/*! 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 29 A point charge of \(-3.0 \times ... [FREE SOLUTION] | 91影视

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Chapter 24: Problem 29

A point charge of \(-3.0 \times 10^{-5} \mathrm{C}\) is placed at the origin of coordinates in vacuum. Find the electric field at the point \(x=5.0 \mathrm{~m}\) on the \(x\) -axis.

Short Answer

Expert verified
The electric field at \( x=5.0 \: \text{m} \) is \( 1.0788 \times 10^4 \: \text{N/C} \), directed towards the origin.

Step by step solution

01

Understand the Formula for Electric Field

The electric field \( E \) created by a point charge \( q \) at a distance \( r \) is given by the formula: \[ E = \frac{k \cdot |q|}{r^2} \] where \( k \) is Coulomb's constant, \( k = 8.99 \times 10^9 \: \text{N}\cdot\text{m}^2/\text{C}^2 \).
02

Identify the Given Values

We know that the charge \( q = -3.0 \times 10^{-5} \: \text{C} \) and the distance \( r = 5.0 \: \text{m} \) on the \( x \)-axis from the charge.
03

Substitute Values Into the Formula

Substitute the known values into the electric field formula: \[ E = \frac{8.99 \times 10^9 \times | -3.0 \times 10^{-5} |}{(5.0)^2} \]
04

Calculate the Electric Field Magnitude

First calculate the magnitude of the charge: \( |-3.0 \times 10^{-5}| = 3.0 \times 10^{-5} \). The expression becomes \[ E = \frac{8.99 \times 10^9 \times 3.0 \times 10^{-5}}{25} \]. Calculate \( E \): \[ E = \frac{2.697 \times 10^5}{25} = 1.0788 \times 10^4 \: \text{N/C} \].
05

Determine the Direction of the Electric Field

Since the charge is negative, the direction of the electric field is towards the charge. Since we are considering the point \( x = 5.0 \: \text{m} \) on the \( x \)-axis, the electric field will point towards the origin where the negative charge is located.

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Key Concepts

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

Coulomb's constant
Coulomb's constant, often denoted by the symbol \( k \), is a fundamental constant in physics that plays a crucial role in the calculation of electric fields due to point charges. It is a measure of the strength of the electric force between two charges separated by a certain distance. In the context of a vacuum, Coulomb's constant has a value of \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \). This constant allows us to understand how forces are transmitted through space by the electric field.
  • Importance in Electrostatics: Coulomb's constant is critical because it quantifies the level of interaction between charged objects.
  • Relation to the Permittivity of Free Space: It is inversely related to the permittivity of free space \( \varepsilon_0 \), defined by \( k = \frac{1}{4\pi \varepsilon_0} \).
When using Coulomb's constant in calculations, it's essential to ensure all measurements are in consistent units 鈥� Newtons for force, meters for distance, and Coulombs for charge.
Point Charge
A point charge is a concept in electrostatics representing an electric charge that is assumed to be located at a single point in space. This simplification makes it easier to apply mathematical formulas to calculate the electric fields and forces.
  • Simplification: Although actual charges have a spatial distribution, they can often be treated as point charges when their size is negligible compared to other dimensions involved in the problem.
  • Applications: Point charges are a fundamental part of many physics problems, helping to simplify complex systems where charge distribution is not uniform.
In this exercise, the point charge of \(-3.0 \times 10^{-5} \, \text{C}\) is placed at the origin, which serves as a reference point for calculating the electric field.
Electric Field Direction
The direction of an electric field is a crucial part of understanding how electric forces act. The electric field direction indicates where a positive test charge placed in the field would move due to the force exerted by the field.
  • Positive vs Negative Charges: For a positive point charge, the electric field radiates outwards, while for a negative point charge, it converges inwards towards the charge.
  • Understanding the Context: In our specific exercise, we have a negative point charge, so the electric field at any given point, like \( x = 5.0 \, \text{m} \), is directed towards the charge itself because it attracts positive charges.
The direction of the electric field is fundamental, as it determines the trajectory and behavior of other charges within the field. This determines how we interpret interactions in both theoretical problems and practical applications.

