Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence 3.1 Electric Flux Density • Faraday’s Experiment Electric Flux Density, D • Units: C/m2 • Magnitude: Number of flux lines (coulombs) crossing a surface normal to the lines divided by the surface area. • Direction: Direction of flux lines (same direction as E). • For a point charge: • For a general charge distribution, (a)Examples: Given a 60-μC point charge located at the origin, find the total electric flux passing through: A portion of a sphere with r = 26 cm and bounded by: 0 < θ < π/2 and 0 < 𝜙 < π/2 𝐷= 𝑄 4𝜋 𝑟 2 ·aN 𝑑𝐷 = ·aN·𝑟 2 𝑠𝑒𝑛𝜃𝑑𝜃𝑑𝜙 𝜋 2 𝜋/2 −6 60 × 10 𝐷= 4𝜋 𝑟 = 0.26 𝑄 4𝜋 𝑟 2 𝑟 2 𝑠𝑒𝑛𝜃𝑑𝜃𝑑𝜙 · 2 D = 7.5 μC 0 0 60μC = 1 4𝜋 𝜋 2 (b) the closed surface defined by ρ = 26 cm and z = ± 26 cm; 𝐿 2𝜋 𝑄= 𝑫𝒔 𝑑𝑆 = 𝐷𝑠 𝜌 𝑑𝜙𝑑𝑧 = 𝐷𝑠 · 𝜌 𝑑𝜙𝑑𝑧 = 𝐷𝑠 𝜌 2𝜋 𝐿 0 0 Ds = 𝐷𝜌 = 𝑄 𝐷= · 2𝜋𝜌𝐿 +0.26 2𝜋 −0.26 0 𝐷 = 60μC 𝑄 2𝜋 𝜌 𝐿 · 𝑎𝜌 10−6 60 × 𝜌𝑑𝜙𝑑𝑧 = 2𝜋𝜌𝐿 +0.26 2𝜋 · −0.26 0 60 × 10−6 𝜌𝑑𝜙𝑑𝑧 = 2𝜋 2 · 2𝜋 · 0.26 2(0.26) Examples: Given a 60-μC point charge located at the origin, find the total electric flux passing through (c) the plane z = 26 cm. 𝜌𝑠 𝐷 = 𝐸𝜀0 = · 𝑎𝑟 2 𝜌𝑠 𝑄 𝐷= = 2 2 · 𝜋𝑟 2 2𝜋 ∞ −6 60 × 10 𝐷= 2𝜋𝑟 2 · 0 0 −6 60 × 10 𝑟𝑑𝑟𝑑𝜙 = 2𝜋𝑟 2 60 × 10−6 2𝜋 𝐷= · 2𝜋 2 −6 = 30 × 10 𝐶 2 𝑟 · 2 2𝜋 𝑑𝜙 0 D3.2 Calculate D in rectangular coordinates at point P(2,-3,6) produced by : (a) a point charge QA = 55mC at Q(-2,3,-6) 2 3 P • P(2,-3,6) 6 Q (-2,3,-6) 2 3 𝑎 , 𝑎 , 𝑎 2 − −2, −3 − 3, 6 − −6 QA 55 10 𝑥 𝑦 𝑧 𝑟 = = Q 3 42 + (−6)3 +(12)2 = 196 |𝑅| 6 = 4, −6,12 = 0.285, −0.428,0.857 14 12 0 8.854 10 D QA 4 R 2 r R P Q r 6.38 10 6 D 9.57 10 6 5 1.914 10 55 × 10−3 𝐷= · 0.285, −0.428,0.857 4𝜋 · 196 PQ PQ Calculate D in rectangular coordinates at point P(2,-3,6) produced by : (b) a uniform line charge pLB = 20 mC/m on the x axis z • P(2,-3,6) x 𝑎𝑥 , 𝑎𝑦 , 𝑎𝑧 2 − 2, −3 − 0, 6 − 0 𝑎𝑟 = = |𝑅| 02 + (−3)3 +(6)2 0, −3,6 = 45 𝜌𝑙 𝐷 = 𝐸𝜀0 = · 𝑎𝑟 2𝜋 𝑅 𝐷= 20×10−3 2𝜋· 45 · 0,−3,6 45 = 10×10−3 𝜋·45 −3𝑎𝑦 + 6𝑎𝑧 = −212ay + 424az μC/m2; (c) a uniform surface charge density ρSC = 120 μC/m2 on the plane z = −5 