Electric field lines are perpendicular to the equipotential lines

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Branden Wilkins
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1 EQUIPOTENTIAL URFACE E = Electric field lines are perpendicular to the equipotential lines Electric field lines are opposite to the direction where maximum variation in the scalar potential occurs E = q R R 2 MATERIAL MEDIA IN TATIC ELECTRIC FIELD Fact tatic Electric field can not exist inside a conductor E 0 : Electric field before the conductor was inserted E : econdary electric field created inside the conductor as a reaction to E 0 E : Resultant electric field after the TRANIENT period following the insertion of the conductor, E = E 0 + E At the end of the transient period E = E 0 inside the conductor At steady state, inside the conductor, E = E 0 + E = E 0 + E 0 = 0

2 Fact 2 E 0 is always perpendicular to the surface of a conductor E = 0 E d = dl = 0 C dl = E dl 4 C i= C i dl = E dl + E dl + C ab bc E dl + cd E dl + da E dl = E dl = 0 bc E 2 =0 due to fact E = E v = E n 2 E is perpendicular to boundary surface E = E v + 0u da E has no tangential component E // = 0

perpendicular to boundary surface dl is // to the surface E dl E dl = 0 = b a = E b dl a = 0

3 Fact 3 Conductor surface is an equipotential surface and conductor body is an equipotential volume Proof As a result of Fact 2, E 0 & E // = 0 = E & E is an equipotential surface E is perpendicular to boundary surface dl is // to the surface E dl E dl = 0 = b a = E b dl a = 0 Fact 4: E = ρ s ε on E = Boundary Condition for a conductor surface d = ρ v d ε 0 d = Q enclosed ε 0

d = E d + E 2 d 2 + E 3 d 3 + E 4 d + 4 E 2 =E 4 =0 due to fact d = E d + E 3 d 3 3 0 => E 3 d 0 3 => d = lim E = Q enclosed h 0 ε 0 E =.

4 d = E d + E 2 d 2 + E 3 d 3 + E 4 d + 4 E 2 =E 4 =0 due to fact d = E d + E 3 d => E 3 d 0 3 => d = lim E = Q enclosed h 0 ε 0 E =. ρ s ε 0 E = ρ s ε on E = Fact 5: No charge can remain within the conductor. The charges will redistribute themselves to the outer surface of the conductorwithin the transition relaxation time. t relaxation,copper 0 9 sec. Electrostatic hielding t relaxation,quartz 0 6 sec. When external E is applied no charge can be induced in the inner conductor surface, in We have E in the cavity.

However, if there are charges total of Q inside the cavity, the exterior region to the shell will be effected.

5 dl = E σ dl + E cavity dl C C σ C cavity 0 = 0 + E cavity dl C cavity => E cavity dl = 0 C cavity => E cavity = 0 A conducting shell protects its interior from the external fields. However, if there are charges total of Q inside the cavity, the exterior region to the shell will be effected. When there is no conductor With existence of the conductor Using Gauss Law For cavity, d = Q ε For σ, E = E 0 + E s = 0 d Q + Q = = 0 ε For out, E = E 0 d Q + Q + Q = ε = Q

6 It can be proven that the distribution of Q out is not effected by the distribution of Q Without conductor With conductor sphere, off-center E = Q 4πεR 2, everywhere 0, 0 < r < a 0, a < r < b E = { Q 4πεR 2, r > b Behaviour of dielectrics in tatic Electric Field => A dipole is created by E external

7 Ex: Electric field due to a dipole: R + = R d 2 z R + = RR d 2 z R + = R d 2 cosθ R = R + d 2 z R = R + d 2 cosθ R = RR + d 2 z q RR E P = q q RR q d 2 z q d 2 z RR d 2 z RR + d R d 3 2 z 2 cosθ R + d 3 2 cosθ E P = R 3 + 3R 2 d 2 cosθ + 3R d 2 2. cos 2 θ + d 2 3. cos 3 θ R 3 3R 2 d 2 cosθ + 3R d 2 2. cos 2 θ d 2 3. cos 3 θ R 3 3R 2 d 2 cosθ + 3R d 2 2. cos 2 θ d 2 3. cos 3 θ R 3 3R 2 d 2 cosθ + 3R d 2 2. cos 2 θ d 2 3. cos 3 θ

8 3 = q 2RR 3R 2 d 2 cosθ + d 2. cos 3 θ 2 d 2 z R3 + R d 2 2. cos 2 θ [ ] = q [ R d. 2. cosθ. R 3 + d. 2. cosθ. R 3 d. z R 3 ] = q [ d. 2. cosθ. R3 = q [ d. 2. cosθ. R3 R + R + 3 d. cosθ. R 3 d. cosθ. R R R 3 d. R 3 d. R z ] z ] = q [ = d. 2. cosθ R + d R 3 [R cosθ z ]] R 3 qd R 3 [2. cosθr + sinθθ ] Polarization ector P P = lim 0 p i where P or Px,y,z is the observation point p i is the elemental dipole moment Coul.m Polarized dielectric P is the polarization vector is the volme average of the vector sum of the elemental dipole moments and its unit is Coul.m/m 3 = Coul/m 2 Microsopic effect of millions of electric dipoles contained in a differential volume d of a polarized dielectric can be represented by a single macroscopic electric dipole moment vector, P d.

