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

Short Answer

Expert verified
The ratio \( \frac{r_2}{r_1} \) is approximately 1.370.

Step by step solution

01

Understand the Problem

We are given two electric field magnitudes, one at distance \( r_1 \) and another at distance \( r_2 \) from a point charge. Our task is to find the ratio \( \frac{r_2}{r_1} \).
02

Recall the Electric Field Formula

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

Write the Equation for Each Distance

At distance \( r_1 \), the electric field is \( 248 \) N/C, so:\[ E_1 = \frac{kQ}{r_1^2} = 248. \]Similarly, at distance \( r_2 \), the electric field is \( 132 \) N/C, so:\[ E_2 = \frac{kQ}{r_2^2} = 132. \]
04

Formulate the Ratio Between Fields

The relationship between the fields and distances can be written as:\[ \frac{E_2}{E_1} = \frac{r_1^2}{r_2^2}. \]
05

Plug-In Values and Solve the Ratio

Substitute the values of \( E_1 = 248 \) and \( E_2 = 132 \) into the equation:\[ \frac{132}{248} = \frac{r_1^2}{r_2^2}. \]Calculate the ratio:\[ \left( \frac{r_2}{r_1} \right)^2 = \frac{248}{132}. \]
06

Simplify the Calculation

Calculate \( \frac{248}{132} \):\[ \frac{248}{132} \approx 1.8788. \]Now, take the square root to find \( \frac{r_2}{r_1} \):\[ \frac{r_2}{r_1} \approx \sqrt{1.8788} \approx 1.370. \]

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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 a fundamental principle which describes how electric charges interact with each other. It states that the electric force between two charged objects is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers. This important law can be expressed with the formula: \[ F = \frac{k \cdot |Q_1 \cdot Q_2|}{r^2} \] where:
  • \( F \) is the electric force between the charges
  • \( k \) is Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \)
  • \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges
  • \( r \) is the distance between the charges
In the context of the given problem, Coulomb's Law provides the foundational understanding for how the electric field is computed from a point charge. Understanding this principle helps us relate the physical setup (distances and fields) to the mathematical expressions in the step-by-step solution.
Point Charge
A point charge is simply an idealized model where the electric charge is considered to be located at a single point in space. This concept is crucial when calculating electric fields because it simplifies our calculations and allows us to use the symmetry in the problem. In our exercise, the problem uses a point charge to create different electric fields at different distances. The electric field (\( E \)) from a point charge is mathematically defined as: \[ E = \frac{kQ}{r^2} \] where:
  • \( k \) is Coulomb's constant, a crucial element for calculations
  • \( Q \) represents the amount of charge
  • \( r \) indicates the distance from the charge
Understanding the point charge helps us effectively use this formula to evaluate how the electric field strength changes with different distances, as seen in the problem scenario.
Distance Ratio
In the given exercise, one key objective is to determine the ratio of two distances (\( r_2/r_1 \)) at which the electric field magnitudes are specified. The concept of having distance ratios is essential when dealing with problems related to electric fields, especially when there's a point charge. To solve for the distance ratio:
  • Remember that the electric field strength varies inversely with the square of the distance. This is captured in the equation \( E = \frac{kQ}{r^2} \).
  • By carefully comparing the fields at two distances, \( r_1 \) and \( r_2 \), we derive that the field magnitude relates to the distance squared as shown in \( \frac{E_2}{E_1} = \frac{r_1^2}{r_2^2} \).
By using this relationship, we solve for the distance ratio as: \[ \left( \frac{r_2}{r_1} \right)^2 = \frac{E_1}{E_2} \] The result is that solving this equation gives us a deeper understanding of how electric fields decrease as we move further away from a point charge, thanks to the inverse square law.

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Most popular questions from this chapter

Review the important features of electric field lines discussed in Conceptual Example 12. Three point charges \((+q,+2 q,\) and \(-3 q)\) are at the corners of an equilateral triangle. Sketch in six electric field lines between the three charges.

An object has a charge of \(-2.0 \mu\) C. How many electrons must be removed so that the charge becomes \(+3.0 \mu \mathrm{C} ?\)

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.

An electrically neutral model airplane is flying in a horizontal circle on a \(3.0-\mathrm{m}\) guideline, which is nearly parallel to the ground. The line breaks when the kinetic energy of the plane is 50.0 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 J. Find the magnitude of the charges.

Water has a mass per mole of \(18.0 \mathrm{g} / \mathrm{mol},\) and each water molecule \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) has 10 electrons. (a) How many electrons are there in one liter \(\left(1.00 \times 10^{-3} \mathrm{m}^{3}\right)\) of water? (b) What is the net charge of all these electrons?

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