/*! 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 70 A charge of \(-\)3.00 nC is plac... [FREE SOLUTION] | 91影视

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Chapter 21: Problem 70

A charge of \(-\)3.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge of 2.00 nC is placed on the \(y\)-axis at \(y =\) 4.00 cm. (a) If a third charge, of 5.00 nC, is now placed at the point \(x =\) 3.00 cm, \(y =\) 4.00 cm, find the \(x\)- and \(y\)-components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

Short Answer

Expert verified
The total force on the 5.00 nC charge is approximately 1.20 碌N at an angle of -85.2掳 from the positive x-axis.

Step by step solution

01

Understand the Scenario

There are three charges in play: Charge \( q_1 = -3.00 \text{ nC} \) at the origin \((0,0)\), charge \( q_2 = 2.00 \text{ nC} \) at \((0, 4 \text{ cm})\), and charge \( q_3 = 5.00 \text{ nC} \) at \((3 \text{ cm}, 4 \text{ cm})\). We need to compute the forces exerted on \( q_3 \) by \( q_1 \) and \( q_2 \).
02

Calculate the Force Between Charges

Use Coulomb's law: \( F = k\frac{|q_1 q_2|}{r^2} \). For \( q_1 \) and \( q_3 \), \( r = 3 \text{ cm} = 0.03 \text{ m} \). For \( q_2 \) and \( q_3 \), \( r = 3 \text{ cm} = 0.03 \text{ m} \) along the x-axis, according to the position difference.Electric constant: \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
03

Calculate Force from Charge q1 on q3

Convert charges to Coulombs: \( q_1 = -3.00 \text{ nC} = -3.00 \times 10^{-9} \text{ C} \), \( q_3 = 5.00 \text{ nC} = 5.00 \times 10^{-9} \text{ C} \). Using \( r_{13} = 3 \text{ cm} = 0.03 \text{ m} \), the force magnitude is:\[ F_{13} = k \frac{|q_1 q_3|}{r_{13}^2} = 8.99 \times 10^9 \frac{|-3 \times 10^{-9} \cdot 5 \times 10^{-9}|}{(0.03)^2} \approx 1.50 \text{ 碌N} \]Direction: attractive towards origin \((3 \text{ cm}, 4 \text{ cm})\).
04

Calculate x and y Components of Force from q1 on q3

Force from \( q_1 \) at an angle, needing calculation:\( \theta = \tan^{-1}(4/3) \)\[ F_{13,x} = F_{13} \cos(\theta) \quad \text{and} \quad F_{13,y} = F_{13} \sin(\theta) \]\[ F_{13,x} = -1.50 \times \frac{3}{5} \approx -0.9 \text{ 碌N}, \quad F_{13,y} = -1.50 \times \frac{4}{5} \approx -1.2 \text{ 碌N} \]
05

Calculate Force from Charge q2 on q3

Convert charges to Coulombs: \( q_2 = 2.00 \text{ nC} = 2.00 \times 10^{-9} \text{ C} \).Using \( r_{23} = 3 \text{ cm} = 0.03 \text{ m} \), the force magnitude is:\[ F_{23} = k \frac{|q_2 q_3|}{r_{23}^2} = 8.99 \times 10^9 \frac{2 \times 10^{-9} \cdot 5 \times 10^{-9}}{(0.03)^2} \approx 1.00 \text{ 碌N} \]Direction: repulsive along the x-axis. \[ F_{23,x} = 1.00 \text{ 碌N}, \quad F_{23,y} = 0 \text{ 碌N} \]
06

Calculate Total x and y Components of Force on q3

Add the components of the forces on \( q_3 \):\[ F_{x} = F_{13,x} + F_{23,x} = -0.9 \text{ 碌N} + 1.0 \text{ 碌N} = 0.1 \text{ 碌N} \]\[ F_{y} = F_{13,y} + F_{23,y} = -1.2 \text{ 碌N} + 0 \text{ 碌N} = -1.2 \text{ 碌N} \]
07

