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Four particles form a square. The charges are \(q_{1}=+Q, q_{2}=q_{3}=q\), and \(q_{4}=-2.00 Q .\) What is \(q / Q\) if the net electrostatic force on particle 1 is zero?

Short Answer

Expert verified
The ratio \(q/Q\) is 1.

Step by step solution

01

Understanding the system

We have four particles placed at the corners of a square. The charges are: \(q_1 = +Q\), \(q_2 = q_3 = q\), and \(q_4 = -2.00Q\). We need to calculate the value of \(q/Q\) such that the net force on \(q_1\) is zero.
02

Identify Forces on Particle 1

Identify all forces acting on particle \(q_1\):1. \(F_{12}\) is the force due to \(q_2\).2. \(F_{13}\) is the force due to \(q_3\).3. \(F_{14}\) is the force due to \(q_4\). These forces act along the lines joining the particles.
03

Calculate Force Due to a Charge

The electrostatic force between two point charges \(q_i\) and \(q_j\), separated by a distance \(r\), is given by:\[ F_{ij} = \frac{{k \, |q_i \, q_j|}}{r^2} \]For now, we can consider the side of the square as \(a\), meaning \(F_{12}\) and \(F_{13}\) are calculated using \(r = a\) and \(F_{14}\) using \(r = \sqrt{2}a\).
04

Forces along the x-axis and y-axis

The force \(F_{12}\) acts along the side of the square, pushing \(q_1\) directly away from \(q_2\) in the x-direction. Similarly, \(F_{13}\) is in the y-direction. Hence:- \(F_{12x} = \frac{kQq}{a^2}\), \(F_{12y} = 0\)- \(F_{13x} = 0\), \(F_{13y} = \frac{kQq}{a^2}\)
05

Calculate Force Due to Diagonal Charge

The diagonal force \(F_{14}\) needs vector decomposition:\[ F_{14x} = F_{14} \cdot \cos(45^\circ) = \frac{\sqrt{2}kQ(2Q)}{(\sqrt{2}a)^2} \cdot \frac{1}{\sqrt{2}} = \frac{kQ^2}{a^2} \]\[ F_{14y} = F_{14} \cdot \sin(45^\circ) = \frac{\sqrt{2}kQ(2Q)}{(\sqrt{2}a)^2} \cdot \frac{1}{\sqrt{2}} = \frac{kQ^2}{a^2} \]
06

Setting Up Equations for Balance

For net force on \(q_1\) to be zero:- In the x-direction: \[ F_{12x} = F_{14x} \rightarrow \frac{kQq}{a^2} = \frac{kQ^2}{a^2} \]- In the y-direction: \[ F_{13y} = F_{14y} \rightarrow \frac{kQq}{a^2} = \frac{kQ^2}{a^2} \]
07

Solve for q/Q

From both directions:\[ \frac{kQq}{a^2} = \frac{kQ^2}{a^2} \rightarrow Qq = Q^2 \rightarrow q = Q \]Thus, the ratio is:\[ \frac{q}{Q} = 1 \]

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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 fundamental in understanding how charged particles interact with one another. It quantifies the electrostatic force between two charges. The law states that the magnitude of the force (\( F \)) between two point charges (\( q_i \) and \( q_j \)) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (\( r \)) between their centers. This relationship is expressed by the formula:

\[F_{ij} = \frac{k |q_i \, q_j|}{r^2}\] Where \( k \) is the electrostatic constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2 / \text{C}^2 \). This formula illustrates a few key points:
  • Proportionality to Charge: The force increases if the magnitude of either charge increases.
  • Inversely Related to Distance: A greater distance between charges reduces the force significantly.
By applying Coulomb's Law, we can derive the electrostatic forces exerted between multiple charge pairs step by step, as done in the provided exercise.
Charge Interaction
Charge interaction is a concept that describes how charged objects exert forces on each other due to their charges. These forces can either be attractive or repulsive, depending on the nature of the charges:
  • Like Charges Repel: Two charges with the same sign (both positive or both negative) will repel each other.
  • Opposite Charges Attract: A positive and a negative charge will attract each other.
In the exercise, four particles form a square with specific charges assigned to each corner. The interaction between these charged particles results in a net electric force acting on each particle. This net force is the vector sum of the individual electrostatic forces exerted by each of the other charges.

In overcoming these interactions, understanding charge relationships allows us to balance forces, as required to solve the given problem: finding the condition when the net force on charge \( q_1 \) is zero.
Vector Decomposition
Vector decomposition is crucial when managing forces acting at angles. In the provided scenario, there are forces acting diagonally, particularly the force due to charge \( q_4 \), which requires decomposition into components along the x and y axes.

For any vector \( F \) at an angle \( \theta \), the components can be calculated as:
  • Horizontal Component: \[ F_x = F \cdot \cos(\theta) \]
  • Vertical Component: \[ F_y = F \cdot \sin(\theta) \]
In the exercise, the diagonal force \( F_{14} \) needed to be split into x and y components using \( 45^\circ \) for both components since a square geometry was involved:
  • \( F_{14x} = \frac{\sqrt{2}kQ(2Q)}{(\sqrt{2}a)^2} \cdot \frac{1}{\sqrt{2}} = \frac{kQ^2}{a^2} \)
  • \( F_{14y} = \frac{\sqrt{2}kQ(2Q)}{(\sqrt{2}a)^2} \cdot \frac{1}{\sqrt{2}} = \frac{kQ^2}{a^2} \)
Decomposing forces is essential in simplifying problems, making it easier to apply conditions of balance, specifically in calculating the forces required to ensure the net force equals zero.

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Most popular questions from this chapter

Four particles are fixed along an \(x\) axis, separated by distances \(d=2.00 \mathrm{~cm} .\) The charges are \(q_{1}=+2 e, q_{2}=-e, q_{3}=\) \(+e\), and \(q_{4}=+4 e\), with \(e=1.60 \times 10^{-19} \mathrm{C}\). In unit-vector notation, what is the net electrostatic force on (a) particle 1 and (b) particle 2 due to the other particles?

The magnitude of the electrostatic force between two identical ions that are separated by a distance of \(5.0 \times 10^{-10} \mathrm{~m}\) is \(3.7 \times 10^{-9} \mathrm{~N}\). (a) What is the charge of each ion? (b) How many electrons are "missing" from each ion (thus giving the ion its charge imbalance)?

Two point charges of \(30 \mathrm{nC}\) and \(-40 \mathrm{nC}\) are held fixed on an \(x\) axis, at the origin and at \(x=72 \mathrm{~cm}\), respectively. A particle with a charge of \(42 \mu \mathrm{C}\) is released from rest at \(x=28 \mathrm{~cm}\). If the initial acceleration of the particle has a magnitude of \(100 \mathrm{~km} / \mathrm{s}^{2}\), what is the particle's mass?

A nonconducting spherical shell, with an inner radius of \(4.0\) \(\mathrm{cm}\) and an outer radius of \(6.0 \mathrm{~cm}\), has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density \(\rho\) is the charge per unit volume, with the unit coulomb per cubic meter. For this shell \(\rho=b / r\), where \(r\) is the distance in meters from the center of the shell and \(b=3.0 \mu \mathrm{C} / \mathrm{m}^{2} .\) What is the net charge in the shell?

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of \(0.108 \mathrm{~N}\) when their center-to-center separation is \(50.0 \mathrm{~cm} .\) The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of \(0.0360 \mathrm{~N}\). Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other?

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