9 The total electrostatic potential Φ = P d ρ ρ 2 = P ρ ρ 2 The infinitesimal electrostatic potential at observation point P, created by the electric dipole moment P d dφ = d R ector from origin to the observation point P R ector from origin to the source point ρ = R R ρ = ρ ρ = R R R R R R R R 3 = R R R R = = R R 2 ρ ρ2 ρ ρ 2 P d ρ ρ 2 ΦR = P d ρ ρ 2 = P R R R d R R 3 Polarization Charges: An equivalent charge distribuiton can be assumed to model the polarization of the dielectric: R = R R = xx + yy + zz xx + yy + zz R = x 2 + y 2 + z 2 R = x 2 + y 2 + z 2 = xx + yy + zz R = R RR R = = R3 R3 x 2 + y 2 + z R 2 R R = xx + yy + zz x x + y y + z z R R = x x x + y y y + z z z R R = R R R R R R = x x x + y y y + z z z x x x + y y y + z z z

10 R R = x x 2 + y y 2 + z z 2 R R = x x + y y + z z x x 2 + y y 2 + z z 2 R R = x x x + y y y + z z z x x 2 + y y 2 + z z R R = R R R R 3 R R = x + y + x y z z x x 2 + y y 2 + z z 2 x x x + y y y + z z z R R = + P R x x 2 + y y 2 + z z R R = + R R = R R 3 R R ΦR = P R d a = ax, y, z F = F x x, y, zx + F y x, y, zy + F z x, y, zz af = a F + a F F a = P R R R a F = F a = af a F ΦR = = Φ P R P R P R P R d

11 Φ = ΦR = P R P R d + d + d = n d ΦR = P R n d + ΦR = ρ sp Electric Flux Density & Dielectric Constant P R d P R d P R d d + ρ sp = P R n [ρ sp ]= Coulomb m 2 ρ vp = P R [ρ vp ]= Coulomb m 3 Assume both free charges & polarized dielectric are given: Free charges: ρ v, ρ s, ρ l externally put Polarization charges: ρ vp, ρ sp For this case, the Gauss Law: ρ vp d d = Q total ε 0 = ε 0 ρ v + ρ vp d = ρ ε v d + ρ 0 ε vp d 0 = ρ ε v d + P d 0 ε 0 = ρ ε v d P d 0 ε 0

12 = ρ ε v d P 0 ε 0 = D d P ε 0 ε 0 d d = ε 0 D = ε 0 εe d P ε 0 d P ε 0 d d d = εe P d ε 0 ε 0 E d = εe P d ε 0 E = εe P ε ε 0 E = P d = ρ ε v d 0 ε 0 P d D ε 0 E + P d = ρ v d d = ρ v d D = ε 0 E + P GGL Integral Form GGL Differential Form D = ρ v [ D ] = [ρ v ] = Coulomb [D m 3 ] = Coulomb m 2 Note: ources of D are only free charges: ρv, ρ s, ρ l Note: ources of E are both free charges and polarization charges

13 Electric Flux Density D : { Displacement vector E : Electric Field Intensity Density Coulomb m 2 olt m D d = ρ v d = Q free For an arbitrary dielectric D = ε 0 E + P For a linear dielectric P = ε 0 χ e E χ e : electric susceptibility χ e 0 Dimensionless Dielectric χ e is independent of Homogeneous pace coordinates Linear E Isotropic Direction of E For a linear dielectric *** D = ε 0 E + P = ε 0 E + ε0 χ e E D = ε 0 + χ e E D = ε 0 ε r E D = εe ε = ε 0 ε Absolute r ε r = + χ e ε r ε ε 0 ε r,vacuum = ε r,air = ε air = ε 0 ε r,air = ε 0 ε 0 Medium is simple if it is homogeneous, linear and isotropic ε r : constant for a simple medium ome material some crystals are anistropic permittivity Relative permittivity always

14 ε r depends on direction E and D are in different directions for anistropic medium Dielectric trength An electric field causes small displacements in the bound charges of the dielectric yielding polarization If the E-field strong enough, it can pull electrons out of their in the atoms/molecules yielding permanent damage in the material Avalanche effect of ionization may occur resulting in very large currents; the dielectric turns into a conductor. This phenomenon is known as Dielectric Breakdown. The maximum electric field that can be applied onto the dielectric before the breakdown occurs is called as the Dielectric trength. Air breaks down for E 3000 /m at atmospheric pressure Massive ionization of air molecules, sparking, corona discharge occur E is larger at the conductor surfaces with larger curvature at the tip of the conductor Wet air close to the tip of the lightning rod can be easily ionized and electric charges in the clouds can be easily discharged to the ground via the lightning rod.

Chapter 4. Electric Fields in Matter

Chapter 4. Electric Fields in Matter 4.1.2 Induced Dipoles What happens to a neutral atom when it is placed in an electric field E? The atom now has a tiny dipole moment p, in the same direction as E.