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Chapter 21: Problem 21

Two point charges are located on the \(y\) -axis as follows: charge \(q_{1}=-1.50 \mathrm{nC}\) at \(y=-0.600 \mathrm{m},\) and charge \(q_{2}=\) \(+3.20 \mathrm{nC}\) at the origin \((y=0) .\) What is the total force (magnitude and direction exerted by these two charges on a third charge \(q_{3}=+5.00 \mathrm{nClocated}\) at \(y=-0.400 \mathrm{m} ?\)

Short Answer

Expert verified
The total force is \( 7.30 \times 10^{-6} \text{ N} \) upwards.

Step by step solution

01

Calculate the Distance between Charges

Determine the distance between each pair of charges. The distance between charge \( q_1 \) at \( y = -0.600 \text{ m} \) and charge \( q_3 \) at \( y = -0.400 \text{ m} \) is \(0.200 \text{ m} \) because \(-0.400 - (-0.600) = 0.200\). The distance between charge \( q_2 \) at the origin and charge \( q_3 \) is \(0.400 \text{ m} \) because \(-0.400 - 0 = -0.400\).
02

Calculate the Force Exerted by Charge q1

Use Coulomb's Law to find the force between \( q_1 \) and \( q_3 \). The formula is \( F = k \frac{|q_1 q_3|}{r^2} \), where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\). Plug in the values: \( F_1 = 8.99 \times 10^9 \frac{|-1.50 \times 10^{-9} \times 5.00 \times 10^{-9}|}{(0.200)^2}\). Calculate to get \( F_1 = 1.6875 \times 10^{-6} \text{ N} \). The direction of this force is towards charge \( q_1 \) (downward on the \( y \)-axis).
03

Calculate the Force Exerted by Charge q2

Similarly, calculate the force between \( q_2 \) and \( q_3 \). Use Coulomb's Law: \( F_2 = k \frac{|q_2 q_3|}{r^2} \). Substitute \( F_2 = 8.99 \times 10^9 \frac{3.20 \times 10^{-9} \times 5.00 \times 10^{-9}}{(0.400)^2} \). Calculate this to get \( F_2 = 8.99 \times 10^{-6} \text{ N} \). The direction of this force is upwards (away from the origin).
04

Determine the Net Force on q3

Combine the forces \( F_1 \) and \( F_2 \) considering their directions. The net force \( F_{net} \) is the difference in magnitude because the forces are in opposite directions: \( F_{net} = F_2 - F_1 = 8.99 \times 10^{-6} - 1.6875 \times 10^{-6} = 7.3025 \times 10^{-6} \text{ N} \). The net force direction is upwards on the \( y \)-axis.

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Key Concepts

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

Electric Force
Electric force is the interaction between charged particles, which can attract or repel each other. It is governed by Coulomb's Law, which states that the force ( \( F \) ) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This force plays a crucial role in physics.
  • The strength of the electric force depends on the magnitudes of the charges involved.
  • Distance between the charges greatly affects the force 鈥� closer charges exert a stronger force.
  • Coulomb's Law formula: \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k \) is the Coulomb's constant.
  • The direction of the force is determined by the nature of the charges: like charges repel and opposite charges attract.
Understanding these basics helps us solve problems involving electric forces, such as calculating the force exerted by multiple charges at different positions.
Point Charges
Point charges are an idealized model to simplify calculations in electric force problems. We consider each charge as being concentrated at a single point, which makes it easier to study their interactions.
  • Point charges represent the simplest form of electric charge, devoid of any complexities related to size or distribution.
  • These are often used in theoretical physics to approximate real-world charged particles when their dimensions are negligible compared to the distance between them.
  • They simplify mathematical calculations by focusing on the distance between the charges rather than the shape or size of the bodies.
Using point charges in exercises helps highlight the effects of their interactions without the need for elaborate geometric calculations.
Vector Addition
Vector addition is a method used to determine the net effect of two or more vector quantities. In the context of electric forces, forces exerted by charges are considered vector quantities due to their magnitude and direction.
  • Vector quantities such as forces have both magnitude and direction, making them different from scalar quantities, which only have magnitude.
  • To find the net force on a charge, calculate each individual force vector first.
  • Combine these vectors by considering their directions; this is typically done using graphical methods like the head-to-tail method or mathematically via components.
In our scenario, understanding vector addition helps us sum forces from different charges correctly, determining the resultant force acting on a point charge.
Physics Problem Solving
Physics problem solving involves a stepwise approach to applying theoretical concepts to practical problems. In solving problems involving electric force and point charges, the process typically follows a structured method.
  • Identify the given information, such as charge values and their positions.
  • Determine which physical laws or formulas are applicable, like Coulomb鈥檚 Law for electric force.
  • Perform calculations step-by-step, addressing one interaction or force at a time.
  • Use logical reasoning to combine and interpret results, especially when dealing with vector quantities.
This approach is essential in breaking down complex problems into manageable parts, facilitating better understanding and accurate solutions.

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Most popular questions from this chapter

A very long, straight wire has charge per unit length \(1.50 \times 10^{-10} \mathrm{C} / \mathrm{m} .\) At what distance from the wire is the electric- field magnitude equal to 2.50 \(\mathrm{N} / \mathrm{C}\)?

Two point charges are placed on the \(x\) -axis as follows: Charge \(q_{1}=+4.00 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+5.00 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) What are the magnitude and direction of the total force exerted by these two charges on a negative point charge \(q_{3}=-6.00 \mathrm{nC}\) that is placed at the origin?

A point charge \(q_{1}=-4.00 \mathrm{nC}\) is at the point \(x=\) \(0.600 \mathrm{m}, y=0.800 \mathrm{m},\) and a second point charge \(q_{2}=+6.00 \mathrm{nC}\) is at the point \(x=0.600 \mathrm{m}, y=0 .\) Calculate the magnitude and direction of the net electric field at the origin due to these two point charges.

A charge \(q_{1}=+5.00 \mathrm{nC}\) is placed at the origin of an \(x y\) -coordinate system, and a charge \(q_{2}=-2.00 \mathrm{nC}\) is placed on the positive \(x\) -axis at \(x=4.00 \mathrm{cm} .\) (a) If a third charge \(q_{3}=\) \(+6.00 \mathrm{nC}\) is now placed at the point \(x=4.00 \mathrm{cm}, y=3.00 \mathrm{cm}\) 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.

In a rectangular coordinate system a positive point charge \(q=6.00 \times 10^{-9} \mathrm{Cis~placed}\) at the point \(x=+0.150 \mathrm{m}, y=0\) and an identical point charge is placed at \(x=-0.150 \mathrm{m}, y=0\) Find the \(x\) - and \(y\) -components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{m}, y=0 ;(\mathrm{c}) x=0.150 \mathrm{m}, y=-0.400 \mathrm{m} ;(\mathrm{d}) x=0\) \(y=0.200 \mathrm{m}\)
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