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/ε0, with Qin/ε , where ε is the permittivity of the material. (Technically, ε0 is called the vacuum permittivity.) Suppose a long, straight wire with linear charge density 250 nC/m is covered with insulation whose permittivity is 2.5ε0.

What is the electric field strength at a point inside the insulation that is 3.0 mm from the axis of the wire?

Explanation:

We shall consider a Gaussian surface inside the insulation in the form of curved wall of a cylinder having radius equal to 3mm and unit length , length being parallel to the axis of wire .

Charge inside the cylinder = 250 x 10⁻⁹ C .

Let E be electric field at the curved surface , perpendicular to surface .

Total electric flux coming out of curved surface

= 2π r x 1 x E

= 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's theorem , total flux coming out

= charge inside / ε ( 250 x 10⁻⁹C  charge will lie inside cylinder )

= 250 x 10⁻⁹ / 2.5 x 8.85 x 10⁻¹²    (  ε = 2.5 ε₀ = 2.5 x 8.85 x 10⁻¹² )

= 11.3 x 10³ weber .

so ,

2 x 3.14 x 3 x 10⁻³ E = 11.3 x 10³

E =  11.3 x 10³ /  2 x 3.14 x 3 x 10⁻³

= .599 x 10⁶ N /C .

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current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete