/*! 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 5 An infinite nonconducting sheet ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An infinite nonconducting sheet has a surface charge density \(\sigma=0.10 \mu \mathrm{C} / \mathrm{m}^{2}\) on one side. How far apart are equipotential surfaces whose potentials differ by \(50 \mathrm{~V}\) ?

Short Answer

Expert verified
8.85 mm

Step by step solution

01

Understanding the Problem

We have an infinite non-conducting sheet with a surface charge density \( \sigma = 0.10 \mu \mathrm{C} / \mathrm{m}^{2} \). We are to find the distance between equipotential surfaces where the potential difference is 50 V.
02

Using Gauss's Law to Find Electric Field

The electric field (E) due to an infinite sheet with surface charge density \( \sigma \) is given by the formula \( E = \frac{\sigma}{2 \varepsilon_0} \), where \( \varepsilon_0 \) is the vacuum permittivity \( 8.85 \times 10^{-12} \mathrm{~C}^2/\mathrm{N} \cdot \mathrm{m}^2 \).
03

Calculate the Electric Field

Substitute \( \sigma = 0.10 \times 10^{-6} \mathrm{~C/m}^2 \) into the formula: \[E = \frac{0.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}} = 5.65 \times 10^3 \mathrm{~N/C}.\]
04

Relate Electric Field to Potential Difference

The potential difference (\( V \)) between two points is related to the electric field (\( E \)) and distance (\( d \)) by the equation \( V = E \cdot d \). Rearrange this to find the distance: \( d = \frac{V}{E} \).
05

Calculate the Distance for Given Potential Difference

Substitute \( V = 50 \mathrm{~V} \) and \( E = 5.65 \times 10^3 \mathrm{~N/C} \) to find \( d \): \[d = \frac{50}{5.65 \times 10^3} \approx 8.85 \times 10^{-3} \mathrm{~m} = 0.00885 \mathrm{~m} = 8.85 \mathrm{~mm}.\]
06

Conclusion

The distance between the equipotential surfaces whose potentials differ by 50 V is 8.85 mm.

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.

Gauss's Law
Gauss's Law is a fundamental principle in the study of electrostatics, helping us understand how electric fields are generated by charges. It states that the net electric flux through any closed surface is equal to the enclosed net charge divided by the permittivity of free space, denoted by \( \varepsilon_0 \).
Gauss's Law can be mathematically expressed as:
  • \( \Phi_E = \frac{Q_{ ext{enclosed}}}{\varepsilon_0} \)
Here, \( \Phi_E \) represents the electric flux and \( Q_{\text{enclosed}} \) is the charge within the surface.
This law simplifies solving problems involving symmetrical charge distributions, like an infinite plane. For an infinite sheet with surface charge density \( \sigma \), the electric field is calculated using the specialized form of Gauss's Law:
  • \( E = \frac{\sigma}{2\varepsilon_0} \)
This formula highlights how the electric field is proportional to the charge density and lessens with increasing permittivity.
Using Gauss's Law in the given exercise helps derive the electric field created by an infinite non-conducting sheet, ultimately aiding in calculating the distance between equipotential surfaces.
Electric Field
The electric field is a vector field surrounding charged particles, describing the force exerted per unit charge on other charges in its vicinity. It’s represented as \( E \) and measured in newtons per coulomb (\( \mathrm{N/C} \)).
An electric field due to a point charge is calculated using Coulomb's Law:
  • \( E = \frac{k \cdot |q|}{r^2} \)
where \( k \) is Coulomb's constant, \( q \) is the point charge, and \( r \) is the distance from the charge.
However, for an infinite sheet, the electric field remains constant and is given by:
  • \( E = \frac{\sigma}{2\varepsilon_0} \)
This consistent field means it doesn’t weaken with distance from the sheet, a unique feature in electrostatics.
In the exercise, this electric field calculation is used to determine how much distance corresponds to a potential difference of 50 V between two equipotential surfaces, demonstrating the relationship between field strength and potential difference.
Equipotential Surfaces
Equipotential surfaces are imaginary surfaces on which every point holds the same electric potential. These surfaces simplify understanding how electric fields influence charges.
Here are some key aspects of equipotential surfaces:
  • They are always perpendicular to electric field lines.
  • No work is done by or against the electric field when moving a charge along such a surface.
  • The spacing between these surfaces reduces as the field strength increases.
In the context of an infinite sheet, equipotential surfaces are parallel planes. The distance between these planes is dictated by the electric field and the potential difference they are separated by.
Using the relation \( V = E \cdot d \), where \( V \) is the potential difference and \( d \) is the distance between surfaces, you can find how far apart two equipotential surfaces are for a given potential difference. This method allows us to find that for an electric field of \( 5.65 \times 10^3 \mathrm{~N/C} \) and a potential difference of \( 50 \mathrm{~V} \), the surfaces are 8.85 mm apart, as calculated in the exercise.

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

(a) If an isolated conducting sphere \(10 \mathrm{~cm}\) in radius has a net charge of \(4.0 \mu \mathrm{C}\) and if \(V=0\) at infinity, what is the potential on the surface of the sphere? (b) Can this situation actually occur, given that the air around the sphere undergoes electrical breakdown when the field exceeds \(3.0 \mathrm{MV} / \mathrm{m}\) ?

Two charged, parallel, flat conducting surfaces are spaced \(d=\) \(1.00 \mathrm{~cm}\) apart and produce a potential difference \(\Delta V=625 \mathrm{~V}\) between them. An electron is projected from one surface directly toward the second. What is the initial speed of the electron if it stops inst at the second surface?

The electric field in a region of space has the components \(E_{y}=\) \(E_{z}=0\) and \(E_{x}=(4.00 \mathrm{~N} / \mathrm{C}) x .\) Point \(A\) is on the \(y\) axis at \(y=3.00 \mathrm{~m}\), and point \(B\) is on the \(x\) axis at \(x=4.00 \mathrm{~m}\). What is the potential difference \(V_{B}-V_{A} ?\)

Suppose that in a lightning flash the potential difference between a cloud and the ground is \(1.0 \times 10^{9} \mathrm{~V}\) and the quantity of charge transferred is \(30 \mathrm{C}\). (a) What is the change in energy of that transferred charge? (b) If all the energy released could be used to accelerate a \(1000 \mathrm{~kg}\) car from rest, what would be its final speed?

A nonuniform linear charge distribution given by \(\lambda=\) \(b x\), where \(b\) is a constant, is located along an \(x\) axis from \(x=0\) to \(x=0.20 \mathrm{~m}\). If \(b=20 \mathrm{nC} / \mathrm{m}^{2}\) and \(V=0\) at infinity, what is the electric potential at (a) the origin and (b) the point \(y=0.15 \mathrm{~m}\) on the \(y\) axis?

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.