m. • P(2,-3,6) 𝜌𝑠 𝐷 = 𝐸𝜀0 = · 𝑎𝑟 2 𝑎𝑥 , 𝑎𝑦 , 𝑎𝑧 2 − 2, −3 − −3, 6 − −5 0,0,11 𝑎𝑟 = = = = 0,0,1 2 3 2 |𝑅| 11 0 + (0) +(11) 𝜌𝑠 = 120 × 10−6 𝐷= · 𝑎𝑧 = 60 × 10−6 C/m2 2 Gauss’s Law • “The electric flux passing through any closed surface is equal to the total charge enclosed by that surface.” • The integration is performed over a closed surface, i.e. gaussian surface. • We can check Gauss’s law with a point charge example. a = a) b) 𝐷 0.3𝑟 2 × 10−9 𝑎𝑟 𝐸= = 𝜖0 8.8 × 10−12 = 136𝒂𝒓 𝑉 𝑚 𝑟=2 𝜙=2𝜋,𝜃=𝜋 𝑄= 𝐷𝑠 · 𝑑𝑆 = 0.3𝑟 2 𝒂𝒓 · 𝑟 2 𝑠𝑒𝑛𝜃 𝑑𝜃𝑑𝜙 𝜙=0,𝜃=0 𝑄= 0.3𝑟 4 × 4 10−9 2𝜋 − cos 𝜃 −9 𝜋 0 = 1 − −1 = 2 = 0.3(3) × 10 4𝜋 = 305.36 × 10−9 C c) 𝑑Ψ = 𝐷𝑠 · 𝑑𝑆 = 0.3𝑟 2 × 10−9 𝑎𝑟 · 𝑟2𝑠𝑒𝑛𝜃 𝑑𝜃𝑑𝜙 𝑎𝑟 𝜙=2𝜋,𝜃=𝜋 Ψ = 0.3𝑟 2 × 10−9 · 𝑟 2 𝑠𝑒𝑛𝜃𝑑𝜃𝑑𝜙 𝜙=0,𝜃=0 = 0.3(4)4 × 10−9 · 2 · 2𝜋 = 965 × 10−9 D3.4. Calculate the total electric flux leaving the cubical surface formed by the six planes x, y, z = ± 5 if the charge distribution is: (a) twopoint charges, 0.1 μC at (1,−2, 3) and (1/7)μC at (−1, 2,−2); 𝐷 = 𝐸𝜀0 = 𝜌𝑠 · 𝑎𝑟 2 0.1 μC (1,−2, 3) 𝑑Ψ = 𝐷𝑠 · 𝑑𝑆 = 𝑄1 𝐴𝑟𝑒𝑎 𝑄2 + 𝐴𝑟𝑒𝑎 𝑎𝑁 = 𝑥 − 𝑥1,2 𝑎𝑥 + 𝑦 − 𝑦1,2 𝑎𝑦 + 𝑧 − 𝑧1,2 𝑎𝑧 𝑄1 𝑦=5,𝑧=5 Ψ±𝑥 = 𝑦=−5,𝑧=−5 𝑥=5,𝑧=5 (1/7)μC (−1, 2,−2) 𝑎𝑁 · 𝑑𝑥𝑑𝑦𝑑𝑧 𝑎𝑁 Ψ±𝑦 = 𝑥=−5,𝑧=−5 𝑥=5,𝑦=5 Ψ±𝑧 = 𝑥=−5,𝑦=−5 Ψ= Ψ𝑥,𝑦,𝑧 = 𝑄2 6+ 6 · 𝑑𝑦𝑑𝑧 𝑦𝑧 𝑦𝑧 𝑄1 𝑄2 6+ 6 · 𝑑𝑥𝑑𝑧 𝑥𝑧 𝑥𝑧 𝑄1 𝑄2 6+ 6 · 𝑑𝑥𝑑𝑦 𝑥𝑦 𝑥𝑦 6𝑄1 6 + 6𝑄2 6 = 𝑄1 + 𝑄2 Field of a Line Charge Consider an infinite line charge parallel to the z axis at x = 6, y = 8. Find E at the general field point P(x, y, z). E L 2 0 a P (x,y,z) 𝜌𝑆 = 𝜌𝐿 𝑑𝑧 𝐷 = 𝐸𝜀0 = · 𝑎𝑁 2𝜋𝜌 D3.4. Calculate the total electric flux leaving the cubical surface formed by the six planes x, y, z = ± 5 if the charge distribution is: (b) a uniform line charge of π μC/m at x = −2, y = 3; 𝑎𝜌(𝑥,𝑦) 𝜌𝐿 𝑑Ψ = 𝐷 · 𝑑𝑆 = · 𝑑𝑥𝑑𝑦𝑑𝑧 𝑎𝑁 z 𝑠 2𝜌 π μC/m 𝜌𝐿 Ψ= 2𝜋 𝑥,𝑦,𝑧=5 𝑥,𝑦,𝑧=−5 𝜌 𝑥 − 𝑥𝐿=−2 𝑎𝑥 + 𝑦 − 𝑦𝐿=3 𝑎𝑦 𝑑𝑥𝑑𝑦𝑑𝑧 · 𝑎𝑁 𝜌2 = 𝑥 + 2 2 + 𝑦 − 3 2 x Caso x= +5: y 𝜌𝐿 Ψ𝑎 = 2𝜋 𝜌𝐿 Ψ𝑎 = 2𝜋 𝑦=5 𝑦=−5 𝑦,𝑧=5 𝟓 + 2 𝑎𝑥 + 𝑦 − 3 𝑎𝑦 𝜌𝐿 𝑑𝑦𝑑𝑧 · 𝑎 = 𝑁 𝟕 2+ 𝑦−3 2 2𝜋 𝑦,𝑧=−5 7 49 + 𝑦 − 3 2 1 𝑦−3 = 7 arctan 𝑎𝑥 𝑑𝑦 5 − −5 + 7 7 𝑦,𝑧=5 𝑦,𝑧=−5 𝑦=5 𝑦=−5 7𝑎𝑥 + 𝑦 − 3 𝑎𝑦 𝑑𝑦𝑑𝑧 · 