Calculate Magnitude and Direction of Total Force

Magnitude:\[ F = \sqrt{F_{x}^2 + F_{y}^2} = \sqrt{(0.1 \text{ 碌N})^2 + (-1.2 \text{ 碌N})^2} \approx 1.20 \text{ 碌N} \]Direction (angle \( \phi \)) downwards from positive x-axis:\[ \phi = \tan^{-1}\left(\frac{F_{y}}{F_{x}}\right) = \tan^{-1}\left(\frac{-1.2}{0.1}\right) \approx -85.2^\circ \]

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Key Concepts

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

Electric Forces
Electric forces are the forces that arise from electric charges. They are governed by Coulomb's Law, which states that the magnitude of the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between their centers. This relationship is given by the formula: \[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \( F \) is the electrostatic force between two charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \).
  • \( q_1 \) and \( q_2 \) are the amounts of the two charges.
  • \( r \) is the separation distance between the charges.
Electric forces can be attractive or repulsive. If the charges are of opposite signs, the force is attractive; if they are of the same sign, they repel. This force plays a critical role in the fields of electricity and physics, explaining interactions at both the atomic and macroscopic levels.
Charge Interactions
Charge interactions are the means by which electric forces manifest between charged particles. When dealing with multiple charges, each pair of charges will interact according to Coulomb's Law, resulting in various force vectors acting on each charge. Important points about charge interactions include:
  • An attractive force occurs when a positive and a negative charge come together.
  • A repulsive force arises when two like charges, either positive-positive or negative-negative, interact.
  • The net force on a charge is the vector sum of all the forces acting upon it from other charges.
For example, in the exercise provided, charge \( q_3 \) experiences forces due to both charge \( q_1 \) and charge \( q_2 \). The resulting total force is found by calculating each individual force as a vector and summing them using vector addition. Understanding these interactions is key to solving problems involving multiple charges and finding resultant electric forces.
Vector Components
In physics, vector components are used to simplify and analyze forces, velocities, and other vector quantities that act in multiple directions. Vectors can be divided into their respective components along the axes of the coordinate system, typically the x- and y-axes in a two-dimensional plane.To find the vector components of a force:
  • The force vector can be split into an \( x \)-component \( F_x \) and a \( y \)-component \( F_y \) using trigonometry.
  • For an angle \( \theta \) with respect to the horizontal, these components are calculated using: \[ F_x = F \cos(\theta) \] \[ F_y = F \sin(\theta) \]
  • Each component is a projection of the original vector onto the corresponding axis, allowing for easier mathematical treatment in calculations.
In the provided problem, the x and y components for forces on charge \( q_3 \) from \( q_1 \) and \( q_2 \) are calculated separately, enabling us to determine the net force acting in each respective direction. Summing these components helps in finding the resultant magnitude and direction of the total force.

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Most popular questions from this chapter

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3q\) charge by each of the other three charges.

A charge \(q_1 = +\)5.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge \(q_2 = -\)2.00 nC is placed on the positive \(x\)-axis at \(x = \)4.00 cm. (a) If a third charge \(q_3 = +\)6.00 nC is now placed at the point \(x =\) 4.00 cm, \(y =\) 3.00 cm, find the \(x\)- and \(y\)-components of the total force exerted on this charge by the other two. (b) Find the magnitude and direction of this force.

A uniform electric field exists in the region between two oppositely charged plane parallel plates. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate, 1.60 cm distant from the first, in a time interval of 3.20 \(\times \space 10^{-6}\) s. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

A proton is traveling horizontally to the right at 4.50 \(\times 10^6\) m\(/\)s. (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.20 cm. (b) How much time does it take the proton to stop after entering the field? (c) What minimum field (magnitude and direction) would be needed to stop an electron under the conditions of part (a)?

What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)? (a) Because air is a good conductor, the positive charge on the bee's surface flowed through the air from bee to plant. (b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee. (c) The plant became electrically polarized as the charged bee approached. (d) Bees that had visited the plant earlier deposited a positive charge on the stem.
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