P20B2016 – Conductor Properties

Charge inside a conductor is free to move and will rearrange until the net electric field inside the conducting material is zero. To illustrate this, we consider a conductor in electrostatic equilibrium with no external electric field as shown below.

In the absence of an external electric field, the atoms and electrons are arranged in such a way that there is no electric field inside the conductor. If there were, then the free electrons would move, and the conductor would not be in electrostatic equilibrium.

If an external electric field is applied, the free electrons rearrange and separate until they create an electric field that cancels the applied electric field as shown in the figure below. Once this induced charge separation cancels the applied electric field, the charges no longer move, the conductor is in electrostatic equilibrium, and the net electric field inside the conductor is zero.

The net electric field inside a conductor in electrostatic equilibrium is zero and therefore we conclude that for ANY closed surface contained within the conducting material, the electric flux through the surface is zero. Indeed, if $E=0$, then we have

\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\oiint_A{E{\mathrm{cos} \left(\theta \right)\ }dA}=0

By Gauss’s Law, this means that for ANY closed surface contained within the conducting material, the net charge enclosed must be zero.

This cannot be achieved by arguing that $Q_{net}=0$ because the surface contains an equal number of positive charges and negative charges, causing the net enclosed charge to be zero. It can only be true for ANY closed surface if there is no charge whatsoever enclosed in the surface, causing $Q_{net}$ to be zero. Thus, we conclude that the charge carried by a conductor must reside on its outer surface.

We consider a metallic sphere of radius $R$ and we differentiate two regions: inside the sphere $\left(r\le R\right)$ and outside the sphere $\left(r>R\right)$. In each region, we choose a sphere radius $r$ centered on the conducting sphere as our Gaussian surface.

Inside the sphere $\left(r\le R\right)$:

The metal sphere is a conductor and therefore the electric field inside the sphere is zero.

\boxed{{\overrightarrow{E}}_I=\overrightarrow{0}}

Outside the sphere $\left(r > R\right)$:

The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{{\overrightarrow{E}}_{II}\cdot d\overrightarrow{A}}&=\iint_A{E_{II}{\mathrm{cos} \left(\theta \right)\ }dA} \\
\\
&=\iint_A{E_{II}{\mathrm{cos} \left(0\right)\ }dA} \\
\\
&=E_{II}{\mathrm{cos} \left(0\right)\ }\iint_A{dA} \\
\\
&=E_{II}\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The Gaussian sphere encloses an amount of charge equal to

Q_{enc}=Q

By Gauss’s Law, we conclude that the magnitude of the electric field created outside the sphere is given by

E_{II}\cdot 4\pi r^2=\frac{Q}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E_{II}=\frac{Q}{4\pi {\varepsilon }_0r^2}

Thus, the electric field created outside the sphere is equal to

\boxed{{\overrightarrow{E}}_{II}=\frac{Q}{4\pi {\varepsilon }_0r^2}\ \hat{r}}

We consider a metallic shell of radius $R$ and we differentiate two regions: inside the shell $\left(r\le R\right)$ and outside the shell $\left(r>R\right)$. In each region, we choose a sphere radius $r$ centered on the conducting sphere as our Gaussian surface.

Inside the shell $\left(r\le R\right)$:

The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{{\overrightarrow{E}}_{I}\cdot d\overrightarrow{A}}&=\iint_A{E_{I}{\mathrm{cos} \left(\theta \right)\ }dA} \\
\\
&=\iint_A{E_{I}{\mathrm{cos} \left(0\right)\ }dA} \\
\\
&=E_{I}{\mathrm{cos} \left(0\right)\ }\iint_A{dA} \\
\\
&=E_{I}\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The Gaussian sphere encloses an amount of charge equal to

Q_{enc}=0

By Gauss’s Law, we conclude that the magnitude of the electric field created outside the sphere of charge is given by

E_I\cdot 4\pi r^2=0\ \ \ \ \Rightarrow \ \ \ \ \ E_I=0

Thus, the electric field created inside the shell is zero and we write

\boxed{{\overrightarrow{E}}_I=\overrightarrow{0}}

Outside the sphere $\left(r > R\right)$:

The electric flux through our Gaussian surface simplifies because everywhere on the surface of the Gaussian sphere the electric field is along $d\overrightarrow{A}$ and has constant magnitude

\begin{aligned}
\oiint_A{{\overrightarrow{E}}_{II}\cdot d\overrightarrow{A}}&=\iint_A{E_{II}{\mathrm{cos} \left(\theta \right)\ }dA} \\
\\
&=\iint_A{E_{II}{\mathrm{cos} \left(0\right)\ }dA} \\
\\
&=E_{II}{\mathrm{cos} \left(0\right)\ }\iint_A{dA} \\
\\
&=E_{II}\cdot 4\pi r^2
\end{aligned}

where the area of the Gaussian sphere is equal to $A=4\pi r^2$.

The Gaussian sphere encloses an amount of charge equal to

Q_{enc}=\sigma \cdot 4\pi R^2

By Gauss’s Law, we conclude that the magnitude of the electric field created outside the shell is given by

E_{II}\cdot 4\pi r^2=\frac{\sigma \cdot 4\pi R^2}{{\varepsilon }_0}\ \ \ \ \Rightarrow \ \ \ \ \ E_{II}=\frac{\sigma \cdot R^2}{{\varepsilon }_0r^2}

Thus, the electric field created outside the sphere is equal to

\boxed{{\overrightarrow{E}}_{II}=\frac{\sigma \cdot R^2}{{\varepsilon }_0r^2}\ \hat{r}}