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In a vacuum, two particles have charges of \(q_{1}\) and \(q_{2}\), where \(q_{1}=+3.5 \mu \mathrm{C}\). They are separated by a distance of \(0.26 \mathrm{m}\), and particle 1 experiences an attractive force of \(3.4 \mathrm{N}\). What is \(q_{2}\) (magnitude and \(\operatorname{sign}\) )?

Short Answer

Expert verified
The magnitude of \(q_2\) is \(7.31 \times 10^{-6} \, \mathrm{C}\) and its sign is negative.

Step by step solution

01

Identify Known Values

We begin by identifying the known values from the problem statement. We know that:- Charge of particle 1, \(q_1 = +3.5 \, \mu\mathrm{C} = 3.5 \times 10^{-6} \, \mathrm{C}\)- Separation distance, \(r = 0.26 \, \mathrm{m}\)- Force experienced by charge 1, \(F = 3.4 \, \mathrm{N}\)
02

Coulomb's Law Formula

The formula for the force between two charges in a vacuum using Coulomb's Law is given by:\[ F = k \frac{|q_1 \cdot q_2|}{r^2} \]where \(k\) is the Coulomb's constant, approximately \(8.988 \times 10^9 \, \mathrm{N \, m^2/C^2}\).
03

Rearrange Formula to Solve for \(q_2\)

We rearrange the formula to solve for \(q_2\):\[ |q_2| = \frac{F \cdot r^2}{k \cdot |q_1|} \]Insert the known values into this expression.
04

Substitute Known Values

Substitute the known values into the rearranged formula:\[ |q_2| = \frac{3.4 \, \mathrm{N} \cdot (0.26 \, \mathrm{m})^2}{8.988 \times 10^9 \, \mathrm{N \, m^2/C^2} \cdot 3.5 \times 10^{-6} \, \mathrm{C}} \]
05

Calculate Magnitude of \(q_2\)

Calculate the value:\[|q_2| = \frac{3.4 \, \mathrm{N} \cdot 0.0676 \, \mathrm{m^2}}{8.988 \times 10^9 \, \mathrm{N \, m^2/C^2} \cdot 3.5 \times 10^{-6} \, \mathrm{C}} = \frac{0.22984 \, \mathrm{Nm^2}}{3.1453 \times 10^{-2} \mathrm{N \, m^2}} \approx 7.31 \times 10^{-6} \, \mathrm{C} \]
06

Determine the Sign of \(q_2\)

Since the force on \(q_1\) is attractive and \(q_1\) is positive, \(q_2\) must be negatively charged (opposite charge attracts). Thus, \(q_2 = -7.31 \times 10^{-6} \, \mathrm{C}.\)

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Key Concepts

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

Understanding Electrostatic Force
Electrostatic force is a fundamental concept in physics that describes the force between charged particles. It's governed by Coulomb's Law, which states: the electrostatic force (\( F \)) between two point charges (\( q_1 \) and \( q_2 \)) is directly proportional to the product of the charges, and inversely proportional to the square of the distance (\( r \)) between them. The equation is given by:
  • \( F = k \frac{|q_1 \cdot q_2|}{r^2} \)
Here, \( k \) is Coulomb's constant (\( 8.988 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \)). This constant makes it possible to connect the charge values to the force. The electrostatic force is either attractive or repulsive.
  • If the charges have opposite signs, the force is attractive.
  • If the charges have the same sign, the force is repulsive.
Understanding this concept helps explain interactions at the atomic level, such as bonding in molecules, and is foundational for fields like chemistry and electronics.
The Charge of a Particle
The charge of a particle is a property that causes it to experience a force when placed in an electric field. Particles can have positive or negative charges, measured in Coulombs (C). In this context, the exercise provides \( q_1 \) as \(+3.5 \, \mu \mathrm{C}\), which translates to \( 3.5 \times 10^{-6} \, \mathrm{C}\). The goal is to find \( q_2 \), given the conditions.

The calculation involves rearranging the formula for Coulomb’s Law to solve for \( q_2 \). The sign of the charge also determines the type of electrostatic interaction:
  • In the given example, \( q_1 \) experiences an attractive force, indicating \( q_2 \) is opposite in charge to \( q_1 \).
    • This essential understanding of particle charge assists in predicting the behavior of charged particles under various circumstances. Various applications include understanding electron flow in circuits and reactions in chemistry.
The Role of Vacuum in Electrostatic Calculations
A vacuum is a space devoid of matter, including air, which has no resistance to the movement of electric fields. In Coulomb's Law context, using a vacuum simplifies calculations, as it removes external influences that might affect the interaction between charges.

A vacuum allows you to use the ideal expression of Coulomb's Law, ensuring that the only variables are the charges themselves and the distance between them:
  • This means that the force calculated is purely due to the charges, without interference from atmospheric molecules.
  • The absence of a medium also avoids additional complications like dielectric constants that would otherwise alter the force magnitude.
This understanding makes it easier to predict and calculate forces in controlled environments. In practical applications, such as in physics experiments or electronic component design, a vacuum is used to ensure accuracy and reliability of results.

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Most popular questions from this chapter

Conceptual Example 13 deals with the hollow spherical conductor in Figure \(18.30 .\) The conductor is initially electrically neutral, and then a charge \(+q\) is placed at the center of the hollow space. Suppose the conductor initially has a net charge of \(+2 q\) instead of being neutral. What is the total charge on the interior and on the exterior surface when the \(+q\) charge is placed at the center?

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 charges are placed between the plates of a parallel plate capacitor. One charge is \(+q_{1}\) and the other is \(q_{2}=+5.00 \mu \mathrm{C} .\) The charge per unit area on each of the plates has a magnitude of \(\sigma=1.30 \times 10^{-4} \mathrm{C} / \mathrm{m}^{2}\) The magnitude of the force on \(q_{1}\) due to \(q_{2}\) equals the magnitude of the force on \(q_{1}\) due to the electric field of the parallel plate capacitor. What is the distance \(r\) between the two charges?

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.

There are four charges, each with a magnitude of \(2.0 \mu\) C. Two are positive and two are negative. The charges are fixed to the corners of a \(0.30-\mathrm{m}\) 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.

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