Lectures 3,4 (Ch. 22) Gauss’s law

1. Lectures 3,4 (Ch. 22)Gauss’s law 1.Flux of • 2. Gauss’s law • 3. Application • Finding of Flux • Finding of Charge • Conductors properties • Finding of E for • symmetric charge distributions • (sphere, cylinder, plane)

2. Flux of a uniform should cross the surface to produce a flux

3. Compare to a flux of water should cross the surface to produce a flux

4. For the closed surface is out of the surface

5. General definition of flux of d

6. Flux of E through the closed surface containing a point charge, q 1. Sphere with q placed in the center 2. Arbitrary surface with arbitrary q position

7. Flux of E through the closed surface containing an ensemble of charges Superposition principle: qN q1 q2 Charges outside the closed surface do not contribute to the flux through this surface For each solid angle flux in =-flux out or

8. Gauss’s Law may be used 1) to find a flux 2) Q enclosed 3) E for some symmetric charge distributions.

9. Find the flux through the closed surfaces: A,B, C, D. Figure 22.15

10. Find the flux through the indicated closed surfaces. Figure 22.16

11. Find the flux through each side. q

12. Find the charge inside E0=500N/C EL=400N/C Q=? =10m

13. 1.E=0 within conductor in electrostatics (otherwise F=qE and hence the charges would be moving, i.e. nonstatic) 2.q =0 in every point inside solid conductor. It follows from the Gauss’s law, indeed, since E=0 one has Conductors properties q=0 Shrinking a Gaussian surface to any point inside a conductor one gets

14. Conductors properties 3. E near the surface Is the surface charge density, i.e. a charge per unit area

16. Conductors properties 4.In the absence of external charge placed into the cavity a) inside a cavity E=0 b) there is no charge on the inner surface of the conductor (“Scooping out theorem” in electrically neutral material)

17. Faraday’s cage Insert uncharged conductor into E - + E=0, qinner=0 + - ☺ Is it safe to be in a car during the thunderstorm?

18. Conductors properties • 5. In the presence of external charge q inside the cavity • Charge –q is induced on the inner surface of the cavity. • It follows from the Gauss’s law. Since E=0 inside a conductor one gets: Hence b) Charge resides on the outer surface of a cavity. It follows from the charge conservation law:

20. Faraday’s ice-pail experiment

21. Calculation of E using the Gauss’s lawThree types of the symmetry for a charge distribution • Spherical A=4 π r2, V=4 π r3/3 • Cylindrical Aside=2 π rL, V= π r2L • Plane A=L1L2, V=AL3

22. Application of Gauss’s law • Look for the symmetry of charge distribution • Use the Gaussian surface of the same symmetry • Then you’ll get here volume charge density surface linear

23. q 2. r>R, • r<R, E=0 • (conductor)

24. Dielectrcic sphere Q 2. r>R,

25. L is the total length of the line Q

26. Q

28. Conducting spherical layer 1. Find a surface charge density on inner and outer sides. 2.Find E(r). Q a Q+q -q q q b

29. Dielectric spherical layer Q a q b

30. Earnshaw’s Theorem, 1842 • A collection of point charges cannot be maintained in a stable stationary equilibrium configuration solely by the electrostatic interaction of the charges. • Suppose a charge q is at stable equilibrium in p. P. • It means if it is moved away from this point in any direction there should be a restoring force (opposite to the direction of a displacement.) • It means that there is a flux of E through the nearby surface surrounding p. P. • According to Gauss’s theorem p. P should be occupied with a charge. • But we supposed that the charge was shifted from this point. P

31. J.J. Thomson’s “plum pudding” model of atom Such atom would be unstable! Sir J.J.Tomson (1856 -1940) Nobel prize for discovery of e,1906

32. Long conducting cylinder 2a 1.r<a E=0 (conductor) 2.r> a the same as for the line of charge Connection between the linear and surface charge density

33. Long dielectric cylinder 2a 1.r>a the same result as for conducting cylinder:

34. Conducting cylindrical shell with a line of charge NB: linear charge density on an inner surface of a shell is on an outer surface 1.r<a the same result as for a line in the center: b a 2.a<r<b E=0 (conductor) 3. r>b

35. Dielectric cylindrical shell with a line of charge b 1.r<a and 3.r>b the same results as for conducting cylindrical shell 2.a<r<b a

36. A conducting slab L2 >>a,L1>>a, r<< L2 ,r<<L1 z 1.|x|<a E=0 (conductor) 2.|x|>a 2EA=2σA/ε0→E= σ/ε0 σ =Q/2S, S= L1 L2 E=Q/2ε0S L1 -a y a L2 x

37. L2 >>a,L1>>a, r<< L2 ,r<<L1 A dielectric slab 1.|x|>a 2EA=σ*A/ε0→E= σ*/2ε0 NB: σ*=Q/S, S= L1 L2 E=Q/2ε0S the same as for conductor 2.|x|<a 2EA=ρA2x/ε0 ρ=Q/S2a= σ*/2a E=σ*x/2aε0 z L1 y -a 0 a L2 x

38. Conclusions 1. Inside of charged bodies Conductors E=0 Dielectrics E ~ r Independent on the symmetry of charge distribution 2.Outside of charged bodies E is the same for conductors and dielectrics. It depends on the symmetry of charge distribution: Sphere E ~ 1/r2 Cylinder E ~1/r (r<<L) Plane E=const (r<< L1 and L2) 3. For finite size bodies at r→∞ E=kQtotal/r2 4. E jumps on the surface of the charged conductor E changes continuously on the surface of dielectric