/*! 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 41 \(\bullet$$\bullet\) A point cha... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet$$\bullet\) A point charge of \(-4.00 \mathrm{nC}\) is at the origin, and a second point charge of \(+6.00 \mathrm{nC}\) is on the \(x\) axis at \(x=0.800 \mathrm{m}\) . Find the magnitude and direction of the electric field at each of the following points on the \(x\) axis: (a) \(x=20.0 \mathrm{cm}\) , (b) \(x=1.20 \mathrm{m},(\mathrm{c}) x=-20.0 \mathrm{cm} .\)

Short Answer

Expert verified
1) At \(x=20.0\ \mathrm{cm}\), \(E = 1.35 \times 10^6\ \mathrm{N/C}\) left. 2) At \(x=1.20\ \mathrm{m}\), \(E = 7.49 \times 10^5\ \mathrm{N/C}\) right. 3) At \(x=-20.0\ \mathrm{cm}\), \(E = 8.99 \times 10^6\ \mathrm{N/C}\) left.

Step by step solution

01

Identifying Known Values

We have two point charges. First charge (\(Q_1\)) is \(-4.00 \mathrm{nC}\) located at the origin (\(x = 0\)). Second charge (\(Q_2\)) is \(+6.00 \mathrm{nC}\) located at \(x = 0.800 \mathrm{m}\).
02

Applying Electric Field Formula

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 \(k = 8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\) (Coulomb's constant).
03

Calculating Electric Field at \(x=0.20\ \mathrm{m}\)

Distance from \(Q_1\) to \(0.20\ \mathrm{m}\) is \(0.20\ \mathrm{m}\) and from \(Q_2\) is \(0.600\ \mathrm{m}\). For \(Q_1\) (negative charge), \(E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{0.20^2}\). For \(Q_2\) (positive charge), \(E_2 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{0.600^2}\). Calculate each, then determine the net electric field. Since \(Q_1\) is negative, \(E_1\) is directed towards the charge and \(E_2\) is directed away.
04

Calculating Electric Field at \(x=1.20\ \mathrm{m}\)

Distance from \(Q_1\) is \(1.20\ \mathrm{m}\) and from \(Q_2\) is \(0.400\ \mathrm{m}\). For \(Q_1\), \(E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{1.20^2}\). For \(Q_2\), \(E_2 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{0.400^2}\). Calculate each. Determine the net electric field with \(E_1\) towards \(Q_1\) and \(E_2\) away from \(Q_2\).
05

Calculating Electric Field at \(x=-0.20\ \mathrm{m}\)

Distance from \(Q_1\) is \(0.20\ \mathrm{m}\) and from \(Q_2\) is \(1.00\ \mathrm{m}\). For \(Q_1\), \(E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{0.20^2}\). For \(Q_2\), \(E_2 = \frac{8.99 \times 10^9 \times 6.00 \times 10^{-9}}{1.00^2}\). Calculate each, and find the net field, noting \(E_1\) is directed toward and \(E_2\) away.
06

Determine Net Electric Field and Direction for Each Point

For each point, sum the individual electric fields, considering their directions (left or right along the x-axis). The net field will be the algebraic sum where the direction is influenced by the larger magnitude of either \(E_1\) or \(E_2\).

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 Charge
A point charge is an idealized model allowing us to simplify complex electric field calculations. This model assumes the charge is concentrated at a single point in space, making it easier to study the influence it has on other charges or spaces around it.
In the provided exercise, we have two point charges:
  • The first point charge (\(Q_1\)) is \(-4.00 \,\mathrm{nC}\), located at the origin of the coordinate system (\(x = 0\)).
  • The second point charge (\(Q_2\)) is \(+6.00 \,\mathrm{nC}\), placed on the \(x\) axis at a distance \(x = 0.800\,\mathrm{m}\).
These charges create electric fields around them, influencing the space and any other charges found along the \(x\)-axis.
Understanding how point charges interact is crucial in solving problems concerning electric fields.
Coulomb's Law
Coulomb's Law is a fundamental principle explaining the force between two charges. According to this law, the electric force (\(F\)) between two charges is directly proportional to the product of their magnitudes (\(|Q_1 \times Q_2|\)) and inversely proportional to the square of the distance (\(r^2\)) between them.
Thus, the equation for Coulomb’s Law is given by:
\[ F = k \frac{|Q_1 \cdot Q_2|}{r^2} \]
where \(k\) is Coulomb's constant (\(8.99 \times 10^9 \mathrm{N \cdot m^2/C^2}\)).

