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# Block01P2Notes.pdf PHYS 242

NC A&T

GPA 3.8

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This 4 page Class Notes was uploaded by DB on Tuesday September 27, 2016. The Class Notes belongs to PHYS 242 at North Carolina A&T State University taught by Prof. Sandin in Fall 2016. Since its upload, it has received 3 views. For similar materials see General Physics II in Physics 2 at North Carolina A&T State University.

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Date Created: 09/27/16

PHYS 242 BLOCK 1, PART 2, NOTES Sections 22.2 to 22.5 Theelectric fΦExthrough a surface is related to the electricfield lines. Although we can draw as many electric field lines as we want to represent the electric field, the electric flux has a definite value. → Special case 1.If E is uniform (that is, constant in both magnitudflat surface of areaA is → N·m 2 perpendicular to E as shown in Fig. 2Φ.E= EA (in C ). Special case 2.If E is uniform over aflat surface of areaA, with an angleφ shΦwn=in Fig. 22.6b, then E EA⊥,whereA⊥= A cos φ. hatisΦE = EA cos φ . Using E cos ⊥,Φ E = ⊥ A . Also, using EA cos φ = E ·Φ = E·A . These are Eqs. (22.1), (22.2), and (22.3). E → That area vectorA has a magnitudeA (in m ) and a direction normal (that is, perpendicular)to the surface. → → For closed surfaces, A →nd dA are always outwardly normal to t→e su→face (SKILL 1). E⊥is the component ofE perpendicular to the surface (and also parallel toA or dA ). → → → → φ is the angle between the directions E and A or E and dA . + if 90˚ > φ ≥ 0 (electric field out) From ΦE = EA cosφ, the electrΦcEfcan b 0 if φ = 90˚ (SKILL 2) . – if 180˚ ≥ φ > 90˚ (electric fieldi n) In general,E and/orφ may not be constant over tΦeE= E cos φ dA gives Eq. (22.5), which is → → the mathematicaldefinitionof electric flux(ΦEE ≡ ∫E cos φ dA⊥= ∫E dA = ∫E·dA . Cover up the solutions and carefully work Examples 22.1, 22.2, and 22.3. 1 |q| In Section 22.3,our text uses E = to find the non-integral form of Gauss’s law (TERM 2) (in 4πε0r2 Qencl vacuum ≈ airΦ:E = ε0 , as well as the three integral forms of Gauss’s law: → → Q encl Q encl Q encl ∫ E ·d A 0 and∫E cos φ dA =0 and ∫E⊥ dA =ε0 Qencl –12 C2 Note that all forms incluε0 “” and recall that the electri0 constantε = N·m24.× 10 The symbol∫means to integrateover a closed surface, so the ΦEein Gauss’s law is the net (that is, total) electric flux through that cloencls the net (that is, total) charge enclosed by that surface. A Gaussian surface is a closed mathematical surface (TERM 3). It does not have to coincide with any material surface. Page 1 of 4 Q + gives net electric flux out From Φ E = εenc, we see thatencl = 0 gives zero net electric f(SKILL 2) . 0 – gives net electric fluxi n Cover up the solution and carefully work Example 22.4. → → F0= q E0 tells us that E = 0 (in the material of a conductor with free charges at rest overall). Section 22.4 explains how thisE = 0 condition tells us that all excess charge is found on the surface of a solidconductor that has its free charges at rest overall (that is, under electrostatic conditions). |q| For uniformly distributed charges: the |charge| per volumeis ρ (rho), ρ V (in C/m ) , the |charge| |q| |q| per area isσ (sigma), σ A (in C/m ) , andthe |charge| per length isλ (lambda),Lλ (in C/m) ,where |q| is the absolute value of the charge (in C) spread uniformly over the e (in m ), area A (in m ), and/or length L (in m). Some Applications of Gauss’s law: 1. Finding electric fields from given symmetric charge distributions: a) Spherical symmetry:Given a sphere of uniform positive charge densityρ and radius R. 1) Use an integral form of Gauss’s law, for exam∫lE cos φ dA =Q encl. ε0 2) Choose a symmetric Gaussian surface: in this case, the surface of a concentric (same center) sphere of radiurs. → → 3) E is away from the enclosed positive charge andso is radially outward. Since dA is always outwardly normal → → → dA is also radially outward. Therefore, the two vs and dA are parallel at all points on theGaussian surface, so φ = 0 and cos φ = cos 0 = 1. Th∫sE cos φ dA =∫ E (1)dA. 4) E is constant by symmetry, so we can take it out of the i: E (1)dA = E∫ dA. 5) Then ∫dA = A = 4πr so that the left side of Gauss’s ∫aE cos φ dA) equals E4πr . Therefore, spherical surface E4πr = Q enclsolves toE = 1 Q enc(telling us that outside of a spherically symmetric charge distribution, the ε0 4πε0 r2 1 |q| electric field looks like4πε0 r2 —as if all the charge were concentrated at the sphere’s center). 6) Forr ≤ R (inside the sphere of charge), there is charge throughout theentire volume of the Gaussian surface (of 4 3 4 3 1 Q encl 1 ρ 3πr ρr radius r). Thus,encl= ρVencl= ρ 3r , so E = 4πε 0 r2 becomesE = 4πε 0 r2 or E = 3ε0 (r ≤ R ) . 