/*! 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 50 A point charge \(q_{1}=-4.00 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Short Answer

Expert verified
The net electric field at the origin is approximately \( 98.9 \text{ N/C} \) at an angle of \(-77.5^\circ\) from the positive x-axis.

Step by step solution

01

Identify Positions of Charges and the Point of Interest

Charge \( q_1 = -4.00 \text{ nC} \) is located at \((x_1, y_1) = (0.600, 0.800) \text{ m}\), and charge \( q_2 = +6.00 \text{ nC} \) is located at \((x_2, y_2) = (0.600, 0.0) \text{ m}\). The electric field is to be calculated at the origin \((0, 0)\).
02

Calculate the Distance to Each Charge

Calculate the distance from the origin to each charge. For \( q_1 \), the distance is \( r_1 = \sqrt{(0.600)^2 + (0.800)^2} \). For \( q_2 \), the distance is \( r_2 = \sqrt{(0.600)^2 + (0)^2} = 0.600 \text{ m}\).
03

Calculate the Electric Field Due to Each Charge

The magnitude of the electric field due to a point charge is given by \( E = \frac{k |q|}{r^2} \), where \( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \). Calculate \( E_1 \) due to \( q_1 \) and \( E_2 \) due to \( q_2 \).
04

Determine the Components of Each Electric Field

Since \( q_1 \) is at \( (0.600, 0.800) \), its electric field, \( E_1 \), can be broken into components: \( E_{1x} = E_1 \cos(\theta_1) \) and \( E_{1y} = E_1 \sin(\theta_1) \), where \( \theta_1 = \tan^{-1}(\frac{0.800}{0.600}) \). Similarly, \( E_{2y} = 0 \) because \( q_2 \) is along the y-axis, and \( E_{2x} = E_2 \) because it is along the x-axis.
05

Calculate the Net Electric Field Components

Find the net components: \( E_{net,x} = E_{1x} + E_{2x} \) and \( E_{net,y} = E_{1y} + E_{2y} \). Consider the signs of charges when calculating the direction of electric fields.
06

Calculate the Magnitude and Direction of the Net Electric Field

The magnitude of the net electric field is given by \( E_{net} = \sqrt{E_{net,x}^2 + E_{net,y}^2} \). The direction is \( \theta = \tan^{-1}\left(\frac{E_{net,y}}{E_{net,x}}\right) \).
07

Solve for Specific Values

Use the given charges and positions to compute specific values: \( r_1 = 1 \text{ m} \), \( E_1 = 8.99 \times 10^9 \times \frac{4.00 \times 10^{-9}}{(1)^2} \), and \( E_2 = 8.99 \times 10^9 \times \frac{6.00 \times 10^{-9}}{(0.600)^2} \). Calculate components and find \( E_{net} \) and direction \( \theta \).
08

Final Calculation and Result

Compute the numerical values for the electric field components and combine them to find the exact magnitude and direction at the origin. Use the results from the previous steps to summarize the net field at the origin.

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 91Ó°ÊÓ!

