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Chapter 18: Problem 31

Two charges are placed on the \(x\) axis. One of the charges \(\left(q_{1}=+8.5 \mu \mathrm{C}\right)\) is at \(x_{1}=+3.0 \mathrm{~cm}\) and the other \(\left(q_{2}=-21 \mu \mathrm{C}\right)\) is at \(x_{1}=+9.0 \mathrm{~cm} .\) Find the net electric field (magnitude and direction) at (a) \(x=0 \mathrm{~cm}\) and (b) \(x=+6.0 \mathrm{~cm}\).

Short Answer

Expert verified
At \( x = 0 \, \mathrm{cm} \), the net electric field is \( 10.827 \times 10^6 \mathrm{N/C} \) to the left. At \( x = +6.0 \, \mathrm{cm} \), it is \( 12.503 \times 10^6 \mathrm{N/C} \) to the left.

Step by step solution

01

Understanding the Problem

We have two point charges on the x-axis and need to determine the net electric field due to these charges at two specific points on the x-axis. The charges are: \( q_1 = +8.5 \, \mu \mathrm{C} \) at \( x_1 = +3.0 \, \mathrm{cm} \) and \( q_2 = -21 \, \mu \mathrm{C} \) at \( x_2 = +9.0 \, \mathrm{cm} \). We will calculate the electric field at \( x = 0 \, \mathrm{cm} \) and \( x = +6.0 \, \mathrm{cm} \).
02

Formulating the Electric Field Equation

The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula: \[ E = \frac{k |q|}{r^2} \] where \( k = 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \) is Coulomb's constant. The direction of \( E \) is away from the charge if the charge is positive and towards the charge if it is negative.
03

Calculating Electric Field at \( x = 0 \, \mathrm{cm} \)

Calculate the electric field due to each charge at \( x = 0 \, \mathrm{cm} \).1. **Charge \( q_1 \):** - Distance \( r_1 = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m} \) - \( E_1 = \frac{k \cdot 8.5 \times 10^{-6}}{0.03^2} = \frac{8.99 \times 10^9 \cdot 8.5 \times 10^{-6}}{0.03^2} = 8.494 \times 10^6 \, \mathrm{N/C} \) directed to the left (negative x-direction).2. **Charge \( q_2 \):** - Distance \( r_2 = 9.0 \, \mathrm{cm} = 0.09 \, \mathrm{m} \) - \( E_2 = \frac{k \cdot 21 \times 10^{-6}}{0.09^2} = \frac{8.99 \times 10^9 \cdot 21 \times 10^{-6}}{0.09^2} = 2.333 \times 10^6 \, \mathrm{N/C} \) directed to the left (negative x-direction).**Net Electric Field:**- \( E = E_1 + E_2 = 8.494 \times 10^6 + 2.333 \times 10^6 = 10.827 \times 10^6 \, \mathrm{N/C} \) to the left.
04

Calculating Electric Field at \( x = +6.0 \, \mathrm{cm} \)

Calculate the electric field due to each charge at \( x = +6.0 \, \mathrm{cm} \).1. **Charge \( q_1 \):** - Distance \( r_1 = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m} \) - \( E_1 = \frac{k \cdot 8.5 \times 10^{-6}}{0.03^2} = 8.494 \times 10^6 \, \mathrm{N/C} \) directed to the right (positive x-direction).2. **Charge \( q_2 \):** - Distance \( r_2 = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m} \) - \( E_2 = \frac{k \cdot 21 \times 10^{-6}}{0.03^2} = 20.997 \times 10^6 \, \mathrm{N/C} \) directed to the left (negative x-direction).**Net Electric Field:**- \( E = E_2 - E_1 = 20.997 \times 10^6 - 8.494 \times 10^6 = 12.503 \times 10^6 \, \mathrm{N/C} \) to the left.

