Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
ELECTRIC FIELD ELECTRIC FLUX GAUSS LAW Today….. • More on Electric Field: – Continuous Charge Distributions • Electric Flux: – Definition – How to think about flux Summary Electric Field Lines Electric Field Patterns Dipole Point Charge Infinite Line of Charge Two types of electric charge: opposite charges attract, like charges repel 1 Q1Q2 ∧ F12 = 2 r12 4πε 0 r12 → Coulomb’s Law: Electric Fields → Charges respond to electric fields: Charges produce electric fields: → F = qE q E=k 2 r Electric Flux Flux: Let’s quantify previous discussion about field-line “counting” Define: electric flux ψ through the closed surface S ψ = ∫ E • dS S “S” is surface of the box Electric Flux ψ = ∫ E • dS S •What does this new quantity mean? • The integral is over a CLOSED SURFACE → → • Since E • dS is a SCALAR product, the electric flux is a SCALAR quantity → • The integration vector dS is normal to the surface and points OUT → → of the surface. E • dS is interpreted as the component of E which is NORMAL to the SURFACE. • Therefore, the electric flux through a closed surface is the sum of the normal components of the electric field all over the surface. • The sign matters!! Pay attention to the direction of the normal component as it penetrates the surface… is it “out of” or “into” the surface? • “Out of” is “+” “into” is “-” How to think about flux • We will be interested in net flux in or out of a closed surface like this box • This is the sum of the flux through each side of the box – consider each side separately • Let E-field point in y-direction → – then E and → → → S are parallel and → E • S = E w2 ELECTRIC FLUX DENSITY We define electric flux ψ D = ε0E in terms of D ψ = ∫ D • dS For example, for an infinite sheet of charge ρs E= an , 2ε 0 ⎛ ρs ⎞ ρs ⎟⎟ a n = D = ε 0 ⎜⎜ an 2 ⎝ 2ε 0 ⎠ For line charge distribution ρL ρL E= aρ , D = aρ 2πε 0 ρ 2πρ For volume charge distribution ρV dv ρV dv E=∫ a D = a , R ∫ 2 2 R 4πε 0 R 4πR Example: Determine D at (4,0,3) if there is a point charge -5π mC at (4,0,0) and a line charge 3 π mC/m along the y axis. Let D = DQ + DL where DQ and DL are flux densities due to the point charge and line charge respectively. Q Q(r − r ) DQ = ε 0 E = aR = 2 ' 3 4πR 4π r − r ' Where r – r’ =(4,0,3) –(4,0,0)=(0,0,3).Hence, − 5π x 10 − 3 (0,0,3 ) mC 2 DQ = = − 0 . 138 a 3 z m 4π 0,0,3 Also, ρL DL = aρ 2πρ aρ ( (4,0,3) − (0,0,0) ) (4,0,3) = = (4,0,3) − (0,0,0) 5 3π (4ax + 3az ) = 0.24ax + 0.18az mC m 2 DL = 2π (5)(5) Thus, D = DQ + DL C µ D = 240 a x + 42 a z m2 GAUSS’S LAW-MAXWELL’S EQUATION Gauss’s Law states that the total electric flux ψ through any closed surface is equal to the total charge, Q enclosed by that surface . ψ = Qenc That is ψ = ∫ dψ = ∫ D • dS S = Total Charge enclosed , where Q = ∫ ρ v dv v ∴ Q = ∫ D • dS = ∫ ρ v dv S v The integration is performed over a closed surface, i.e. gaussian surface. By applying Divergence Theorem ∫ D • dS = ∫ ∇ • Ddv Where, ρV = ∇ • D APPLICATION OF GAUSS’S LAW A. Point Charge Since D is everywhere normal to gaussian surface, that is D = Dr ar From Gauss Law, ψ = Qenc Q = ∫ D • dS = Dr ∫ dS = Dr 4πr Where ∫ D • dS = 2π ∫φ θ π 2 2 2 r sin θ d θ d φ = 4 π r ∫ =0 =0 Q ∴D = ar 2 4π r B. Infinite Line of Charge Suppose the infinite of uniform charge ρ L C m lies along the z-axis. Determine D at point P? D = Dρ aρ Q = ∫ ρ L dl but , Q = ∫ D • dS = Dρ ∫ dS = Dρ 2πρL ρL D= aρ 2πρ C. Infinite Sheet of Charge Consider the infinite sheet of uniform charge is ρS C/m2 D = DZ aZ ∫ρ ∫ D • dS ρ dS = D ( A + A ) ρ A = D (A + A ) Q = S dS , Q = S Z S D = Z ρS 2 aZ ρS Therefore , E = = aZ ε 0 2ε 0 D D. Uniformly Charged Sphere Gaussian surface for a uniformly charged sphere Q enc ∫ρ = = ρ v dv = ρ 2π v π ∫ ∫ ∫ φ θ =0 = ρ r v ∫ v dv r 2 sin θ drd θ d φ =0 r=0 4 πr 3 3 And ψ = ∫ D • dS = D = D 2π r π = D r 4π r ∫ dS r sin θ d θ d φ ∫ ∫ φ θ =0 r 2 =0 2 Hence , ψ = Qenc gives 4π r ρv D r 4π r = 3 Or .. 3 2 r D = ρ v ar 3 for 0 > r ≤ a r ≥ a, For Q enc ∫ρ = = ρ v dv = ρ 2π v r 2 dv sin θ drd θ d φ =0 r=0 4 πa 3 3 , While ψ = v a ∫ θ∫ ∫ φ =0 = ρ π ∫ v ∫ D • dS D r 4π r 2 = D r 4π r 4 = πa 3ρ 3 2 v Or .. a3 D = ρ va r 2 3r for r ≥ a D everywhere is given by: ⎧r a ρ ⎪3 v r D=⎨ 3 a ⎪ 2 ρ v ar ⎩ 3r 0<r ≤a r≥a