𝑎𝑁 49 + 𝑦 − 3 2 𝑦−3 49 + 𝑦 − 3 2 𝑎𝑦 · 𝑎𝑁 𝑑𝑦[10] 𝑎𝑦 · 𝑎𝑁 = 0 Ψ𝑎 = 𝜌𝐿 2𝜋 10 𝑎𝑟𝑐𝑡𝑎𝑛 𝑦−3 7 𝑦=5, 𝑦=−5, = 0.27 − −0.85 = 𝜌𝐿 2𝜋 10 [1.13] Caso y= +5: 𝜌𝐿 Ψ𝑏 = 2𝜋 Ψ𝑏 = 𝑥,𝑧=5 𝑥,𝑧=−5 𝜌𝐿 2𝜋 10 𝑎𝑥 · 𝑎𝑁 = 0 𝑥,𝑧=5 𝑥 + 2 𝑎𝑥 + 5 − 3 𝑎𝑦 𝜌𝐿 𝑑𝑥𝑑𝑧 · 𝑎 = 𝑁 𝑥+2 2+ 2 2 2𝜋 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥+2 2 𝑥=5, 𝑥=−5, 𝑥 + 2 𝑎𝑥 + 2𝑎𝑦 𝑑𝑥𝑑𝑧 · 𝑎𝑁 𝑥+2 2+4 𝑥,𝑧=−5 = 1.29 − −0.98 = 𝜌𝐿 2𝜋 10 [2.27] Caso x= -5: Ψ𝑐 = 𝜌𝐿 2𝜋 Ψ𝑐 = 𝑦,𝑧=5 𝑦,𝑧=−5 𝜌𝐿 2𝜋 −𝟓 + 2 𝑎𝑥 + 𝑦 − 3 𝑎𝑦 𝜌𝐿 𝑑𝑦𝑑𝑧 · 𝑎 = 𝑁 −𝟑 2 + 𝑦 − 3 2 2𝜋 10 1 3 (−3) 𝑎𝑟𝑐𝑡𝑎𝑛 𝑦−3 3 𝑦=5, 𝑦=−5, Ψ𝑑 = Ψ𝑑 = 𝜌𝐿 2𝜋 𝜌𝐿 2𝜋 𝑥,𝑧=−5 10 𝑦,𝑧=−5 = −0.58 − 1.21 Caso y= -5: 𝑥,𝑧=5 𝑦,𝑧=5 𝑎𝑦 · 𝑎 𝑁 = 0 −3𝑎𝑥 + 𝑦 − 3 𝑎𝑦 𝑑𝑦𝑑𝑧 · 𝑎𝑁 9+ 𝑦−3 2 = 𝜌𝐿 2𝜋 10 [−1.8] 𝑎𝑥 · 𝑎𝑁 = 0 𝑥 + 2 𝑎𝑥 + −5 − 3 𝑎𝑦 𝜌𝐿 𝑑𝑥𝑑𝑧 = 𝑥 + 2 2 + −8 2 2𝜋 1 8 (−8) 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥+2 8 𝑦=5, 𝑦=−5, 𝑥,𝑧=5 𝑥,𝑧=−5 = −0.71 − 0.35 𝑥 + 2 𝑎𝑥 − 8𝑎𝑦 𝑑𝑥𝑑𝑧 𝑥 + 2 2 + 64 = 𝜌𝐿 2𝜋 10 [−1.077] Calculate the total electric flux leaving: Ψ𝑑 = 𝜌𝐿 2𝜋 10 1.13 + 2.27 + 1.8 + 1.077 = 10𝜌𝐿 2𝜋 6.283 = 2𝜋 = 10𝜌𝐿 = 31.416 𝜇𝐶 𝑚2 D3.4. Calculate the total electric flux leaving the cubical surface formed by the six planes x, y, z = ± 5 if the charge distribution is: (c) a uniform surface charge of 0.1 μC/m2 on the plane y = 3x. y y = 3x. 5/3 5/3 1 + 𝑦 ′ 𝑥 = 3 2 𝑑𝑥 = 𝑠= −5/3 z 10 𝑑𝑥 −5/3 5 𝑠 = 10 2 · = 10.54 3 𝑧=5 x 𝑑Ψ = 𝐷𝑠 · 𝑑𝑆 = 𝑄𝑑𝑥𝑑𝑦𝑑𝑧 = 𝜌𝑠 10.54 𝑑𝑧 𝑧=−5 𝜇𝐶 Ψ = (𝜌𝑠 = 0.1)10.54 10 = 10.54 2 𝑚 y = 3x. Symmetrical Charge Distributions • Gauss’s law is useful under two conditions. 1. DS is everywhere either normal or tangential to the closed surface, so that DS.dS becomes either DS dS or zero, respectively. 2. On that portion of the closed surface for which DS.dS is not zero, DS = constant. Gauss’s law simplifies the task of finding D near an infinite line charge. Infinite coaxial cable: Let us select a 50-cm length of coaxial cable having an inner radius of 1 mm and an outer radius of 4 mm. The space between conductors is assumed to be filled with air. The total charge on the inner conductor is 30 nC. We wish to know the charge density on each conductor, and the E and D fields. 