Electrostatic - Chapter 1 & 2 - PDF

# Chapitre 1 ## Loi de Coulomb et distributions des charges ### 1.1 La loi de Coulomb Charles Auguste de Coulomb effectue une série d'expériences et de mesures qui lui ont permis de déterminer avec précision les propriétés de la force électrostatique exercée par une charge ponctuelle q1 sur une autr...

# Chapitre 1 ## Loi de Coulomb et distributions des charges ### 1.1 La loi de Coulomb Charles Auguste de Coulomb effectue une série d'expériences et de mesures qui lui ont permis de déterminer avec précision les propriétés de la force électrostatique exercée par une charge ponctuelle q1 sur une autre charge q2 (figure 1.1). A partir de cette expérience, Coulomb décrit la force électrostatique comme une force : * Radiale (dirigée selon la droite qui joint les deux charges). * Proportionnelle au produit des charges (attractive si elles sont de signes opposés sinon répulsive). * Varie comme l'inverse du carré de la distance entre les deux charges. 1. la balance de Coulomb ; il s'agit d'un fil de torsion métallique avec deux boules électrisés considérés comme des charges ponctuelles. 2. la charge est dite ponctuelle si ses dimensions sont négligeables. The image shows two point charges, q1 and q2, a distance r apart. q1 exerts a force on q2, shown as Fq1/q2. ### FIGURE 1.1 * La force électrostatique de deux charges ponctuelles q1 et q2. For two point charges q1 and q2 a distance of r apart, the mathematical expression of the Coulomb force that expresses the properties above is: $$F_{q1/q2}=\frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\vec{u}$$ The expression can be read as: the force Fq1/q2 is the force exerted on charge q2 due to the presence of charge q1. The constant $K = \frac{1}{4\pi\epsilon_0} = 9.10^9N.m^2.C^{-2}$ and $\epsilon_0$ is the permittivity of vacuum $\epsilon_0 = 8.85.10^{-12}F.m^{-1}$ ### Remarques * The electrostatic force is a Newtonian force (law in $1/r^2$). * This expression of force is only valid for stationary charges in a vacuum. * This force obeys the principle of action and reaction. ### Ordre de grandeur 1. **Example:** Comparison between gravitational forces and electrostatic forces. The ratio of the magnitudes of the two forces is $$\frac{F_E}{F_G} = 4.9.10^{42}$$ We observe that at the atomic scale, gravitational forces are negligible compared to electrostatic forces. However, on the scale of the universe, gravitational forces dominate. This leads us to conclude that celestial bodies are electrically neutral. Furthermore, gravitational forces are always attractive, while electrostatic forces cancel each other out. 2. **Example:** What is the Coulomb force between two charges of 1C located 1km apart? $$F_E=\frac {1}{4\pi\epsilon_0}\frac {1}{({10^3})^2}= 9.10^3 N$$ This is equivalent to: $$\frac{F_E}{g}=900 kg$$ In this example, the electrostatic force is equivalent to the weight of one ton. # Chapitre 2 ## Flux du champ et théorème de Gauss ### 2.1 Flux du champ électrostatique à travers une surface élémentaire We propose here to evaluate the flux of the electrostatic field of a point charge through an elementary area dS. The point M represents the mean point for the elementary surface dS and n is the normal to dS. The flux element of the field E through dS is defined as follows $$d\Phi=E.