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Basic Electrostatics From Molecular to Con/nuum Physics I WS 11/12 Emiliano Ippoli/| October, 2011 Wednesday, October 12, 2011 Review Mathematics ... Physics • Basic thermodynamics • Temperature, ideal gas, kinetic gas theory, laws of thermodynamics • Statistical thermodynamics • Canonical ensemble, Boltzmann statistics, partition functions, internal and free energy, entropy • Basic electrostatics • Classical mechanics • Newtonian, Lagrangian, Hamiltonian mechanics • Quantum mechanics • Wave mechanics • Wave function and Born probability interpretation • Schrödinger equation • Simple systems for which there is an analytical solution • Free particle • Particle in a box, particle on a ring • Rigid rotator • Harmonic oscillator • Basics • Uncertainty relation • Operators and expectation values • Angular momentum • Hydrogen atom • Energy values, atomic orbitals • Electron spin • Quantum mechanics of several particles (Pauli principle) • Many electron atoms • Periodic system: structural principle • Molecules • Two-atomic molecules (H2+,H2, X2) • Many-atomic molecules Emiliano Ippoliti Wednesday, October 12, 2011 Chemistry Informatics ... ... 2 Coulomb’s law Let us consider two point-like electric charges q and Q at position x1 and x2, respectively. The force on q due to Q is then: qQ FQ→q = k 3 r r r = r = xq − xQ 1. proportional to the strength of charges; 2. inversely proportional to the square of the separation; 3. directed along the line connecting the charges; 4. repulsive for like charges and attractive for opposite charges. Emiliano Ippoliti Wednesday, October 12, 2011 3 Units In SI units: 1 k= ≈ 9 × 10 9 Nm 2C-2 4πε 0 an electron carries a charge e equal to 1.6 x 10−19 C In Gaussian units: k =1 an electron carries a charge e equal to 4.803 x 10−10 statcoulomb 1 statcoulomb = 3.3356 10−10 C Emiliano Ippoliti Wednesday, October 12, 2011 4 Electric field Take a very small test charge q (small so that it does not disturb the charge distribution whose field we're measuring), and measure the force on the test charge as a function of position x. Then the electric field is defined as: F ( x) E ( x ) = lim q→0 q Q x Q ê r E(x) = 3 = 2 4πε 0 x 4πε 0 r Emiliano Ippoliti Wednesday, October 12, 2011 5 Superposition principle It is an experimental fact that electrostatics is linear, so that the electric fields produced by a collection of point charges {qi} at positions {xi} simply add: 1 x − xi E(x) = qi 3 ∑ 4πε 0 i =1 x − xi N that can be rewritten as: E(x) = 1 ⎡ ⎤ 3 (3) d x ′ ⎢ ∑ qiδ ( x ′ − xi ) ⎥ ∫ 4πε 0 ⎣i =1 ⎦ ρ ( x′ ) N Charge density Emiliano Ippoliti Wednesday, October 12, 2011 6 x − x′ 3 x − x′ Dirac delta function The Dirac delta function is a mathematically convenient way of representing singularities such as point charges. It is really not a function but a “distribution.” However, we will ignore this at this level. One way of defining the delta function in one dimension is: ⎧ 1/ w ⎪ δ ( x ) = lim ⎨ w→0 ⎪ 0 ⎩ Emiliano Ippoliti Wednesday, October 12, 2011 7 if - w / 2 < x < w / 2 otherwise Dirac delta function Properties +∞ 1. ∫ δ ( x ) dx = 1. −∞ +∞ 2. ∫ δ ( x − a ) f ( x ) dx = f ( a ). −∞ +∞ 3. ∫ δ ′ ( x − a ) f ( x ) dx = − f ′ ( a ) [ integrate by parts and use 2 ]. −∞ 4. Let f ( x ) have simple zeros at { xi } , i.e. f ( x ) ≈ f ′ ( x ) ( x − xi ) for x near xi , then δ ⎡⎣ f ( x ) ⎤⎦ = ∑ i 1 δ ( x − xi ) . f ′ ( xi ) 5. In three dimensions δ ( x) = δ (x)δ (y)δ (z). This simple formula hold only with cartesian coordinates. (d ) 6. In d dimensions, δ ( x ) has dimensions of L− d . Emiliano Ippoliti Wednesday, October 12, 2011 8 Gauss’ law For a single charge q: E ⋅ n̂ = q cosθ 4πε 0 r 2 so that the flux of the electric field through the area element da is: E ⋅ n̂ da = q da cosθ q = dΩ 2 4πε 0 r 4πε 0 Then, integrating over the entire surface: q ∫ S E ⋅ n̂ da = ε 0 Emiliano Ippoliti Wednesday, October 12, 2011 9 solid angle subtended by da (i.e., r2 dΩ = cosθ da) ( ∫ S dΩ = 4π ) Differential form of the Gauss’ law For a collection of charges {qi} inside the surface: 1 1 ∫ S E ⋅ n̂ da = ε 0 ∑i qi = ε 0 ∫ V 3 ρ(x)d x