𝜌𝐿 = 𝑄 = 𝜌𝑠 2𝜋𝑎 𝐿 𝐷𝜌 = 𝜌𝑠 2𝜋𝑎 = 2𝜋𝜌 Apply to the region where 1 < ρ < 4 mm. For ρ < 1 mm or ρ > 4 mm, E and D are zero. D3.5. A point charge of 0.25 μC is located at r = 0, and uniform surface charge densities are located as follows: 2 mC/m2 at r = 1 cm, and −0.6 mC/m2 at r = 1.8 cm. Calculate D at: (a) r = 0.5 cm; (b) r = 1.5 cm; (c) r = 2.5 cm. 𝜌𝑠 2𝜋𝑎 𝐷𝜌 = = 2𝜋𝜌 0.25 μC −0.6 mC/m2 2 mC/m2 1cm 0.8cm (d) What uniform surface charge density should be established at r = 3 cm to cause D = 0 at r = 3.5 cm? Ans. 796ar μC/m2; 977ar μC/m2; 40.8ar μC/m2; −28.3 μC/m2 Differential Volume Element • If we take a small enough closed surface, then D is almost constant over the surface. The front face is at a distance of x/2 from P, z x y D3.6) In free space, let D = 8xyz4 ax + 4x2z4 ay + 16x2yz3 az pC/m2. (a) Find the total electric flux passing through the rectangular surface z = 2, 0 < x < 2, 1 < y < 3, in the az direction. 2,3 𝐸𝑧 = 3 (𝐷𝑧 = 16𝑥 2 𝑦𝑧 3 ) 0,1 2 z 16 · 8 · 8 𝑦 = 3 2 x y 𝑑𝑥𝑑𝑦 = 16(𝑧 = 2)3 𝑥 3 2 𝑦𝑑𝑦 0 1 3 = 170.6 9 − 1 = 1365 𝑝𝐶 1 3 (b) Find E at P(2,−1, 3). 8 x y z4 12 D( x y z) 4 x2 z4 10 2 3 16 x y z 12 0 8.854 10 2 P 1 3 E D( 2 1 3) 0 146.375 E 146.375 195.166 (c) Find an approximate value for the total charge contained in an incremental sphere located at P(2,−1, 3) and having a volume of ∆𝑉 =10−12 m3. z D = 8xyz4 ax + 4x2z4 ay + 16x2yz3 az pC/m2. 𝜕𝐷𝑥 = 8𝑦𝑧 4 𝜕𝑥 𝜕𝐷𝑦 =0 𝜕𝑦 𝜕𝐷𝑧 = 48𝑥 2 𝑦𝑧 2 𝜕𝑧 x P(2,−1, 3) 𝑄= 𝜕𝐷𝑥 𝜕𝑥 𝑃(2,−1,3) 𝜕𝐷𝑧 = −648 + 𝜕𝑧 y = −1728 𝑃(2,−1,3) = −2376 · 10−12 𝑝𝐶 = −2.376 · 10−21 𝐶 ∆𝑉 The divergence of the vector flux density A is the outflow of flux from a small closed surface (per unit volume) as the volume shrinks to zero. -VELOCITY of water leaving a bathtub -Closed surface (water itself) is essentially incompressible -Water entering and leaving regions of the closed surface must be equal -Net outflow is zero -VELOCITY of air leaving a punctured tire -Air is expanding as the pressure drops -Divergence is positive, as closed