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, $Q_{\text{in}}/{ϵ}_{0,}$with $Q_{\text{in}}/{ϵ}_1$, where $\mathrm{ϵ}$is the permittivity of the material. (Technically, ${ϵ}_0$is called the vacuum permittivity.) Suppose a long, straight wire with linear charge density $250\mathrm{nC}/\mathrm{m}$is covered with insulation whose permittivity is $2.5{ϵ}_0$. What is the electric field strength at a point inside the insulation that is $1.5$$\mathrm{mm}$from the axis of the wire?

The amount of electric field that flows through some kind of closed surface is called as the electric flux. The electric field through a surface is related to the charge inside the surface, as per Gauss's law. Because of electric field is uniform here, we can calculate the electric flux with equation (24.3).

Additionally, the electric field is proportional to the charge${\mathrm{\Phi }}_E$by

${\mathrm{\Phi }}_e=EA$ $\left(1\right)$

The electric field which is related to the charge $Q_{\text{in}}$by

${\mathrm{\Phi }}_e=\frac{Q_{\text{in}}}{\epsilon }$ $\left(2\right)$

The wire with charge density $\lambda$and length $L$is

$Q_{\text{in}}=\pi rL$

Since the wire's top and bottom edges are proportional to the direction field, the flux into them is zero, whereas its flux through the wire's wall is greatest. As a conclusion, we gain flux via the wire by

$\begin{array}{r}{\mathrm{\Phi }}_{\mathrm{e}}={\mathrm{\Phi }}_{\text{top}}+{\mathrm{\Phi }}_{\text{bottom}}+{\mathrm{\Phi }}_{\text{wall}}\end{array}$

$\begin{array}{r}\\ =0+0+EA\\ \end{array}$

$\begin{array}{r}=2\pi rLE\end{array}$

From equations $\left(1\right)$and $\left(2\right)$, we calculate the electric field by

${\mathrm{\Phi }}_{\mathrm{e}}=2\pi rLE=\frac{Q_{\text{in}}}{ϵ}=\frac{\lambda L}{ϵ}$

So, the electric field at distance$r$is

$E=\frac{\lambda }{2\pi ϵr}$

We are given that $ϵ=2.5{ϵ}_o$and $r=1.5\mathrm{mm}=1.5\times {10}^{-3}\mathrm{m}$. So, we plug the values for $r$, $ϵ$and $\lambda$to get $E$by

$\begin{array}{r}E=\frac{\lambda }{2\pi \left(2.5{ϵ}_o\right)r}\end{array}$ $=\frac{250\times {10}^{-9}\mathrm{C}/\mathrm{m}}{2\pi \left(2.5\right)\left(8.82\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}\cdot {\mathrm{m}}^2\right)\left(1.5\times {10}^{-3}\mathrm{m}\right)}$

$=1.2\times {10}^6\mathrm{N}/\mathrm{C}$