**Concept:**The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, \( Q_{\text{in}}/\varepsilon_0 \), with \( Q_{\text{in}}/\varepsilon \), where \( \varepsilon \) is the permittivity of the material. (Technically, \( \varepsilon_0 \) is called the vacuum permittivity.) Suppose that a \( 40 \, \text{nC} \) point charge is surrounded by a thin, \( 32\text{-cm}\)-diameter spherical rubber shell and that the electric field strength inside the rubber shell is \( 2500 \, \text{N/C} \).---**Exercise: Part A****Question:**What is the permittivity of rubber?**Input Box:**\[ \varepsilon = \, \]\[ \text{C}^2/\text{N} \cdot \text{m}^2 \]**Options:**- Submit- Request Answer**Explanation:**You are asked to determine the permittivity (\( \varepsilon \)) of the rubber based on the given scenario using the principles of electrostatics and Gauss's law.

Answered: Image /qna-images/answer/903b085f-4b73-45cb-b8bb-11d2c63c76ba.jpg

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin /ɛ0, with Qin /e, where e is the permittivity of the material. (Technically, eo is called the vacuum permittivity.) Suppose that a 40 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C. Part A What is the permittivity of rubber? ? C /N • m? Submit Request Answer

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin /ɛ0, with Qin /e, where e is the permittivity of the material. (Technically, eo is called the vacuum permittivity.) Suppose that a 40 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C. Part A What is the permittivity of rubber? ? C /N • m? Submit Request Answer

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Related questions

Q: A uniform electric field of magnitude E = 38 N/C points along the x-axis. A circular loop of radius…

A: Given data :- Electric field along x axis Ex = 38 N/C Radius of a circular loop R = 26 cm = 0.26 m…

Q: The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an…

Q: 1. Consider a spherical dielectric shell of inner and outer radii a and b respectively. Assume that…

Q: Newton's law of gravity and Coulomb's law are both inverse-square laws. Consequently, there should…

A: Newton's law of gravity and Coulomb's law are both inverse-square laws. Consequently, there should…

Q: A bolt of lightning is associated with the motion of what type of charge? O the phenomenon of…

A: Bolt of lightning is associated with the motion of

Q: A parallel plate capacitor of width w, length L, and separation d has a solid dielectric slab of…

A: Given Width = W Length = L Permitivity = Potential difference =

Q: 7p = xpl + Ypj ++++++++ 2a -> X 0,0 2a If the material is a conductor, what will be the electric…

Q: Gauss’s law also helps us locate charge on conductors. There are several distinct pieces we need for…

A: (a) This is true because conductors are characterized by their ability to conduct electricity,…

Q: An electron moves in uniform circular motion in a circle of radius r equal to 0.1 cm around an…

Q: Consider a point charge Q sitting at the center of a spherical, linear dielectric shell with a…

A: The objective of the question is to find the bound volume charge densities in the dielectric and the…

Q: A 8.20 μC charged 1.00 g cork ball is suspended vertically on a 1.100 m long light string in the…

A: Given: The charge is q=8.20 μC. The mass of the ball is m=1 g. The electric field is E= 8.50x105…

Q: Problem 9: Consider the three-dimensional conductor in the figure, which has a cavity on the inside.…

Q: Problem 7: The formula for the molecular polarizability a = 4reg R of a metal sphere is useful for…

Q: An electric field of magnitude 457 V/m passing through a flat square plate of length 0.556 m on a…

A: Electric field (E) = 457 V/m Side of the plate (l) = 0.556 m Area of the plate = A A = l2 A =…

Q: AU SOI the velocity components of an electron moving between two charged parallel plates are vr and…

Q: Problem 2: A coaxial cable is used to send video signals and other data streams from one device to…

A: Given that:The diameter of the inner conducting wire: The magnitude of the electric field at a…

Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

**Concept:**

The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, \( Q_{\text{in}}/\varepsilon_0 \), with \( Q_{\text{in}}/\varepsilon \), where \( \varepsilon \) is the permittivity of the material. (Technically, \( \varepsilon_0 \) is called the vacuum permittivity.) Suppose that a \( 40 \, \text{nC} \) point charge is surrounded by a thin, \( 32\text{-cm}\)-diameter spherical rubber shell and that the electric field strength inside the rubber shell is \( 2500 \, \text{N/C} \).

---

**Exercise: Part A**

**Question:**
What is the permittivity of rubber?

**Input Box:**
\[ \varepsilon = \, \]
\[ \text{C}^2/\text{N} \cdot \text{m}^2 \]

**Options:**
- Submit
- Request Answer

**Explanation:**
You are asked to determine the permittivity (\( \varepsilon \)) of the rubber based on the given scenario using the principles of electrostatics and Gauss's law.

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Given and required quantities

VIEW

Permittivity of rubber

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 2 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.

