Gauss’sLaw 1 P05 - The first Maxwell Equation A very useful computational technique This is important!

Review: field line diagrams

Gauss’s Law – The Idea The total “flux” of field lines penetrating any of these surfaces is the same and depends only on the amount of charge inside 3 P05 -

Gauss’s Law – The Equation 00 closed surface S in E q E  dA E  dA     Electric flux  E (the surface integral of E over closed surface S) is proportional to charge inside the volume enclosed by S P  0  8.85 x 10 −12 C 2 /(Nm 2 )

Electric Flux  E Case I:E is constant vector field perpendicular to planar surface S of area A  E   E  dA  E   EA P Our Goal:Always reduce problem to this

Electric Flux  E Case II:E is constant vector field directed at angle  to planar surface S of area A  E   E  dA  E  EA cos  P05 - 6

PRS Question: Flux Thru Sheet P05 - 7

Gauss’sLaw 00 closed surface S in E q E  dA E  dA     Note:Integral must be over closed surface We will need the charge INSIDE P05 - 9

Open and Closed Surfaces A rectangle is an open surface — it does NOT contain a volume A sphere is a closed surface — it DOES contain a volume P

Area Element dA:Closed Surface For closed surface, dA is normal to surface and points outward ( from inside to outside)  E > 0 if E points out  E < 0 if E points in P

Electric Flux  E Case III:E not constant, surface curved EdEEdAE dEEdEEdAE dE P S dAdA

Example:Point Charge Open Surface P

Example:Point Charge Closed Surface P

PRS Question: Flux Thru Sphere P

Electric Flux:Sphere Point charge Q at center of sphere, radius r E field at surface:  P  0r4 0r E  Q r ˆ Electric flux through sphere: 4r24r2 0 S rˆ  dArˆrˆ  dArˆ Q E  dA E  dA    E   dA  dArˆdA  dArˆ 40r040r0 dA  Q 4  r 2  Q 2 2   S 40 r40 r  Q Q

Arbitrary Gaussian Surfaces 00 surface S closed E  dA  QE  dA  Q E    For all surfaces such as S 1, S 2 or S 3 P

Applying Gauss’s Law 1.Identify regions in which to calculate E field. 2.Choose Gaussian surfaces S:Symmetry 00 closed surfaceS in E E  dA E  dA   3.Calculate  E P S 4.Calculate q in, charge enclosed by surface S 5.Apply Gauss’s Law to calculate E: q    E  dA E  dA

Problem solving steps: (Lea Textbook Summary)

Choosing Gaussian Surface Choose surfaces where E is perpendicular & constant. Then flux is EA or -EA. Choose surfaces where E is parallel. Then flux is zero OR EA P  EA E Example: Uniform Field Flux is EA on top Flux is –EA on bottom Flux is zero on sides

Symmetry & Gaussian Surfaces P Use Gauss’s Law to calculate E field from highly symmetric sources SymmetryGaussian Surface SphericalConcentric Sphere CylindricalCoaxial Cylinder PlanarGaussian “Pillbox”

PRS Question: Should we use Gauss’ Law? P

Gauss: Spherical Symmetry +Q uniformly distributed throughout non-conducting solid sphere of radius a. Find E everywhere P

Gauss: Spherical Symmetry Symmetry is Spherical P  E  E r ˆ Use Gaussian Spheres

Gauss: Spherical Symmetry Region 1: r > a Draw Gaussian Sphere in Region 1 (r > a) Note:r is arbitrary but is the radius for which you will calculate the E field! P

Gauss: Spherical Symmetry Region 1: r > a Total charge enclosed q in = +Q S  E    E  dA  E   dA  EA   E 4 r 2  E 4 r 2   4  r 2 E  q in 00 E  Q Q  2 40r40r40r40r E  Q  E  Q r ˆ  P

Gauss: Spherical Symmetry Gauss’s law: q  3 Q  r Q a aa rr in Region 2: r < a Total charge enclosed: E 3 40a40a E  Q r  E  Q r r ˆ 40a40a   E  4  r 2   q in  r Q P OR q in  V V aa

PRS Question: Field Inside Spherical Shell P

Gauss: Cylindrical Symmetry Infinitely long rod with uniform charge density Find E outside the rod. P

Gauss: Cylindrical Symmetry Symmetry is Cylindrical  E  E r ˆ Use Gaussian Cylinder P Note:r is arbitrary but is the radius for which you will calculate the E field!  is arbitrary and should divide out l

Gauss: Cylindrical Symmetry   Total charge enclosed: q in  E  2  r    q in    0 S E  E  dA  E  dA  EAE  E  dA  E  dA  EA  20r20r E   E  r ˆ 20r20r  P

Gauss: Planar Symmetry Infinite slab with uniform charge density  Find E outside the plane P

Gauss: Planar Symmetry Symmetry is Planar  E   E x ˆ Use Gaussian Pillbox xˆxˆ Gaussian Pillbox P Note:A is arbitrary (its size and shape) and should divide out

Gauss: Planar Symmetry  A A Total charge enclosed: q in NOTE:No flux through side of cylinder, only endcaps  E  2 A   q in   A00  A00 S  E    E  dA  E   dA  EA Endcaps   E  E  x A  0 ˆ xto right -x ˆ to left  E 2 E 2   E  20E  20 P

PRS Question: Slab of Charge P

Group Problem:Charge Slab Infinite slab with uniform charge density  Thickness is 2d (from x=-d to x=d). Find E everywhere. xˆxˆ P

PRS Question: Slab of Charge P