\vec{n}. dS=E. dS. cos\alpha$$ and as $$||E||=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2} $$ We get $$d\Phi=\frac{q}{4\pi\epsilon_0}\frac{dS.cos\alpha}{r^2}$$ Since dS.cos\alpha is, by definition, the elementary solid angle under which the point O sees dS, so we get: $$d\Phi = \frac {q}{4\pi\epsilon_0} d\Omega$$ dΩ has an algebraic value whose sign depends on the choice of orientation of the normal vector to the surface (the choice must respect conventions). ### FIGURE 2.1 * Flux of the electrostatic field through an elementary surface. ### 2.2 Flux du champ électrostatique à travers une surface finie non fermée (ouverte) The total flux of the electrostatic field created by a single point charge through a surface (S) that is not closed is: $$\Phi=\int_S d\Phi = \frac{q}{4\pi\epsilon_0}\int_S dS $$ Let Ω = ∫∫s dΩ be the solid angle under which the point O sees the surface (S), then: $$\Phi = \frac{q}{4\pi\epsilon_0} \Omega$$ ### FIGURE 2.2 * Flux of the electrostatic field through an open surface. ### 2.3 Flux du champ électrostatique à travers une surface finie fermée #### 2.3.1 La charge ponctuelle est à l'extérieur de la surface fermée Let's take a closed surface (S) and a charge q (positive) located at O, this last is outside of S. We consider the cone with vertex O generating a solid angle dΩ. This last cuts the surface (S) in elementary surfaces dS₁ and dS₂ with normal vectors n₁ and n₂ (oriented outwards according to the convention). Let E₁ and E₂ represent the electrostatic field created by q on dS₁ and dS₂. The total flux is: $$ d\Phi = d\Phi_1 + d\Phi_2$$ The flux dФ₁ of E₁ through dS₁ is negative (n₁ and E₁ have opposite directions--see figure 2.3) while the flux dФ₂ of E₂ through dS₂ (in magnitude these fluxes are equal). The total flux going through a closed surface is then $$d \Phi = d\Phi_1 + d\Phi_2 = 0$$ ### FIGURE 2.3 * Flux of the electrostatic field of a charge outside a closed surface. #### 2.3.2 La charge ponctuelle est à l'intérieur de la surface fermée The flux of the vector Ē exiting dS is: $$d\Phi=\frac{q}{4\pi\epsilon_0} d\Omega$$ For the whole surface(S), we have $$ \Phi=\iint_S d\Phi = \iint_S \frac{q}{4\pi\epsilon_0} d\Omega = \frac{q}{4\pi\epsilon_0} \Omega$$ The solid angle Ω under which O (inside S) sees all the surface S is equal to 4π (Ω(space) = 4π steradians) so: $$ \Phi= \frac{q}{4\pi\epsilon_0} 4\pi = \frac{q}{\epsilon_0}$$ ### FIGURE 2.4 * Flux of the electrostatic field of a charge inside a closed surface. ### 2.4 Théorème de Gauss We can generalize the result above by considering the electrostatic field E₁ created by a collection of charges qint inside the closed surface S with the flux of this field exiting from this surface (for the charges outside qext the flux of the field is null). For the collection of charges inside, we have: $$ \Phi = \int_S E_1 dS = \frac{Q_{int}}{\epsilon_0}$$ Where $Q_{int}$ $(Q_{int} = \sum q_{int})$ represents the total electric charge contained in volume V limited by the closed surface S. ### 2.5 Lignes du champ électrostatique #### 2.5.1 Définition An electrostatic field line is defined as any curve that is tangent at each point to the vector E defined at that point. Generally, a field line is oriented in the direction of the field E. ### FIGURE 2.5 * Electrostatic field line (in red). #### 2.5.2 Équation des lignes de champ The field E is tangent to the field line at point M. Let M' be a neighboring point located on this field line. In Cartesian coordinates, E has the components (Ex, Ey, Ez) and MM' has the components (dx, dy, dz). ### FIGURE 2.6 * Electrostatic field line (in red). By the definition of the field line, E and MM' are collinear, which can be written mathematically by: $$\frac{dx}{E_x}=\frac{dy}{E_y}=\frac{dz}{E_z}$$ In cylindrical coordinates: $$\frac{dr}{E_r}=\frac{rd\theta}{E_\theta}=\frac{dz}{E_z}$$ In spherical coordinates: $$\frac{dr}{E_r}=\frac{rd\theta}{E_\theta}=\frac{rsin\theta d\phi}{E_\phi}$$ These equations represent the differential equations for the field lines. By integrating these equations, we can obtain the equations of the field lines. ### Example des lignes de champ #### Champ uniforme ### FIGURE 2.7 * Uniform electric field lines. #### Champ créé par une charge ponctuelle ### FIGURE 2.8 * Electrostatic field lines of a point charge. #### Champ créé par des charges de signe opposé ### FIGURE 2.9 * Electrostatic field lines of opposite charges.