The divergence theorem states: ∫ S A ⋅ n̂ da = ∫ V 3 ∇⋅A d x Then ∫ S Emiliano Ippoliti Wednesday, October 12, 2011 E ⋅ n̂ da = 10 3 1 ∫V ∇ ⋅ E d x = ε 0 ρ 3 ∫V ρ ( x ) d x ⇒ ∇ ⋅ E = ε 0 Electrostatic potential By using the identity: ⎛ 1⎞ ê r ∇⎜ ⎟ = − 2 ⎝ r⎠ r ⎛ q ⎞ we can write the electric field of a point charge: E = ∇ ⎜ ⎝ 4πε 0 r ⎟⎠ and the field of a set of charges: 1 1 E=− qi ∇ ≡ −∇Φ ( x ) ∑ 4πε 0 i x − xi where Φ is the electrostatic potential: Φ( x ) = Emiliano Ippoliti Wednesday, October 12, 2011 11 1 qi 1 = ∑ 4πε 0 i x − xi 4πε 0 ∫ ρ ( x′ ) 3 d x′ x − xi Meaning of Φ If there are no charges at infinity, so that Φ ( ∞ ) = 0, then qΦ ( x ) is the work required to bring a charge q from ∞ to x (the other charges being held fixed). More generally, the work to bring q from A to B is B W = − ∫ F ⋅ dl = − ∫ qE ⋅ dl = q ⎡⎣ Φ ( B ) − Φ ( A ) ⎤⎦ B A A The work done depends only on the end points (A, B), not on the path; hence the net work in going around a closed path is zero. In this case one says the electric field is conservative. Emiliano Ippoliti Wednesday, October 12, 2011 12 Curl of the electric field For any smooth function Φ: ∇ × ∇Φ = 0 where the curl ∇ × of a vector V is defined as: ( ) ⎛ ∂Vz ∂Vy ⎞ ⎛ ∂Vy ∂Vx ⎞ ⎛ ∂Vx ∂Vz ⎞ ∇×V = ⎜ − ê x + ⎜ − ê y + ⎜ − ê z ⎟ ⎟ ⎟ ⎝ ∂z ⎝ ∂y ⎝ ∂x ∂z ⎠ ∂x ⎠ ∂y ⎠ Therefore Emiliano Ippoliti Wednesday, October 12, 2011 ∇×E =0 13 Stokes’ theorem From the Stokes’ theorem: ∫ C A ⋅ dl = ∫ S ∇ × A ⋅ n̂ da where S is any surface bounded by the closed contour C, we can derive the previous statement that the work in going for a closed path is zero: ∫ C Emiliano Ippoliti Wednesday, October 12, 2011 14 E ⋅ dl = ∫ S ∇ × E ⋅ n̂ da = 0 Lines of forces The lines of force (also called the field lines) provide a method for graphing the electric field. • They are everywhere tangent to the electric field E and therefore for a point charge are tangent to the force exerted by the field on the particle. • They begin on positive charges and terminate on negative charges. • The local density of the field lines is proportional to the strength of the electric field. • The electric field lines do not cross (otherwise the field would not be unique at that point). The lines of force are not particle trajectories! The particle trajectories are obtained by solving F = ma with F = qE . Emiliano Ippoliti Wednesday, October 12, 2011 15 Lines of forces Examples 1 CHARGE 2 CHARGES The equipotentials are contours of constant electrostatic potential. They are analogous to the contours on a topographic map. They are perpendicular to the lines of force. Emiliano Ippoliti Wednesday, October 12, 2011 16 Dipole A dipole is a model of two point charges q and -q at positions x ′and x′′, separated by a infinitesimal displacement d = x ′ − x ′′ The potential in the point P at position x will then be: ⎤ 1 ⎡ q q 1 p ⋅ ( x − x′ ) Φ( x ) = → ⎢ − ⎥ ⎯d⎯⎯ 3 →0 4πε 0 ⎣ x − x ′ x − x′ + d ⎦ 4πε 0 x − x ′ where the dipole moment p = qd. Emiliano Ippoliti Wednesday, October 12, 2011 17 Dipole The electrostatic potential can also be written as Φ( x ) = 1 p cosθ 2 4πε 0 r where θ is the angle between the dipole moment and the observation point P. The electric field is then: 1 3n̂ ( p ⋅ n̂ ) − p E(x) = 3 4πε 0 x − x′ where n is the unit vector directed from x′to x. x Emiliano Ippoliti Wednesday, October 12, 2011 18 Poisson equation Starting from the Coulomb’s law we have derived the two differential field equations of electrostatics: ∇×E =0 ∇ ⋅ E = ρ / ε0 The most general solution of the first equation can be written: E = −∇Φ Inserting it in the second equation, we find that Φ must satisfy: ∇ Φ = −ρ / ε0 2 Poisson equation In a region of space with no sources (ρ = 0) this reduces to: ∇ Φ=0 2 Emiliano Ippoliti Wednesday, October 12, 2011 19 Laplace equation References 1. A. Dorsey. Basic Electrostatics. http://www.phys.ufl.edu/ ~dorsey/phy6346-00/lectures/lect01.pdf 2. D.J. Griffiths. Introduction to Electrodynamics. 3th Eds. Benjamin Cummings, New Jersey, 1999. 3. J.D. Jackson. Classical Electrodynamics. 3th Eds. John Wiley & Son, New York, 1998. Emiliano Ippoliti Wednesday, October 12, 2011 20