/*! 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 28 At a certain distance from a poi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 \(\mathrm{V}\) and \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?

Short Answer

Expert verified
Distance is approximately 0.415 m, charge is about \( 2.30 \times 10^{-10} \mathrm{C} \), field points away from charge.

Step by step solution

01

Understanding the Relationship between Potential and Electric Field

The electric potential \( V \) at a distance \( r \) from a point charge \( Q \) is given by \( V = \frac{kQ}{r} \), where \( k \) is Coulomb's constant (\( 8.99 \, \times \, 10^9 \, \mathrm{N} \, \mathrm{m}^2/\mathrm{C}^2 \)). The electric field \( E \) is given by \( E = \frac{kQ}{r^2} \). We have \( V = 4.98 \, \mathrm{V} \) and \( E = 12.0 \, \mathrm{V}/\mathrm{m} \).
02

Relating V and E to Find Distance

Using the given equations, we have two equations: \(\frac{kQ}{r} = 4.98 \) and \( \frac{kQ}{r^2} = 12.0 \). Dividing the potential equation by the electric field equation \( \frac{kQ/r}{kQ/r^2} = \frac{4.98}{12.0} \), we can solve for \( r \) as: \( r = \frac{4.98}{12.0} \approx 0.415 \, \mathrm{m} \).
03

Finding the Magnitude of the Point Charge

Using \( V = \frac{kQ}{r} \), substitute \( r = 0.415 \, \mathrm{m} \) and \( V = 4.98 \, \mathrm{V} \) to find \( Q \). Rearranging gives \( Q = \frac{Vr}{k} = \frac{4.98 \, \times \, 0.415}{8.99 \, \times \, 10^9} \approx 2.30 \, \times \, 10^{-10} \, \mathrm{C} \).
04

Determining the Direction of the Electric Field

The electric field points away from positive charges and towards negative charges. Since only the magnitudes are provided and not whether the field or potential is positive or negative, with no additional charges nearby, we typically assume the point charge to be positive, hence the field points away from the charge.

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 a hypothetical charge that is considered to be concentrated at a single point in space. This notion is incredibly useful in physics, especially when studying the behavior of electric fields and potentials, as it simplifies calculations. Point charges are assumed to have no physical size, which allows physicists to analyze their effects without worrying about the complexities that come with real objects with volumes.

Point charge is a vital concept in electromagnetism because:
  • It allows us to easily calculate electric fields and potentials.
  • It lays the foundation for understanding more complex charge distributions.
  • It is a simplified model that still delivers high precision in predicting the real-world behavior of more complex systems.
In practical terms, when dealing with a point charge, the electric potential and field can be described using simple equations. These relationships reveal how the charge influences the space around it, giving insight into how it interacts with other charges in its vicinity.
Coulomb's Law
Coulomb's Law is a fundamental principle that describes how two point charges interact in an electric field. It states that the force between two point charges is directly proportional to the product of the magnitudes of the charges, and inversely proportional to the square of the distance between them.

The law is mathematically expressed as:\[ F = k \frac{Q_1 Q_2}{r^2} \]where:
  • \( F \) is the magnitude of the force between the charges,
  • \( Q_1 \) and \( Q_2 \) are the amounts of the two charges,
  • \( r \) is the distance between the charges, and
  • \( k \) is Coulomb’s constant, approximately \( 8.99 \times 10^9 \mathrm{N} \mathrm{m}^2/\mathrm{C}^2 \).
Coulomb's Law is instrumental for understanding electrostatic interactions and is the electrostatic equivalent of Newton's law of universal gravitation. The law explains:
  • How electric forces can both attract and repel, unlike gravitational forces which only attract.
  • The importance of distance as even a small increase can greatly reduce the force, which is why charged objects need to be close to significantly interact.
  • The role of the medium between the charges, as this law assumes a vacuum or air, but different materials can influence the force.
Electric Field Direction
The direction of an electric field is conceptually simple but crucial to understanding the nature of electric interactions. It is defined as the direction that a positive test charge would move if placed within the field. This means that the electric field always points away from positive charges and toward negative charges.

Understanding the direction of electric fields helps in:
  • Predicting how electrons will move, as they are negatively charged and thus move opposite to the field direction.
  • Designing electric circuits and components by informing the placement of positive and negative terminals.
  • Understanding the behavior of various materials in electric fields, particularly conductive versus insulative materials.
Let's consider our exercise: if we assume that the point charge is positive, the electric field will point away from it. This aligns with the observations about how electric fields operate around point charges, providing straightforward predictions about interactions with other charges or test charges placed in the vicinity.

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

Self-Energy of a Sphere of Charge. A solid sphere of radius \(R\) contains a total charge \(Q\) distributed uniformly through- out its volume. Find the energy needed to assemble this charge by bringing infinitesimal charges from far away. This energy is called the "self-energy" of the charge distribution. (Hint: After you have assembled a charge \(q\) in a sphere of radius \(r\) , how much energy would it take to add a spherical shell of thickness \(d r\) having charge \(d q ?\) Then integrate to get the total energy.)

A small particle has charge \(-5.00 \mu \mathrm{C}\) and mass \(2.00 \times 10^{-4} \mathrm{kg}\) . It moves from point \(A\) , where the electric potential is \(V_{A}=+200 \mathrm{V},\) to point \(B\) , where the electric potential is \(V_{B}=+800 \mathrm{V}\) . The electric force is the only force acting on the particle. The particle has speed 5.00 \(\mathrm{m} / \mathrm{s}\) at point \(A .\) What is its speed at point \(B ?\) Is it moving faster or slower at \(B\) than at \(A ?\) Explain.

An infinitely long line of charge has linear charge density \(5.00 \times 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton (mass \(1.67 \times 10^{-27} \mathrm{kg}\) . charge \(+1.60 \times 10^{-19} \mathrm{C} )\) is 18.0 \(\mathrm{cm}\) from the line and moving directly toward the line at \(1.50 \times 10^{3} \mathrm{m} / \mathrm{s}\) (a) Calculate the proton's initial kinetic energy. (b) How close does the proton get to the line of charge? (Hints See Example \(23.10 . )\)

A positive charge \(q\) is fixed at the point \(x=0, y=0,\) and a negative charge \(-2 q\) is fixed at the point \(x=a, y=0\) . (a) Show the positions of the charges in a diagram. (b) Derive an expression for the potential \(V\) at points on the \(x\) -axis as a function of the coordinate \(x\) . Take \(V\) to be zero at an infinite distance from the charges. (c) At which positions on the \(x\) -axis is \(V=0 ?(\text { d) Graph }\) \(V\) at points on the \(x\) -axis as a function of \(x\) in the range from \(x=-2 a\) to \(x=+2 a\) (e) What does the answer to part \((b)\) become when \(x \gg a ?\) Explain why this result is obtained.

Alpha particles \(\left(\text { mass }=6.7 \times 10^{-27} \mathrm{kg}, \text { charge }=+2 e\right)\) are shot directly at a gold foil target. We can model the gold nucleus as a uniform sphere of charge and assume that the gold does not move. (a) If the radius of the gold nucleus is \(5.6 \times\) \(10^{-15} \mathrm{m}\) , what minimum speed do the alpha particles need when they are far away to reach the surface of the gold nucleus? (lanore relativistic effects.) (b) Give good physical reasons why we can ignore the effects of the orbital electrons when the alpha particle is (i) outside the electron orbits and (ii) inside the electron orbits.

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.