/*! 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 46 Two point charges are on the \(x... [FREE SOLUTION] | 91Ó°ÊÓ

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Two point charges are on the \(x\) axis. Charge 1 is \(+q\) and is located at \(x=-1.0 \mathrm{m} ;\) charge 2 is \(+2 q\) and is located at \(x=1.0 \mathrm{m}\). Make sketches of the equipotential surfaces for this system (a) out to a a distance of about \(2.0 \mathrm{m}\) from the origin and (b) far from the origin. In each case, indicate the direction in which the potential increases.

Short Answer

Expert verified
Near the origin, the potential surfaces are asymmetric due to charge differences; far away they resemble a single charge system.

Step by step solution

01

Understand the Problem

We have two charges along the x-axis: a charge of "+q" at x = -1.0 m, and a charge of "+2q" at x = 1.0 m. Our task is to sketch the equipotential surfaces for these point charges at two different distances: near (up to 2.0 m from the origin) and far from the origin.
02

Review Equipotential Surfaces

Equipotential surfaces are surfaces where the electric potential is constant. For point charges, these surfaces are spherical. The potential increases as we approach positive charges and decreases as we move away. Equipotential surfaces are perpendicular to electric field lines.
03

Sketch Equipotential Surfaces Near the Origin

For part (a), sketch equipotential surfaces close to the charges up to 2.0 m from the origin. Near the charges, the equipotential surfaces will be more closely spaced. Expect asymmetric shapes because the magnitude of the charges differs. The surfaces near `+q` at x = -1.0 m will be less dense compared to those near `+2q` at x = 1.0 m.
04

Determine Direction of Potential Increase Near Origin

Potential increases as we move closer to the point charges. Therefore, near their locations, the potential increases as we move inward toward either charge, particularly towards the charge `+2q` which has a larger magnitude.
05

Sketch Equipotential Surfaces Far from the Origin

For part (b), consider points far from the origin. Here, the equipotential surfaces will appear roughly spherical centered around the combined charge system (+3q) at a distance. Because they are far, the distinct contribution of each charge diminishes and appears like a single charge system.
06

Determine Direction of Potential Increase Far From Origin

At far distances, the potential will increase as you approach the combined charge. The surfaces appear further apart as we move away, indicating smaller potential differences between them.

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Key Concepts

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

Point Charges
Point charges are fundamental in understanding electric potential and fields. They are hypothetical charges located at a single point in space. Think of them as tiny particles that create an electric influence around themselves.

Each point charge has a specific magnitude and sign. The sign indicates whether the charge is positive or negative, affecting how it interacts with other charges. - Positive charges repel each other - Unlike charges attract

In the problem we are examining, two point charges "+q" and "+2q" are positioned along the x-axis. The charge at x = -1.0 m is lesser than the one at x = 1.0 m, meaning "+2q" has a stronger influence. This strength results in a larger electric contribution at its location, impacting the shape and density of equipotential surfaces nearby.
Point charges are essential for calculating electric potentials because they allow us to predict how electricity behaves in a given space.
Equipotential Surfaces
Equipotential surfaces are crucial for visualizing electric potential in a field. An equipotential surface is defined as a region in space where the electric potential value is constant. This means that if you move along this surface, you do not gain or lose any electric potential energy.

In our scenario, these surfaces are shaped based on the influence of the point charges. Because we have two charges, their equipotential surfaces are initially complex and non-spherical near the charges:
  • Near "+q" and "+2q", expect asymmetrical and more condensed surfaces
At greater distances, however, the distinct influence of each separate charge diminishes, and the surfaces start to form spherical patterns centered around the combined effect of both charges.

It's interesting to note that the configuration of these surfaces indicates energy levels. Moving from one surface to another requires work if not perpendicular to the field, essentially changing the electric potential as you go.
Electric Field
The electric field represents the force per unit charge and is fundamental for understanding electric interactions. This invisible force field surrounds charged objects and influences other charges.
The concept of the electric field is central to sketching and understanding equipotential surfaces. Electric field lines tell us about the field's direction and strength: - Field lines point away from positive charges - Stronger fields are depicted by closely spaced lines

Generally, equipotential surfaces are perpendicular to these field lines. This means that moving parallel to an equipotential surface involves no work being done by or against the field. This perpendicular relationship helps us understand the increase or decrease in potential energy when moving through different parts of the field.
For our problem, consider these field lines to wrap around the two charges we are analyzing. Near the charges, these field patterns are more distinct and varied, requiring keen observation to predict how the potential and field will change as you move through space.

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Most popular questions from this chapter

Rutherford's Planetary Model of the Atom In 1911 , Ernest Rutherford developed a planetary model of the atom, in which a small positively charged nucleus is orbited by electrons. The model was motivated by an experiment carried out by Rutherford and his graduate students, Geiger and Marsden. In this experiment, they fired alpha particles with an initial speed of \(1.75 \times 10^{7} \mathrm{m} / \mathrm{s}\) at a thin sheet of gold. (Alpha particles are obtained from certain radioactive decays. They have a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27} \mathrm{kg}\).) How close can the alpha particles get to a gold nucleus (charge \(=+79 e\) ), assuming the nucleus remains stationary? (This calculation sets an upper limit on the size of the gold nucleus. See Chapter 31 for further details.)

A uniform electric field of magnitude \(4.1 \times 10^{5} \mathrm{N} / \mathrm{C}\) points in the positive \(x\) direction. Find the change in electric potential energy of a \(4.5-\mu C\) charge as it moves from the origin to the points (a) \((0,6.0 \mathrm{m}) ;\) (b) \((6.0 \mathrm{m}, 0) ;\) and (c) \((6.0 \mathrm{m}, 6.0 \mathrm{m})\).

Electron Escape Speed An electron is at rest just above the surface of a sphere with a radius of \(2.7 \mathrm{mm}\) and a uniformly distributed positive charge of \(1.8 \times 10^{-15} \mathrm{C}\). Like a rocket blasting off from the Earth, the electron is given an initial speed \(v_{e}\) radially outward from the sphere. If the electron coasts to infinity, where its kinetic energy drops to zero, what is the escape speed, \(v_{e} ?\)

A parallel-plate capacitor has plates with an area of \(0.012 \mathrm{m}^{2}\) and a separation of \(0.88 \mathrm{mm} .\) The space between the plates is filled with a dielectric whose dielectric constant is \(2.0 .\) (a) What is the potential difference between the plates when the charge on the capacitor plates is \(4.7 \mu \mathrm{C}\) ? (b) Will your answer to part (a) increase, decrease, or stay the same if the dielectric constant is increased? Explain. (c) Calculate the potential difference for the case where the dielectric constant is \(4.0 .\)

A parallel-plate capacitor has plates separated by \(0.75 \mathrm{mm}\). If the electric field between the plates has a magnitude of (a) \(1.2 \times 10^{5} \mathrm{V} / \mathrm{m}\) or (b) \(2.4 \times 10^{4} \mathrm{N} / \mathrm{C},\) what is the potential difference between the plates?

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