Similar questions

In empty space there is (-∞, 0) semi-infinite linear uniform and constant charge density ρl = 3 [C / m] on the z-axis. Calculate the electrostatic field that this charge density will create at the point C (0,0,4). ke = 1 / 4πε Write numerically the components of the electrostatic field in terms of the given quantities.
Satisfaction of Maxwell's curl equations for a specified electric field. For the electric field E = Ege¯k* cos (2 × 10°t – y) a̟ in free space (J = 0), find the value(s) of k for which the field satisfies both of Maxwell's curl equations.
A very long coaxial cable has an inner conductor shell with a radius of 1 mm and a thickness of 1 mm. The outer conductor shell has a radius of 4 mm and a thickness of 2 mm. If there are volume charge distributions of e and -p. in the inner and outer conductor shells respectively, i) determine D everywhere, ii) using Maxwell's equation, show charge distribution everywhere.
An electric field of magnitude 464 V/m passing through a flat square plate of length 0.644 m on a side makes an angle of 63.6 degrees with the surface of plate. Determine the electric flux passing through the surface of the square plate. (Hint: the angle given here is the angle that the field makes with the surface, not with the area vector. In what direction does the area vector of a surface point relative to that surface?)
Consider the following electrostatic problem. A perfectly conductive spherical conductor of radius a=5cm is covered by a dielectric layer extending to a radius of b%3D8cm in free space. Relative electrical permittivity of the dielectric iS E 4, and Electric Flux Density expression in the medium is given as (D = 0.05r-u,: Cim, for ra), where r is radial distance from the center of sphere. a) What is the total charge on the conductor surface? b) Determine Electric field intensity for each region (r
g A parallel plate,air capacitor of plate. separation 2 'mm and capacitance 1. MF is charged to 200 V A dielectric of relative Permittivity 50 is now inserted so as to fill the space: between the plates. () Find the new value of capacitance (ii) Find the polarisation charge on one of the boundaries of the dielectric, (iii) Find the magnitude of the polarisation of the electric (iv) What is the magnitude of the electric field inside the dielectric?
By using Gauss's Law, show that the electric field in the space between the two conductors of a coaxial cable is given by: E(r) τρι 2πει where "-p" is the linear line charge density (Coulombs / meter) defined for the inner conductor.
B/sl O 8:25 Int+96277923 .1023/q قبل دقيقتين 19=3 1| \mu \jC ) is allached by an insulating string to the surface of a very large conductor with a surface charge density of \( \sigma \) = 36.4 \( \mu \)C. Given that the string makes an angle with the surface equal to 30 degrees, find the tension ( in N) in the string. use ɛo = 8.8542x10-12 F-m-1. String + 0 Select one: А. 12.33 B. 24.67 С. 37.00 D. 14.24 8
Q No. 3: Consider a 10 nC charge located at the origin, what will be Electric field intensity at (10, n/2, T/4). Represent the Electrical field in Cartesian coordinates.
R1 S V = 10V L= 10H R2 At time t=o, the switch S is closed. What is the current through the battery just after the switch is closed? R1=23N and R2=542
You are an astronaut, living for a long time interval in the International Space Station (ISS). During your off-duty hours, you have run out of books to read and video games to play. So, your mind wanders to your hobby of music. The last book you read discussed Gauss's law, and you get an inspiration. You plan to attach two nonconducting spheres of radius r = 1.50 cm together using a light insulating string of length L and linear mass density μ = 5.00 x 10-3 kg/m, with the string attached at the surface of each sphere. Then, using the electrical system on the ISS, you will be able to electrify each sphere to a charge of Q = 75.0 μC, uniformly spread over the surface of the sphere. The combination will then be allowed to float freely in the ISS. The spheres will repel, creating a tension in the string. When you pluck the string, you wish it to play a perfect middle C, at 262 Hz. Determine the length of the string (in cm) that you need. (Assume the frequency of 262 Hz is the fundamental…
Consider an infinitely long wire of charge carrying a positive charge density of A. The electric field due to λ this line of charge is given by E= 2kef= -, where is a unit vector directed radially outward Σπερμ from the infinitely long wire of charge. Hint #3 a. Letting the voltage be zero at some reference distance (V(ro) = 0), calculate the voltage due to this infinite line of charge at some distance r from the line of charge. Give your answer in terms of given quantities (A,ro,r) and physical constants (ke or Eo). Use underscore ("_") for subscripts and spell out Greek letters. Hint for V(r) calculation 3 V(r) = b. There is a reason we are not setting V(r → ∞o) = 0 as we normally do (in fact, in general, whenever you have an infinite charge distribution, this "universal reference" does not work; you need a localized charge distribution for this reference to work). Which of the following best describes what happens to potential as roo? (That is, what is V(ro), with our current…