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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 to understanding electric fields and the forces between charged particles. It 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 them. The formula is given by:\[ F = k \frac{|q_1 q_2|}{r^2} \]where:
  • \( F \) is the magnitude of the force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \mathrm{N \cdot m^2/C^2} \).
  • \( q_1 \) and \( q_2 \) are the values of the charges.
  • \( r \) is the distance between the centers of the two charges.
It's important to note that the force is attractive if the charges are of opposite signs and repulsive if the charges are of the same sign. This law helps us calculate not only the force between charges but also understand the interaction of charged objects in different setups.
Point Charge
A point charge can be imagined as a charge located at a single point in space. In reality, all charges have dimensions, but when they are small enough compared to the distances involved in a problem, we can model them as point charges for simplicity. The electric field generated by a point charge is considered radially symmetric and always extends in uniform spherical patterns around the charge.The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is defined by the equation:\[ E = \frac{k |q|}{r^2} \]
  • The electric field emanates outwards from a positive point charge.
  • It converges inward towards a negative point charge.
This field represents how a point charge can exert a force on nearby charges. It's a crucial concept for solving problems involving charge distributions and helps simplify complex interactions among multiple charges.
Electric Field Direction
The direction of an electric field is defined as the direction of the force that would be experienced by a positive test charge placed in the field. This direction depends on the sign of the charges involved:
  • For a positive charge, the electric field points away from the charge.
  • For a negative charge, the electric field points towards the charge.
Knowing the direction of the electric field is crucial for predicting how other charges will move. In problems like those involving multiple charges, you consider the contributions to the electric field from each charge separately, then use vector addition to find the net electric field. The net direction is determined by the relative magnitudes and directions of individual electric fields contributed by each charge. This concept helps us not only envision how the charge interacts with its surroundings but also assists in solving complex configurations of charges.

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Most popular questions from this chapter

Interactive Solution \(\underline{18.15}\) at provides a model for solving this type of problem. Two small objects, \(\mathrm{A}\) and \(\mathrm{B},\) are fixed in place and separated by \(3.00 \mathrm{~cm}\) in a vacuum. Object \(\mathrm{A}\) has a charge of \(+2.00 \mu \mathrm{C},\) and object \(\mathrm{B}\) has a charge of \(-2.00 \mu \mathrm{C}\). How many electrons must be removed from \(\mathrm{A}\) and put onto \(\mathrm{B}\) to make the electrostatic force that acts on each object an attractive force whose magnitude is \(68.0 \mathrm{~N} ?\)

Consider three identical metal spheres, \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\). Sphere A carries a charge of \(+5 q .\) Sphere \(\mathrm{B}\) carries a charge of \(-q\). Sphere \(\mathrm{C}\) carries no net charge. Spheres \(\mathrm{A}\) and \(\mathrm{B}\) are touched together and then separated. Sphere \(\mathrm{C}\) is then touched to sphere \(A\) and separated from it. Last, sphere \(C\) is touched to sphere \(B\) and separated from it. (a) How much charge ends up on sphere \(\mathrm{C}\) ? What is the total charge on the three spheres (b) before they are allowed to touch each other and (c) after they have touched?

Two particles are in a uniform electric field whose value is \(+2500 \mathrm{~N} / \mathrm{C}\). The mass and charge of particle 1 are \(m_{1}=1.4 \times 10^{-5} \mathrm{~kg}\) and \(q_{1}=-7.0 \mu \mathrm{C},\) while the corresponding values for particle 2 are \(m_{2}=2.6 \times 10^{-5} \mathrm{~kg}\) and \(q_{2}=+18 \mu \mathrm{C} .\) Initially the particles are at rest. The particles are both located on the same electric field line, but are separated from each other by a distance \(d\). When released, they accelerate, but always remain at this same distance from each other. Find \(d\).

A charge \(+q\) is located at the origin, while an identical charge is located on the \(x\) axis at \(x=+0.50 \mathrm{~m}\). A third charge of \(+2 q\) is located on the \(x\) axis at such a place that the net electrostatic force on the charge at the origin doubles, its direction remaining unchanged. Where should the third charge be located?

Consult Concept Simulation 18.1 at for insight into this problem. Three charges are fixed to an \(x, y\) coordinate system. A charge of \(+18 \mu \mathrm{C}\) is on the \(y\) axis at \(y=+3.0 \mathrm{~m}\). A charge of \(-12 \mu \mathrm{C}\) is at the origin. Last, a charge of \(+45 \mu \mathrm{C}\) is on the \(x\) axis at \(x=+3.0 \mathrm{~m}\). Determine the magnitude and direction of the net electrostatic force on the charge at \(x=+3.0 \mathrm{~m}\). Specify the direction relative to the \(-x\) axis.
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