Chapter 24: Q. 51 (page 685)

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, $Q_{in} / ϵ_{0}$ , with $Q_{in} / ϵ$ , where $ϵ$ is the permittivity of the material. (Technically, $ϵ_{0}$ is called the vacuum permittivity.) Suppose that a $50 nC$ point charge is surrounded by a thin, $32 - cm$ -diameter spherical rubber shell and that the electric field strength inside the rubber shell is $2500 N / C$ . What is the permittivity of rubber $?$

Expert verified

Permittivity of rubber is $6.2 \times 10^{- 11} C^{2} / N \cdot m^{2}$ .

Step by step solution

The amount of electric field that flows through a closed surface is known as the electric flux.

The electric field travelling through a surface is proportional to the charge within the surface, according to Gauss' law.

Electric flux,

$Φ_{e} = EA = \frac{Q_{in}}{ε}$

$ε = \frac{Q_{in}}{EA}$

Area is,

$A = 4 {πr}^{2}$

Radius,

$r = \frac{d}{2}$

$= \frac{32 cm}{2} = 16 cm$

Substitute all of the values.

$ε = \frac{Q_{in}}{4 {πr}^{2} E}$

The following is the inert charge:

$Q_{in} = (50 nC) (\frac{1 \times 10^{- 9} C}{nC}) = 50 \times 10^{- 9} C$

Then,

$ε = \frac{Q_{in}}{4 {πr}^{2} E}$

$= \frac{50 \times 10^{- 9} C}{4 π (0.16 m)^{2} (2500 N / C)}$

$= 6.2 \times 10^{- 11} C^{2} / N \cdot m^{2}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.