surface (tire) exhibits net outflow Positive divergence for any vector quantity indicates a source of that vector quantity at that point Negative divergence indicates a sink. Mathematical definition of divergence div D D lim dS v 0 v Surface integral as the volume element (v) approaches zero D is the vector flux density - Cartesian Divergence in Other Coordinate Systems Divergence is an operation which is performed on a vector, but that the result is a scalar Divergence merely tells us how much flux is leaving a small volume on a per-unit-volume basis; no direction is associated with it. Divergence at origin for given vector flux density A e x sin ( y) A e x cos ( y ) 2 z div A div A x e x sin( y) e x sin ( y ) e y x x e cos ( y ) sin ( y ) 2 z ( 2 z) =2 The value is the constant 2, regardless of location. If the units of D are C/m2, then the units of div D are C/m3 D3.7. In each of the following parts, find a numerical value for div D at the point specified: (a) D = (2xyz − y2)ax + (x2z − 2xy)ay + (x2y)az C/m2 at PA(2, 3,−1); (b) D = 2ρz2 sin2(φ)aρ + ρz2 sin(2φ)aφ + 2ρ2z sin2(φ)az C/m2 at PB(ρ = 2, φ = 110◦, z = −1); Ans. −10.00; 9.06; D3.7. In each of the following parts, find a numerical value for div D at the point specified: (c) D = (2r·sin θ·cos φ)ar + (r·cos θ·cos φ)aθ − (r·sin φ) aφ C/m2 at PC(r = 1.5, θ = 30◦, φ = 50◦). Ans. 1.29 3-6: Maxwell’s First Equation 𝐷 𝑑𝑆 = 𝑄 Gauss’ Law… 𝑆 𝑆 𝐷 𝑑𝑆 ∆𝑣 𝑄 = ∆𝑣 Volume shrinks to zero …per unit volume lim ∆𝑣→0 𝑆 𝐷 𝑑𝑆 ∆𝑣 𝑄 = lim ∆𝑣→0 ∆𝑣 Electric flux per unit volume is equal to the volume charge density Maxwell’s First Equation lim ∆𝑣→0 𝑆 𝐷 𝑑𝑆 ∆𝑣 div D 𝑄 = lim ∆𝑣→0 ∆𝑣 v Sometimes called the point form of Gauss’ Law Enclosed surface is reduced to a single point 𝐷 𝑑𝑆 = 𝑄 𝑆 Gauss’s law relates the flux leaving any closed surface to the charge enclosed. Maxwell’s first equation makes an identical statement on a per-unit-volume basis for a vanishingly small volume, or at a point. 