7) For r ≥ R (outside the sphere of charge), there is charge only from r = 0 to r = R (the charge radius). Thus, 4 3 3 4 3 1 Q encl 1 ρ 3πR ρR Q encl= ρVencl= ρ3πR , so E = 4πε0 r2 becomesE = 4πε 0 r2 or E = 3ε0r (r ≥ R ) . The magnitude ofE has the shape of Fig. 22.22. Page 2 of 4 b) Cylindrical symmetry: Given a very long rod of positive charge of radius R, much larger length L, volume charge densityρ, and linear charge densityλ. Q encl 1) Use an integral form of Gauss’s law, for ∫xE cos φ dA =ε0 . 2) Choose a symmetric Gaussian surface: in this case, the surface of aaxis) cylinder of radius r and lengthl (l <<L). 3) Break the integral into thre∫E cos φ dA = ⌡E cos φ dA + ⌡E cos φ dA + ⌡E cos φ dA . left end side right end → → 4) Over both ends: E is away from the enclosed + charge and so is radially outward. dA is always outwardly normal and so is parallel to the axis of the coaxial cylinder. Thereforrs and dA are perpendicular , so φ = 90˚ and cos φ = cos 90˚ = 0. Th⌡E cos φ dA = 0 = ⌡E cos φ dA . (That is, there is no electric left end right end fluxΦ Ethrough the ends.) 5) Over the side: E is away from the enclosed + charge and so is radially outward. dA is always outwardly → → normal and so is also radially outward. Therefore,the two vectors E and dA are parallel, so φ = 0 and cos φ = cos 0 = 1. Thus, ⌡E cos φ dA =E(1)dA . side side 6) E is constant by symmetry, so we can take it out of the integral: EdA .A = E side side 7) Then ⌡dA = A side 2πrl. Thus, the left side of Gauss∫E cos φ dA ) equals 0 + E2πrl + 0 = E2πrl. side 8) Forr ≥ R (outside the cylindrical charge distrencl ρπR l or λl. 2 2 9) Thus,∫E cos φ dA = enclbecomes E2πrl =ρπR l or λl. Solving, E = or λ (r ≥ R ) . 0 ε0 2ε0r 2πε0r 10) Forr ≤ R (inside the cylindrical charge distrencl ρ times what?, soE = what? c) Flat symmetry: Given a very large flat horizontal conductor with its free charges at rest overall. It has a uniform negative surface charge denσ on its bottom surface and no other excess charge. o Q encl 1) Use an integral form of Gauss’s law, for ∫xE cos φ dA =ε0 . 2) Choose a symmetric Gaussian surface: in this case,the surface of a cylinder with its verticalaxis perpendicular to the bottom surface of the conductor. The top and part of theside of the Gaussian cylinder are in the materialof the conductor. The bottom of the Gaussian cylinder is below the flat conductor’s bottom sura.ce and has areA 3) Break the integral into thre∫E cos φ dA = ⌡E cos φ dA + ⌡E cos φ dA + ⌡E cos φ dA . top side bottom 4) In the material of a conductor with free charges at rest overall, E = 0, so ⌡E cos φ dA = ⌡0 cos φ dA = 0. top top (That is, there is no electΦEcthrough the top because there is no electric field there.) Page 3 of 4 5) Part of the side is in the conductor whereE = 0. The rest of the side is outside theconductor where E is toward → → the enclosed – charge, makingE parallel to the side.dA is always outwardly normal and so is perpendicularto the → → ⌠ side. Therefore,the twovectorsE and dA areperpendicularto one another, so cosφ = cos 90˚ =0 so ⌡E cos φ dA side ⌡ = 0 dA = 0. (That is, there is no elΦEtthrough the side either.) side → 6) Over the bottom of →he Gaussian surface:E is toward the enclosed negative chargeand so is toward the bottom surface of the conductor.dA is always outwardly normal and so is away from the bottomconductor.f the → → Therefore, the two vectorsE and dA are antiparallel (opposite) to one another , soφ = 180˚ and cos φ = cos 180˚ = –1. Thus, ⌡E cos φ dA = ⌡E(–1)dA . bottom bottom 7) E is constant by symmetry, so we can take it out of theE(–1)dA = –E ⌡dA . bottom bottom 8) Then ⌡dA = A. Thus, the left side of Gauss’∫E cos φ dA ) equals 0 + 0 + (–EA) = –EA. bottom Q encl –σA σ 9) Since encl –σA,o∫E cos φ dA =ε0 becomes –EA = ε0 , or E ε0 . 2. Finding a charge distribution from a known electric field: Figure 22.23c shows the cross section of a conductor (with its free chargesat rest overall).There is a cavityin the conductor that contains a chargeq that is insulated from the conductor. (The chargeq is shown as plus, butit could Q just as well be minus.) In the material of the conductor, E = 0, so G∫ E ·d Aaw=( encl) becomes ε0 0 = enc. Therefore,Q = 0 (andΦ = 0) for all Gaussian surfaces that are completelyin the materialof the 0 encl E conductor. Since Qencl= 0, the excess charcw on the cavity wall, (that is, the inner surface of the conductor) is qcw= –q and the excess charge is zero in the bulk of the material of the conbulk= 0). Therefore, any excess charge is only the conductor’s surfaces (the cavity wall and the conductor’s outer surface), that is, qtotal on conductcw + qouter surfaceILL 7) . For example, suppose that the total excess charge on the conductor is –9 nC and there is +6 nC on the cavity wall. (Thattotal on conductornC and cw= +6 nC.) First, the excess charge in the bulk of the conductor’s matbulk= 0).zeroq Second,qcw= –q tells us the insulated charge in the cavity must beq = –6 nC. Third,total on conductcw + qouter surfaceomes –9 nC = +6 nC + qouter surface the charge on the outer surface iouter surface15 nC. Try Conceptual Example 22.11, covering up its solution (“interior” = “bulk”). Page 4 of 4

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