Key Concepts

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

Point Charges
Point charges are small charged particles considered to be at a single point in space. They serve as useful idealized models in physics to study electric fields and forces. In the given exercise, we are dealing with two point charges, \( q_1 \) and \( q_2 \), each with a specific charge and location. These charges create an electric field around them, due to the electric forces they exert on other charges nearby. Understanding the position of these charges is essential because the electric field strength depends on the distance between the point of interest and the charges.
For example, \( q_1 = -4.00 \text{ nC} \) is located at \((0.600, 0.800)\) meters, while \( q_2 = +6.00 \text{ nC} \) is located at \((0.600, 0.0)\) meters. Calculating the electric field involves determining the individual contribution from each charge at a specific point, like the origin.
Electric Field Components
The electric field produced by a point charge at a certain distance can be quantified as a vector with both magnitude and direction. To effectively analyze the net field from multiple charges, it's vital to break down their fields into components. Each component aligns with either the x-axis or y-axis.
If we look at \( q_1 \), situated at \((0.600, 0.800)\), its electric field vector can be split into horizontal (x-component) and vertical (y-component) parts. The x-component, \( E_{1x} \), and the y-component, \( E_{1y} \), can be calculated using trigonometric functions as follows:
  • \( E_{1x} = E_1 \cos(\theta_1) \)
  • \( E_{1y} = E_1 \sin(\theta_1) \)
\( \theta_1 \) is the angle formed with the horizontal, which is determined by the position of the charge. In this case, \( \theta_1 = \tan^{-1}\left(\frac{0.800}{0.600}\right) \).
For \( q_2 \), which lies on the x-axis at \((0.600, 0)\), the electric field has no y-component, so \( E_{2y} = 0 \), making the x-component \( E_{2x} \) the same as the whole electric field \( E_2 \). These components are crucial for calculating the net electric field.
Electric Field Direction
To determine the direction of the net electric field from multiple charges, we combine the directional contributions from each charge. This involves analyzing the angles at which the electric fields act.
Using the individual components found for \( q_1 \) and \( q_2 \), we compute the net electric field components \( E_{net,x} \) and \( E_{net,y} \):
  • \( E_{net,x} = E_{1x} + E_{2x} \)
  • \( E_{net,y} = E_{1y} + E_{2y} \)
The direction of the net electric field is the angle \( \theta \) relative to the x-axis, given by:
\[ \theta = \tan^{-1}\left(\frac{E_{net,y}}{E_{net,x}}\right) \]
Knowing \( \theta \) tells us where the resultant field is directed in the plane. Keep in mind, the sign of the charge affects the direction of the field lines. Positive charges push away field lines, while negative charges pull them closer.
Coulomb's Law
Coulomb's law is a fundamental principle that describes the electric force between two point charges. The law states that the electric force (F) between two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
\[ F = \frac{k |q_1 q_2|}{r^2} \]
where:
  • \( F \) is the magnitude of the force.
  • \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is Coulomb's constant.
  • \( |q_1| \) and \( |q_2| \) are the absolute values of the charges.
  • \( r \) is the distance between the charges.
This law provides the groundwork for calculating the electric field, since the field is essentially the force per unit charge. In the exercise, Coulomb's law helps us determine the electric fields created by \( q_1 \) and \( q_2 \) at the origin. Applying the formula for each point charge, we find their respective electric fields, which are then used to compute the total field experienced at the given point.

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

Two very large horizontal sheets are 4.25 \(\mathrm{cm}\) apart and carry equal but opposite uniform surface charge densities of magnitude \(\sigma .\) You want to use these sheets to hold stationary in the region between them an oil droplet of mass 324\(\mu\) that carries an excess of five electrons. Assuming that the drop is in vacuum, (a) which way should the electric field between the plates point, and (b) what should \(\sigma\) be?

CP At \(t=0\) a very small object with mass 0.400 \(\mathrm{mg}\) and charge \(+9.00 \mu \mathrm{C}\) is traveling at 125 \(\mathrm{m} / \mathrm{s}\) in the \(-x\) -direction. The charge is moving in a uniform electric field that is in the +y-direction and that has magnitude \(E=895 \mathrm{N} / \mathrm{C}\) . The gravitational force on the particle can be neglected. How far is the particle from the origin at \(t=7.00 \mathrm{ms} ?\)

A dipole consisting of charges \(\pm e, 220 \mathrm{nm}\) apart, is placed between two very large (essentially infinite) sheets carrying equal but opposite charge densities of 125\(\mu \mathrm{C} / \mathrm{m}^{2}\) . (a) What is the maximum potential energy this dipole can have due to the sheets, and how should it be oriented relative to the sheets to attain this value? (b) What is the maximum torque the sheets can exert on the dipole, and how should it be oriented relative to the sheets to attain this value? (c) What net force do the two sheets exert on the dipole?

Two very small \(8.55-\) g spheres, 15.0 \(\mathrm{cm}\) apart from center to center, are charged by adding equal numbers of electrons to each of them. Disregarding all other forces, how many electrons would you have to add to each sphere so that the two spheres will accelerate at 25.0 \(\mathrm{g}\) when released? Which way will they accelerate?

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A\) ? (b) A proton is moving in the uniform electric field of part (a). At point \(A,\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s},\) east. What is the speed of the proton at point \(B\) ?

See all solutions

Recommended explanations on Physics Textbooks

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.