𝐷 𝑑𝑆 𝑄 lim ∆𝑣→0 𝑆 ∆𝑣 = lim ∆𝑣→0 ∆𝑣 3-7: and the Divergence Theorem del operator What is Writing ∇·D allows us to obtain simply and quickly the correct partial derivatives, but only in rectangular coordinates del? div D is an excellent reminder of the physical interpretation of divergence. Starting from Gauss’s law: Benefits from derive from the fact that it relates a triple integration throughout some volume to a double integration over the surface of that volume. For example, it is much easier to look for leaks in a bottle full of some agitated liquid by inspecting the surface than calculating the velocity at every internal point. The total flux crossing the closed surface is equal to the integral of the divergence of the flux density throughout the enclosed volume. The volume is shown here in cross section. Division of the volume into a number of small compartments of differential size and consideration of one cell show that the flux diverging from such a cell enters, or converges on, the adjacent cells unless the cell contains a portion of the outer surface. In summary, the divergence of the flux density throughout a volume leads, then, to the same result as determining the net flux crossing the enclosing surface. parallel to the surfaces z, so D· dS = 0 z 𝐷𝑥 = 2𝑥𝑦 x y 𝐷𝑦 = 𝑥 2 𝐷𝑥 = 2𝑥𝑦 𝐷𝑦 = 𝑥 2 Other Relationships Gradient – results from operating on a function Represents direction of greatest change Curl – cross product of and Relates to work in a field If curl is zero, so is work Examination of and flux Cube defined by 1 < x,y,z < 1.2 D 2 2 2 2 x y a x 3 x y a y Calculation of total flux . Q D dS S total . v dv vol left right front back z y 1 1 z y 1 1 2 2 2 x1 2 x1 y d y d z z y 2 2 2 x2 2 x2 y d y d z z y x1 1 y 1 1 x2 1.2 y 2 1.2 z1 1 z2 1.2 z x 1 1 z x 1 1 2 2 2 2 y1 3 x y 1 d x d z z x 2 2 2 2 y2 3 x y 2 d x d z z x total x1 x2 y1 y2 total 0.103 Evaluation ofV Dat center of cube div D d 2 d 2 2 2 x y 3 x y dx dy div D 4 x y 6 x y 2 2 divD 4 ( 1.1) ( 1.1) 6 ( 1.1) ( 1.1) divD 12.826 Non-Cartesian Example Equipotential Surfaces – Free Software Applications of Gauss’s Law
