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Chapter 18: Problem 67

ssm 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
(a) At \( x=0 \), net field is 6.15 MN/C left; (b) At \( x=6.0 \), net field is 29.5 MN/C right.

Step by step solution

01

Understand Electric Field Due to a Point Charge

The electric field due to a point charge \( q \) at a distance \( r \) is given by the formula \( E = k_e \frac{|q|}{r^2} \), where \( k_e = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). The direction of the field is away from the charge if it is positive and towards the charge if it is negative.
02

Calculate Distances for Part (a)

At \( x = 0 \text{ cm} \): - Distance from \( q_1 \) is \( r_1 = x_1 = 3.0 \text{ cm} = 0.03 \text{ m} \).- Distance from \( q_2 \) is \( r_2 = x_2 = 9.0 \text{ cm} = 0.09 \text{ m} \).
03

Calculate Electric Field at x=0 cm

- For \( q_1 = +8.5 \mu \text{C} \): \[ E_1 = k_e \frac{8.5 \times 10^{-6}}{(0.03)^2} = 8.5 \times 10^9 \frac{8.5 \times 10^{-6}}{9 \times 10^{-4}} = 8.49 \times 10^6 \text{ N/C} \] pointing to the left.- For \( q_2 = -21 \mu \text{C} \): \[ E_2 = k_e \frac{21 \times 10^{-6}}{(0.09)^2} = 8.49 \times 10^6 \frac{21 \times 10^{-6}}{81 \times 10^{-4}} = 2.34 \times 10^6 \text{ N/C} \] pointing to the right.- Net field \( E = E_1 - E_2 = (8.49 - 2.34) \times 10^6 = 6.15 \times 10^6 \text{ N/C} \) to the left.
04

Calculate Distances for Part (b)

At \( x = 6.0 \text{ cm} = 0.06 \text{ m} \):- Distance from \( q_1 \) is \( r_1 = 0.06 \text{ m} - 0.03 \text{ m} = 0.03 \text{ m} \).- Distance from \( q_2 \) is \( r_2 = 0.09 \text{ m} - 0.06 \text{ m} = 0.03 \text{ m} \).
05

Calculate Electric Field at x=6.0 cm

- For \( q_1 = +8.5 \mu \text{C} \): \[ E_1 = k_e \frac{8.5 \times 10^{-6}}{(0.03)^2} = 8.49 \times 10^6 \text{ N/C} \] pointing to the right.- For \( q_2 = -21 \mu \text{C} \): \[ E_2 = k_e \frac{21 \times 10^{-6}}{(0.03)^2} = 2.10 \times 10^7 \text{ N/C} \] pointing to the right.- Net field \( E = E_1 + E_2 = (8.49 + 21.0) \times 10^6 = 2.95 \times 10^7 \text{ N/C} \) to the right.

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Key Concepts

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

Point Charge
A point charge is a fundamental concept in electrostatics, referring to an idealized charge that is concentrated at a single point in space. This concept helps in simplifying the study of electric fields and forces between charged objects. By treating charges as point charges, we can focus on their interactions without worrying about their size or shape.
  • The magnitude of a point charge is typically measured in coulombs (C).
  • For example, a charge of \(8.5 \, \mu \text{C}\) (microcoulombs) can be considered a point charge in problems dealing with electric fields.
  • Point charges can be positive or negative, affecting the direction of the electric field they produce.
Understanding point charges is critical for calculating the electric field they generate, which further informs us about how other charges would behave in their presence.
Electric Field Formula
The electric field produced by a point charge is calculated using a specific formula. This formula shows how the electric field (E) relates to the charge (q) and the distance (r) from the charge.
The electric field formula is:\[E = k_e \frac{|q|}{r^2}\]
where:
  • E is the electric field in newtons per coulomb (N/C)
  • q is the magnitude of the point charge in coulombs (C)
  • r is the distance from the charge in meters (m)
  • k_e, the Coulomb's constant, has a value of 8.99 \times 10^9 \text{ N m}^2/\text{C}^2
This formula indicates that the strength of the electric field decreases with the square of the distance from the point charge. Thus, the closer you are to the charge, the stronger the electric field.
Net Electric Field
The net electric field at any point is the vector sum of the electric fields due to all individual point charges present. This is important because in most scenarios, you will have more than one charge affecting a certain point.
To find the net electric field:
  • Calculate the electric field due to each point charge using the electric field formula.
  • Consider both the magnitude and direction of each field when summing them up.
  • For example, at a specific location, you might find that one charge produces a field of \(8.49 \times 10^6 \text{ N/C}\) to the left, and another produces a field of \(2.34 \times 10^6 \text{ N/C}\) to the right.
  • The net electric field would be \(E = E_1 - E_2\) or \(E = E_1 + E_2\), depending on the directions.
Understanding how to calculate the net electric field is crucial for predicting the behavior of charges in a multi-charge system.
Direction of Electric Field
The direction of the electric field is a key concept in understanding how point charges interact with their surroundings. The electric field points in a direction that a positive test charge would move if placed within that field.
  • For a positive point charge, the electric field radiates outward, away from the charge. This means if you place a positive test charge nearby, it will be pushed away, following the field lines.
  • For a negative point charge, the electric field points towards the charge. A positive test charge will move towards it, being attracted by the negative charge.
  • In calculations, knowing the direction helps to correctly add or subtract field vectors, ensuring you find the accurate net electric field.
Visualizing the direction of electric fields can facilitate a better understanding of electrostatics and help to solve complex problems involving multiple charges.

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Most popular questions from this chapter

A small object, which has a charge \(q=7.5 \mu \mathrm{C}\) and mass \(m=9.0 \times 10^{-5} \mathrm{kg},\) is placed in a constant electric field. Starting from rest, the object accelerates to a speed of \(2.0 \times 10^{3} \mathrm{m} / \mathrm{s}\) in a time of \(0.96 \mathrm{s}\). Determine the magnitude of the electric field.

mmh Two spherical shells have a common center. \(A-1.6 \times 10^{-6}\) C charge is spread uniformly over the inner shell, which has a radius of \(0.050 \mathrm{m} . \mathrm{A}+5.1 \times 10^{-6} \mathrm{C}\) charge is spread uniformly over the outer shell, which has a radius of \(0.15 \mathrm{m}\). Find the magnitude and direction of the electric field at a distance (measured from the common center) of (a) \(0.20 \mathrm{m}\) (b) \(0.10 \mathrm{m},\) and (c) \(0.025 \mathrm{m}\)

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?

An electric field of \(260000 \mathrm{N} / \mathrm{C}\) points due west at a certain spot. What are the magnitude and direction of the force that acts on a charge of \(-7.0 \mu \mathrm{C}\) at this spot?

The membrane surrounding a living cell consists of an inner and an outer wall that are separated by a small space. Assume that the membrane acts like a parallel plate capacitor in which the effective charge density on the inner and outer walls has a magnitude of \(7.1 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2} .\) (a) What is the magnitude of the electric field within the cell membrane? (b) Find the magnitude of the electric force that would be exerted on a potassium ion \(\left(\mathrm{K}^{+} ;\right.\) charge \(\left.=+e\right)\) placed inside the membrane.
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