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A very large plastic sheet carries a uniform charge density of \(-6.00 \mathrm{nC} / \mathrm{m}^{2}\) on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by \(1.00 \mathrm{~V}\). What type of surfaces are these?

Short Answer

Expert verified
The electric potential increases as you move away from the negatively charged plastic sheet, and this does not depend on the choice of reference point. The spacing between equipotential surfaces that differ by \(1.00 V\) is calculated to be approximately \(0.113m\). These surfaces are equipotential surfaces.

Step by step solution

01

Understand the Electric Potential Created by the Sheet

The plastic sheet has a negative charge. Electric potential is high near positive charges and low near negative charges. Therefore, as you move away from the sheet, the electric potential increases.
02

Dependence on the Choice of Reference Point

The choice of reference point doesn't change the fact that the electric potential increases as you move away from the negatively charged sheet, it can only shift the potential values, but the increase remains.
03

Determine the Spacing Between Equipotential Surfaces

The potential difference between equipotential surfaces that differ by \(1.00 V\) due to a charge density of \( \sigma = -6.00 \ nC/m^2 \) can be found using the equation \(d =\frac{V}{|\frac{\sigma}{2\epsilon_0}|}\). Substituting \(V = 1.00 V\) and \(\sigma = -6.00 \ nC/m^2\) (remember to convert nC to C), and \(\epsilon_0 = 8.85 \times 10^{-12} \ m^{-3}kg^{-1}s^4A^2\), the spacing would be \(d =\frac{1.00}{|\frac{-6.00 \times 10^{-9}}{2\times8.85 \times 10^{-12}}|}\)
04

Type of Surfaces

Equipotential surfaces are surfaces over which the electric potential is constant. Therefore, the type of the surface would be an equipotential surface.

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Key Concepts

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

Equipotential Surfaces
When we talk about equipotential surfaces, we refer to hypothetical planes in space where every point on that plane has the same electric potential. These surfaces are invaluable tools for visualizing electric fields and understanding the behavior of electric potentials in various configurations. In the case of a charged plastic sheet, imagine it as a vast ocean with the charge spreading out evenly like the water's surface. Equipotential surfaces near the sheet would resemble planes parallel to the sheet.

Since electric field lines are always perpendicular to equipotential surfaces, we know that for a large sheet with a uniform charge distribution, the electric field lines would be straight lines emanating from the sheet. This attribute simplifies the process of calculating potential differences and crafting an intuitive understanding of the physics involved. While moving perpendicular to the charged sheet, one would transition between these invisible yet detectable equipotential planes, experiencing no change in potential until crossing into an adjacent surface, at which point a discrete potential difference is noted.

These surfaces are particularly handy when dealing with the motion of charged particles, as no work is done by or against the electric field when a charge moves along an equipotential surface. Therefore, analyzing these surfaces, as in the provided exercise, gives students a clearer picture of the electric landscape created by the plastic sheet.
Charge Density
Charge density is a measure of the amount of electric charge per unit area. In the provided exercise example, we're given a uniform charge density on one face of the plastic sheet. This uniformity facilitates a consistent and predictable electric field, which would be harder to determine if the charge were distributed unevenly.

Understanding charge density is crucial because it defines the strength and direction of the electric field that emerges from a surface. The negative sign in our example, \(-6.00 \text{nC/m}^2\), indicates that we're dealing with an abundance of electrons. Importantly, the direction of electric potential change, as one moves away from a negative charge density, will always be towards increasing potential - confirming that electric potential is indeed higher where negative charges are less prevalent.

When students learn to envision charge densities, they can form mental pictures of how charges populate different materials and objects, leading to a more substantial grasp of electrostatic interactions in various contexts. Remember, charge density is not just a quantity but a starting point for visualizing the entire ecosystem of electric forces.
Electric Field
The electric field is a vector field that surrounds electric charges and exerts force on other charges within the field. It can be thought of as an invisible force field that permeates space, becoming evident through its action on charges. For students visualizing this concept, it might be helpful to think of the electric field as the wind that you cannot see but can observe its effects on the leaves of a tree.

The exercise deals with an electric field created by a charged surface. Here, the field emanates perpendicularly from the charged sheet and diminishes with distance. The electric field's direction, by convention, points away from positive charges and towards negative charges. Since our sheet is negatively charged, the electric field vectors would be directed inward, towards the sheet. This means any positive charge in the field would be attracted to the sheet, and any negative charge would be repelled.

By delving into the properties of electric fields, students enhance their ability to predict electrical interactions and the kinematics of charged particles in different scenarios. The field's strength is directly related to the charge density: a higher charge density results in a stronger electric field and vice versa. Thus, the concepts of electric field and charge density are intimately connected and essential for a comprehensive understanding of electromagnetism.

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Most popular questions from this chapter

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\). Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere; (d) the clectric ficld at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

In a certain region of space, the electric potential is \(V(x, y, z)=A x y-B x^{2}+C y,\) where \(A, B,\) and \(C\) are positive constants. (a) Calculate the \(x-y^{-},\) and \(z\) -components of the electric field. (b) At which points is the electric field equal to zero?

A solid conducting sphere has net positive charge and radius \(R=0.400 \mathrm{~m}\). At a point \(1.20 \mathrm{~m}\) from the center of the sphere, the electric potential due to the charge on the sphere is \(24.0 \mathrm{~V}\). Assume that \(V=0\) at an infinite distance from the sphere. What is the electric potential at the center of the sphere?

Three equal \(1.20 \mu \mathrm{C}\) point charges are placed at the corners of an equilateral triangle with sides \(0.400 \mathrm{~m}\) long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

An alpha particle with kinetic energy \(9.50 \mathrm{MeV}\) (when far away) collides head-on with a lead nucleus at rest. What is the distance of closest approach of the two particles? (Assume that the lead nucleus remains stationary and may be treated as a point charge. The atomic number of lead is \(82 .\) The alpha particle is a helium nucleus, with atomic number \(2 .\) )

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