Question

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/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 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?

Answer 1

The permittivity of rubber is
`\epsilon = 8.703 *10^(-11)`

Step-by-step explanation:

From the question we are told that

The magnitude of the point charge is
`q_1 = 70 \ nC = 70 *10^(-9) \ C`

The diameter of the rubber shell is
`d = 32 \ cm = 0.32 \ m`

The Electric field inside the rubber shell is
`E = 2500 \ N/ C`

The radius of the rubber is mathematically evaluated as

`r = (d)/(2) = (0.32)/(2) = 0.16 \ m`

Generally the electric field for a point is in an insulator(rubber) is mathematically represented as

`E = (Q)/( \epsilon ) * (1)/(4 * \pi r^2)`

Where
`\epsilon` is the permittivity of rubber

=>
`E * \epsilon * 4 * \pi * r^2 = Q`

=>
`\epsilon = (Q)/(E * 4 * \pi * r^2)`

substituting values

`\epsilon = (70 *10^(-9))/(2500 * 4 * 3.142 * (0.16)^2)`

`\epsilon = 8.703 *10^(-11)`