from a point charge, we can determine how the electric field changes with distance.
In this particular exercise, you need to find the ratio of two distances \( r_2 \) and \( r_1 \) from the charge. The electric fields are provided at these distances, which lets us establish a relationship.
  • The electric field is inversely proportional to the square of the distance — doubling the distance reduces the field strength by a factor of four.
  • By expressing the ratio \( \frac{E_2}{E_1} \) for fields at \( r_2 \) and \( r_1 \), it can help us determine \( \frac{r_2}{r_1} \).
Solving this ratio is essential as it provides insight into how quickly the electric field diminishes with increasing distance.
electric field calculation
Electric field calculations are necessary to understand the influence charges have in their surroundings. The electric field of a point charge is calculated using:
  • \[ E = k \frac{|q|}{r^2} \]
where:
  • \( E \) is the electric field,
  • \( k \) is the Coulomb's constant \( (8.99 \times 10^9 \, \text{N m}^2/\text{C}^2) \),
  • \( q \) is the magnitude of the charge, and
  • \( r \) is the distance from the charge.
For any given point, the strength of the field is greater closer to the charge and decreases as one moves further away.
In this exercise, two electric fields at different distances are given. By setting up their ratio and solving for the distance ratio, we can examine how electric field calculations highlight the spatial impact of a charge. Calculating the electric field at varying distances is vital for practical applications in electric fields, such as in designing circuits and understanding electromagnetic forces.

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Most popular questions from this chapter

A charge of \(q=+7.50 \mu \mathrm{C}\) is located in an electric field. The \(x\) and \(y\) components of the electric field are \(E_{x}=6.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\) and \(E_{y}=8.00 \times 10^{3} \mathrm{~N} / \mathrm{C}\), respectively. (a) What is the magnitude of the force on the charge? (b) Determine the angle that the force makes with the \(+x\) axis.

There are four charges, each with a magnitude of \(2.0 \mu \mathrm{C}\). Two are positive and two are negative. The charges are fixed to the corners of a 0.30 -m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. Find the magnitude of the net electrostatic force experienced by any charge.

An electrically neutral model airplane is flying in a horizontal circle on a 3.0 -m guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is \(50.0 \mathrm{~J}\). Reconsider the same situation, except that now there is a point charge of \(+q\) on the plane and a point charge of \(-q\) at the other end of the guideline. In this case, the line breaks when the kinetic energy of the plane is \(51.8 \mathrm{~J}\). Find the magnitude of the charges.

Suppose you want to neutralize the V gravitational attraction between the earth and the moon by placing equal amounts of charge on each. (a) Should the charges be both positive, both negative, or one positive and the other negative? Why? (b) Do you need to know the distance between the earth and the moon to find the magnitude of the charge? Why or why not?

Two charges are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{~cm}\), and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{~cm}\). Two other charges are located on the \(y\) axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{3}=+5.0 \mathrm{~cm}\), and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{~cm} .\) Find the net electric field (magnitude and direction) at the origin.

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