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Poisson's equation 2nd 4
This document discusses Poisson's and Laplace's equations which relate electric potential to charge density. Poisson's equation applies to regions with charge density, while Laplace's equation applies to charge-free regions. The equations are derived and their applications are demonstrated, including calculating electric fields and potentials. Examples are provided of solving Laplace's equation for different boundary conditions. The document also covers capacitance of parallel plate capacitors, with and without a dielectric, as well as resistance and combinations of resistors in series and parallel.
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Poisson's equation 2nd 4
- 1. Unit-2 Poission’s equations JETGI Mr. HimanshuDiwakar Assistant Professor JETGI 1
- 2. Poisson’s and LaplaceEquations A useful approach to the calculation of electric potentials Relates potential to the charge density. The electric field is related to the charge density by the divergence relationship The electric field is related to the electric potential by a gradient relationship Therefore the potential is related to the charge density by Poisson's equation In a charge-free region of space, this becomes Laplace's equation JETGI 2
- 3. Potential of aUniform Sphere of Charge outside inside JETGI 3
- 4. Poisson’s and LaplaceEquations Poisson’s Equation From the point form of Gaus's Law Del_dot_D ρv Definition D D εE and the gradient relationship E DelV− Del_D Del_ εE( ) Del_dot_ εDelV( )− ρv Del_DelV ρv− ε Laplace’s Equation if ρ v 0 Del_dot_D ρ v Del_Del Laplacian The divergence of the gradient of a scalar function is called the Laplacian. JETGI 4
- 5. LapR x x V xy, z,( ) d d d d y y V x y, z,( ) d d d d + z z V x y, z,( ) d d d d + := LapC 1 ρ ρ ρ ρ V ρ φ, z,( )d d ⋅ d d ⋅ 1 ρ 2 φ φ V ρ φ, z,( )d d d d ⋅+ z z V ρ φ, z,( )d d d d +:= LapS 1 r 2 r r 2 r V r θ, φ,( )d d ⋅ d d ⋅ 1 r 2 sin θ( )⋅ θ sin θ( ) θ V r θ, φ,( )d d ⋅ d d ⋅+ 1 r 2 sin θ( )2 ⋅ φ φ V r θ, φ,( )d d d d ⋅+:= Poisson’s and Laplace Equations JETGI 5
- 6. Given V x y,z,( ) 4 y⋅ z⋅ x 2 1+ := x y z 1 2 3 := εo 8.854 10 12− ⋅:= V x y, z,( ) 12=Find: V @ and ρv at P LapR x x V x y, z,( ) d d d d y y V x y, z,( ) d d d d + z z V x y, z,( ) d d d d + := LapR 12= ρv LapR εo⋅:= ρv 1.062 10 10− ×= Examples of the Solution of Laplace’s Equation D7.1 JETGI 6
- 7. Examples of theSolution of Laplace’s Equation Example 7.1 Assume V is a function only of x – solve Laplace’s equation V Vo x⋅ d JETGI 7
- 8. Examples of theSolution of Laplace’s Equation Finding the capacitance of a parallel-plate capacitor Steps 1 – Given V, use E = - DelV to find E 2 – Use D = εE to find D 3 - Evaluate D at either capacitor plate, D = Ds = Dn an 4 – Recognize that ρs = Dn 5 – Find Q by a surface integration over the capacitor plate C Q Vo ε S⋅ d JETGI 8
- 9. CAPACITANCE • The ratioof electric charge to electric potential of a conductor or a device is called capacitance • Capacitance C = Q/V • Unit is farad (F) • 1 farad = 1 coulomb / 1 volt JETGI 9
- 10. PRINCIPLE OF ACAPACITOR • Capacitor is based on the principle that the capacitance of an isolated charged conductor increases when an uncharged earthed conductor is kept near it and the capacitance is further increased by keeping a dielectric medium between the conductors. JETGI 10
- 11. CAPACITANCE OF APARALLEL PLATE CAPACITOR Electric field between the plates, E = σ/ε0 But σ=Q/A ∴E=Q/Aε0 Potential difference between the two plates , V = Ed = Qd/A ε0 Capacitance, C = Q/V C=A ε0/d JETGI 11
- 12. CAPACITANCE OF APARALLEL PLATE CAPACITOR WITH A DIELECTRIC SLAB When a dielectric slab is kept between the plates COMPLETELY filling the gap E’ = E0/K where K is the dielectric constant of the medium. Potential difference V’ = E’d = E0d/K=Qd/K ε0A Capacitance C’ = Q/V’ = K ε0A/d = KC ∴when a dielectric medium is filled between the plates of a capacitor, its capacitance is increased K times. JETGI 12
- 13. DIELECTRIC STRENGTH • Dielectricstrength of a dielectric is the maximum electric field that can be applied to it beyond which it breaks down. JETGI 13
- 14. PRACTICE PROBLEMS • Calculatethe number of electrons in excess in a body with 1 coulomb of negative charge. • Q = ne • Q = 1C • e = 1.6 X 10-19 C • n = Q/e= 1/(1.6 X 10-19 C) = 6.25 X 1018 JETGI 14
- 15. Resistance • The electricalresistance of an electrical conductor is a measure of the difficulty to pass an electric current through that conductor. • The inverse quantity is electrical conductance, and is the ease with which an electric current passes. JETGI 15
- 16.
- 17. RESISTORS IN SERIESAND PARALLEL nRRRR +++= ...21eq RESISTORS IN SERIES The magnitude of the charge is constant. Therefore, the flow of charge, current I is also constant. The potential of the individual resistors are in general different. The equivalent resistance of resistors in series equals the sum of their individual resistances.
- 18. RESISTORS INPARALLEL The upper plates of the capacitors are connected together to form an equipotential surface – they have the same potential. The lower plate also have equal potential. The charges on the plates may not necessarily be equal. RESISTORS IN SERIES AND PARALLEL 1 21 eq 1 ... 11 − +++= nRRR R
- 19. RESISTORS IN SERIES ANDPARALLEL JETGI 19