diff --git a/docs/physics/electrostatics/4-conductors.md b/docs/physics/electrostatics/4-conductors.md new file mode 100644 index 0000000..42b6af8 --- /dev/null +++ b/docs/physics/electrostatics/4-conductors.md @@ -0,0 +1,19 @@ +# Chapter 4 - Conductors and Static Electric Fields + +## Section 4.1 - Introduction + +We will focus primarily on electric fields and charges. For the purposes for this section, we will assume insulators are perfect. + +## Section 4.2 - Electrostatic Properties of a Conductor + +In a metal or conductor, there are plentiful charges not bound to a particular atom and are thus free to move throughought the material. + +We note that there is no electric fiend inside a conductor, as charges internal to the material would move under the force it generates until they find a configuration that eliminates the field. This may happen, but not in electrostatics. + +Additionally, as the field is zero, it follows from Maxwell's equations that there is no charge inside a conductor. However, charge may be present at the surface. For sufficiently symnetric charges, this charge may be calculated. + +Consider any two points internal to the conductor. The voltage between said points is defined as $\int_A^B \vb{E} \vdot \dd{\vb{l}}$. Since $\vb{E} = 0$ inside the conductor, the volage difference must be zero. Thus, any two points in or on the surface (TODO: Why on the surface?) of a conductor must be at the same potential. + +The electric field at the surface of a conductor is perpendicular to its surface. Consider some displacement $\dd{\vb{l}}$. Now, $\vb{E} \vdot \dd{\vb{l}} = \vb{E}_s \vdot \dd{\vb{l}}_s + \vb{E}_p \vdot \dd{\vb{l}}_p = \dd{V_s} + \dd{V_p}$, in terms of parallel and perpendicular components. The paralell voltage difference is zero, so the electric field must be zero. + +Consider the surface of a conductor with surface charge density $\sigma_e$. A cyliner with one end inside and one end outside said surface, with its axis normal to said surface, will be a Gaussian "pillbox", which will show that with V being the volume of the pillbox, $\int_V \div{\vb{E}} \dd{V} = \frac{Q_e}{\epsilon_0} = \frac{A\sigma_e}{\epsilon_0}$. Thus, $\sigma_e = \epsilon_0 E$. \ No newline at end of file