The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that

Question

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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?

Physics

1 answer:

serg [3.5K] · Answer 1 · 2026-04-02T21:07:13+03:00

Response:

Reasoning:

We will utilize a Gaussian surface that resembles the curved wall of a cylinder, with a radius of 3mm and a length of 1 unit directed parallel to the wire axis.

The charge within this cylinder amounts to 250 x 10⁻⁹ C.

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

The total electric flux leaving the curved surface

is calculated as 2π r x 1 x E

or 2 x 3.14 x 3 x 10⁻³ E

According to Gauss's law, the total flux is given by the charge within divided by ε (the charge inside the cylinder being 250 x 10⁻⁹C)

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

resulting in 11.3 x 10³ weber.

Thus,

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.