/*! 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 40 Two +400-nC point charges are in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 24: Problem 40

Two +400-nC point charges are in vacuum separated by \(20.0 \mathrm{~cm}\). Determine the electric field at a point midway between the charges.

Short Answer

Expert verified
The electric field at the midpoint is 0 N/C.

Step by step solution

01

Identify Given Values

We are given two point charges, each with a charge of \( +400 \text{ nC} \), separated by a distance of \( 20.0 \text{ cm} \). The point where we need to find the electric field is midway, which means it is distanced equally between both charges.
02

Electric Field Equation

Recall the equation for the electric field due to a point charge: \( E = \frac{k \cdot |q|}{r^2} \). Here, \( k \) is Coulomb's constant \( 8.99 \times 10^9 \text{ Nm}^2/ ext{C}^2 \), \( q \) is the charge, and \( r \) is the distance from the charge.
03

Calculate Distance to Midpoint

The distance from each charge to the midpoint is half of the total distance between the charges, which is \( r = \frac{20.0 \text{ cm}}{2} = 10.0 \text{ cm} = 0.1 \text{ m} \).
04

Calculate Electric Field from Each Charge

Use the electric field formula for each charge at the midpoint. Since both charges are identical: \[ E = \frac{8.99 \times 10^9 \cdot 400 \times 10^{-9}}{(0.1)^2} = \frac{8.99 \times 10^9 \cdot 400 \times 10^{-9}}{0.01} \] After simplifying, \[ E = 3.596 \times 10^5 \text{ N/C} \].
05

Vector Nature of Electric Fields

Determine the direction of the electric field. Since both charges are positive, the electric fields at the midpoint due to each charge will push away from the charges in opposite directions, effectively cancelling each other out.
06

Resultant Electric Field

As both fields cancel each other out due to symmetry and equal magnitude, the total electric field at the midpoint is \[ E_{total} = 0 \text{ N/C} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in physics that describes the electric force between two stationary charged particles. It states that the force (\( F \)) between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The mathematical expression of Coulomb's Law is:
  • \( F = k \frac{|q_1 \, q_2|}{r^2} \)
where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, ext{Nm}^2/ ext{C}^2 \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the separation distance between the charges.
This principle helps us understand interactions at the atomic and subatomic levels, explaining how charged particles attract or repel each other.
This is crucial for analyzing situations like the one in our original exercise.
Electric Charges
Electric charges are fundamental properties of matter responsible for electric forces and fields. They come in two types: positive and negative. Like charges repel each other, while opposite charges attract. The unit of charge is the coulomb (C).
In our exercise, both charges are positive, each possessing a charge of \( +400 \, ext{nC} \), or nanocoulombs, which is \( 400 \times 10^{-9} \, ext{C} \). This is a relatively small charge that generates an electric field, influencing other charged particles in its vicinity.
Understanding how electric charges behave and interact helps us grasp the basics of electrical circuits and electronics.
Electric Field Equation
The electric field equation quantifies the strength of an electric field generated by a point charge. This field represents the force a charge would experience if placed in the field. The equation is defined as:
  • \( E = \frac{k \cdot |q|}{r^2} \)
This equation shows that the electric field (\( E \)) is proportional to the magnitude of the charge (\( q \)) and inversely proportional to the square of the distance (\( r \)) from the charge.
In our exercise, this principle is applied to calculate the electric field at a point equidistant from two identical charges. By plugging in the values, we find that the field due to each charge is significant, yet they cancel each other due to their opposing directions.
Vector Nature of Electric Fields
Electric fields are vector quantities, meaning they have both magnitude and direction. The direction of an electric field is defined as the direction a positive test charge would move if placed in the field.
In the context of our problem, since both charges are positive, the electric field due to each charge at the midpoint between them points away from the respective charge. This results in two fields of equal magnitude but opposite directions.
When fields are equal and opposite, as in this exercise, they cancel out, leading to a net electric field of zero at the midpoint. This demonstrates how the vector nature of fields is crucial for determining net effects in scenarios with multiple charges.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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, (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 $$ v=\sqrt{\frac{\left(8.2 \times 10^{-8} \mathrm{~N}\right)(r)}{m}}=\sqrt{\frac{\left(8.2 \times 10^{-8} \mathrm{~N}\right)\left(5.3 \times 10^{-11} \mathrm{~m}\right)}{9.1 \times 10^{-31} \mathrm{~kg}}}=2.2 \times 10^{6} \mathrm{~m} / \mathrm{s} $$

Two point charges, one \(+400.0 \mathrm{nC}\) and the other \(-400.0 \mathrm{nC}\), located \(20.00 \mathrm{~cm}\) to the right of the first, are in vacuum. Determine the electric field (magnitude and direction) at a point midway between the charges.

Find the ratio of the Coulomb electric force \(F_{E}\) to the gravitational force \(F_{G}\) between two electrons in vacuum. From Coulomb's Law and Newton's Law of gravitation, The electric force is much stronger than the gravitational force. $$ \begin{array}{c} F_{E}=k \frac{q^{2}}{r^{2}} \text { and } F_{G}=G \frac{m^{2}}{r^{2}} \\ \text { Therefore, } \quad \begin{aligned} \frac{F_{E}}{F_{G}} &=\frac{k q_{*}^{2} / r^{2}}{G m^{2} / r^{2}}=\frac{k q_{i}^{2}}{G m^{2}} \\ =& \frac{\left(9.0 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{\left(6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\right)\left(9.1 \times 10^{-31} \mathrm{~kg}\right)^{2}}=4.2 \times 10^{42} \end{aligned} \end{array} $$

Compute the force on each of two electrons when they are separated in vacuum by a distance corresponding to the approximate size of an atom \((0.100 \mathrm{~nm})\).
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Wave Optics

Read Explanation

Modern Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.