For electric fields, this law helps determine the field strength (\(E\)) generated by a single point charge at a certain distance.
The electric field from a point charge is computed using:
\[ E = \frac{k |Q|}{r^2} \]
In the exercise, this formula was applied for each point charge to calculate the electric fields at different points on the \(x\)-axis.
Understanding the calculations can aid students in solving similar problems involving the interaction of point charges.
Vector Addition
When dealing with electric fields, we often encounter multiple fields interacting at a single point. These fields add together vectorially, meaning we consider both their magnitude and direction.
To find the net electric field at a given point, we must perform vector addition for each field's contribution.
Consider the example at \(x = 0.20\,\mathrm{m}\):
  • Calculate the electric field contributed by each point charge using \(E = \frac{k |Q|}{r^2}\).
  • Then, consider the direction each field is pointing.
  • Since the point charge \(Q_1\) is negative, its field points towards the charge.
  • On the other hand, \(Q_2\) being positive causes the field to point away from it.
The net field is determined by summing the fields, taking into account their directions (which could be on opposite sides of the \(x\)-axis).
Mastering vector addition is crucial for determining overall field effects.
Electric Field Direction
Understanding electric field direction is key to correctly mapping forces and fields around point charges.
An electric field generated by a point charge has a distinct direction:
  • From a positive charge, electric fields radiate outwards.
  • For a negative charge, these fields point inward, towards the charge.
In our exercise, we determine electric field direction at different points by considering:
  • The sign of the charge (\(Q\)): negative or positive.
  • The position of the charges along the \(x\)-axis, relative to the point of interest.
  • For instance, at \(x = 0.20\,\mathrm{m}\), the electric field due to \(Q_1\) (negative) points left towards the origin, while \(Q_2\) (positive) points right.
This determination is essential for calculating net electric field and understanding charge interactions.

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

\(\bullet\) A positively charged rubber rod is moved close to a neutral copper ball that is resting on a nonconducting sheet of plastic. (a) Sketch the distribution of charges on the ball. (b) With the rod still close to the ball, a metal wire is briefly connected from the ball to the earth and then removed. After the rubber rod is also removed, sketch the distribution of charges (if any) on the copper ball.

\(\bullet$$\bullet\) Three point charges are arranged along the \(x\) axis. Charge \(q_{1}=-4.50 \mathrm{nC}\) is located at \(x=0.200 \mathrm{m},\) and charge \(q_{2}=+2.50 \mathrm{nC}\) is at \(x=-0.300 \mathrm{m} .\) A positive point charge \(q_{3}\) is located at the origin. (a) What must the value of \(q_{3}\) be for the net force on this point charge to have magnitude 4.00\(\mu \mathrm{N} ?\) (b) What is the direction of the net force on \(q_{3} ?\) (c) Where along the \(x\) axis can \(q_{3}\) be placed and the net force on it be zero, other than the trivial answers of \(x=+\infty\) and \(x=-\infty\) ?

\(\bullet\) A neutral conductor completely encloses a hole inside of it. You observe that the outer surface of this conductor carries a charge of \(-12 \mu \mathrm{C}\) (a) Can you conclude that there is a charge inside the hole? If so, what is this charge? (b) How much charge is on the inner surface of the conductor?

\(\bullet$$\bullet\) Two small plastic spheres are given positive electrical charges. When they are 15.0 \(\mathrm{cm}\) apart, the repulsive force between them has magnitude 0.220 \(\mathrm{N} .\) What is the charge on each sphere (a) if the two charges are equal? (b) if one sphere has four times the charge of the other?

\(\bullet\) (a) A closed surface encloses a net charge of 2.50\(\mu \mathrm{C}\) . What is the net electric flux through the surface? (b) If the electric flux through a closed surface is determined to be \(1.40 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) how much charge is enclosed by the surface?

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.