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Most popular questions from this chapter

If two equal point charges, each of \(1 \mathrm{C}\), were separated in air by a distance of \(1 \mathrm{~km}\), what would be the force between them?

In the Bohr model of the hydrogen atom, an electron \((q=-e)\) circles a proton \(\left(q^{\prime}=e\right)\) in an orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). The attraction between the proton and electron furnishes the centripetal force needed to hold the electron in orbit. Find \((a)\) the force of electrical attraction between the particles and \((b)\) the electron's speed. The electron mass is \(9.1 \times 10^{-31} \mathrm{~kg}\). The electron and proton are essentially point charges. Accordingly, (a) \(\quad F_{E}=k_{0} \frac{q . q^{\prime}}{r^{2}}=\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(5.3 \times 10^{-11} \mathrm{~m}\right)^{2}}=8.2 \times 10^{-8} \mathrm{~N}=82 \mathrm{nN}\) (b) The force found in \((a)\) is the centripetal force, \(m v^{2} / r\). Therefore, $$8.2 \times 10^{-8} \mathrm{~N}=\frac{m v^{2}}{r}$$ from which it follows that $$ v=\sqrt{\frac{\left(8.2 \times 10^{-8} \mathrm{~N}\right)(r)}{m}}=\sqrt{\frac{\left(8.2 \times 10^{-8} \mathrm{~N}\right)\left(5.3 \times 10^{-11} \mathrm{~m}\right)}{9.1 \times 10^{-31} \mathrm{~kg}}}=2.2 \times 10^{6} \mathrm{~m} / \mathrm{s} $$

A 0.200-g ball in air hangs from a thread in a uniform vertical electric field of \(3.00 \mathrm{kN} / \mathrm{C}\) directed upward. What is the charge on the ball if the tension in the thread is \((a)\) zero and \((b) 4.00 \mathrm{mN} ?\)

Two charged metal plates in vacuum are \(15 \mathrm{~cm}\) apart as drawn in Fig. \(24-7\). The electric field between the plates is uniform and has a strength of \(E=3000 \mathrm{~N} / \mathrm{C}\). An electron \(\left(q=-e, m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\right)\) is released from rest at point \(P\) just outside the negative plate. ( \(a\) ) How long will it take to reach the other plate? \((b)\) How fast will it be going just before it hits? The electric field lines show the force on a positive charge. (A positive charge would be repelled to the right by the positive plate and attracted to the right by the negative plate.) An electron, being negative, will experience a force in the opposite direction, toward the left, of magnitude $$F_{E}=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)(3000 \mathrm{~N} / \mathrm{C})=4.8 \times 10^{-16} \mathrm{~N}$$ Because of this force, the electron experiences an acceleration toward the left given by $$a=\frac{F_{E}}{m}=\frac{4.8 \times 10^{-16} \mathrm{~N}}{9.1 \times 10^{-31} \mathrm{~kg}}=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}$$ In the motion problem for the electron released at the negative plate and traveling to the positive plate, $$v_{i}=0 \quad x=0.15 \mathrm{~m} \quad a=5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}$$ (a) From \(x=v_{i} t+\frac{1}{2} a t^{2}\) we have $$t=\sqrt{\frac{2 x}{a}}=\sqrt{\frac{(2)(0.15 \mathrm{~m})}{5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}}}=2.4 \times 10^{-8} \mathrm{~s}$$ (b) \(\quad v=v_{i}+a t=0+\left(5.3 \times 10^{14} \mathrm{~m} / \mathrm{s}^{2}\right)\left(2.4 \times 10^{-8} \mathrm{~s}\right)=1.30 \times 10^{7} \mathrm{~m} / \mathrm{s}\) As you will see in Chapter 41 , relativistic effects begin to become important at speeds above this. Therefore, this approach must be modified for very fast particles.

What is the force of repulsion between two argon nuclei that are separated in vacuum by \(1.0 \mathrm{~nm}\left(10^{-9} \mathrm{~m}\right) ?\) The charge on an argon nucleus is \(+18 e\)
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