21. Gauss’s Law Electric Field Lines Electric Flux & Field Gauss’s Law Using Gauss’s Law Fields of Arbitrary Charge Distributions Gauss’s Law & Conductors

Huge sparks jump to the operator’s cage, but the operator is unharmed Huge sparks jump to the operator’s cage, but the operator is unharmed. Why? Ans: Gauss’ law  E = 0 inside cage

21.1. Electric Field Lines Vector gives E at point Field line gives direction of E Spacing gives magnitude of E Electric field lines = Continuous lines whose tangent is everywhere // E. They begin at + charges & end at  charges or . Their density is  field strength or charge magnitude.

Field Lines of Electric Dipole Field is strong where lines are dense. Direction of net field tangent to field line

Field Lines

21.2. Electric Flux & Field 8 lines out of surfaces 1, 2, & 3. But 22 = 0 out of 4. 16 lines out of surfaces 1, 2, & 3. But 0 out of 4. 8 lines out of surfaces 1, 2, & 3. But 0 out of 4. 8 lines out of surfaces 1 & 2. 16 lines out of surface 3. 0 out of 4.  8 lines out of surface 1. 8 lines out of surface 2. 44 = 0 lines out of surface 3. 1: 4 2: 8 3: 4 4: 0 Count these. Number of field lines out of a closed surface  net charge enclosed.

Electric Flux A flat surface is represented by a vector where A = area of surface and Electric flux through flat surface A : [  ]  N m2 / C. E,  Open surface: can get from 1 side to the other w/o crossing surface. Direction of A ambiguous. A,  Closed surface: can’t get from 1 side to the other w/o crossing surface. A defined to point outward.

GOT IT? 21.1. The figure shows a cube of side s in a uniform electric field E. What is the flux through each of the cube faces A, B, and C with the cube oriented as in (a) ? Repeat for the orientation in (b), with the cube rotated 45 °.

21.3. Gauss’s Law Gauss’s law: The electric flux through any closed surface is proportional to the net charges enclosed.  depends on units. For point charge enclosed by a sphere centered on it: SI units  = vacuum permittivity Field of point charge: Gauss’s law:

Gauss & Coulomb For a point charge: E  r 2 A  r 2   indep of r. Outer sphere has 4 times area. But E is 4 times weaker. So  is the same For a point charge: E  r 2 A  r 2   indep of r. Principle of superposition  argument holds for all charge distributions Gauss’ & Colomb’s laws are both expression of the inverse square law. For a given set of field lines going out of / into a point charge, inverse square law  density of field lines  E in 3-D.

GOT IT? 21.2. A spherical surface surrounds an isolated positive charge, as shown. If a 2nd charge is placed outside the surface, which of the following will be true of the total flux though the surface? it doesn’t change; it increases; it decreases; it increases or decreases depending on the sign of the second charge. Repeat for the electric field on the surface at the point between the charges. Opposite charges Same charges

21.4. Using Gauss’s Law Useful only for symmetric charge distributions. Spherical symmetry: ( point of symmetry at origin ) 

Example 21.1. Uniformily Charged Sphere A charge Q is spreaded uniformily throughout a sphere of radius R. Find the electric field at all points, first inside and then outside the sphere.   True for arbitrary spherical (r).

Example 21.2. Hollow Spherical Shell A thin, hollow spherical shell of radius R contains a total charge of Q. distributed uniformly over its surface. Find the electric field both inside and outside the sphere. Reflection symmetry  E is radial.  Contributions from A & B cancel.

GOT IT? 21.3. A spherical shell carries charge Q uniformly distributed over its surface. If the charge on the shell doubles, what happens to the electric field inside and outside the shell? Stays 0. Doubles.

Example 21.3. Point Charge Within a Shell A positive point charge +q is at the center of a spherical shell of radius R carrying charge 2q, distributed uniformly over its surface. Find the field strength both inside and outside the shell. 

Tip: Symmetry Matters Spherical charge distribution inside a spherical shell is zero  E = 0 inside shell E  0 if either shell or distribution is not spherical. Q = qq = 0 But E  0 on or inside surface

Line Symmetry Line symmetry: r = perpendicular distance to the symm. axis. Distribution is independent of r//  it must extend to infinity along symm. axis. 

Example 21.4. Infinite Line of Charge Use Gauss’ law to find the electric field of an infinite line charge carrying charge density  in C/m. (radial field) No flux thru ends  c.f. Eg. 20.7  True outside arbitrary radial (r).

Example 21.5. A Hollow Pipe A thin-walled pipe 3.0 m long & 2.0 cm in radius carries a net charge q = 5.7 C distributed uniformly over its surface. Fine the electric field both 1.0 cm & 3.0 cm from the pipe axis, far from either end.  at r = 1.0 cm at r = 3.0 cm

Plane Symmetry Plane symmetry: r = perpendicular distance to the symm. plane. Distribution is independent of r//  it must extend to infinity in symm. plane. 

Example 21.6. A Sheet of Charge An infinite sheet of charge carries uniform surface charge density  in C/m2. Find the resulting electric field.   E > 0 if it points away from sheet.

21.5. Fields of Arbitrary Charge Distributions Dipole : E  r 3 Point charge : E  r 2 Line charge : E  r 1 Surface charge : E  const

Conceptual Example 21.1. Charged Disk Sketch some electric field lines for a uniformly charged disk, starting at the disk and extending out to several disk diameters. infinite-plane-charge-like point-charge-like

Making the Connection Suppose the disk is 1.0 cm in diameter& carries charge 20 nC spread uniformly over its surface. Find the electric field strength 1.0 mm from the disk surface and 1.0 m from the disk. (a) Close to disk : (b) Far from disk :

GOT IT? 21.4. A uniformly charged sheet measures 1m on each side, and you are told the total charge Q. What expression would you use to get approximate values for the field magnitude 1 cm from the sheet (but not near an edge) and 1 km from the sheet? (a) (b)

21.6. Gauss’s Law & Conductors Electrostatic Equilibrium Conductor = material with free charges E.g., free electrons in metals. Neutral conductor Uniform field External E  Polarization  Internal E Total E = 0 : Electrostatic equilibrium ( All charges stationary ) Induced polarization cancels field inside Microscopic view: replace above with averaged values. Net field

Charged Conductors Excess charges in conductor tend to keep away from each other  they stay at the surface.  = 0 thru this surface More rigorously: Gauss’ law with E = 0 inside conductor  qenclosed = 0  For a conductor in electrostatic equilibrium, all charges are on the surface.

Example 21.7. A Hollow Conductor An irregularly shaped conductor has a hollow cavity. The conductor itself carries a net charge of 1 C, and there’s a 2 C point charge inside the cavity. Find the net charge on the cavity wall & on the outer surface of the conductor, assuming electrostatic equilibrium. E = 0 inside conductor  = 0 through dotted surface qenclosed = 0 Net charge on the cavity wall qin = 2 C qout qin +2C Net charge in conductor = 1 C = qout + qin  charge on outer surface of the conductor qout = +3 C

GOT IT? 21.5. A conductor carries a net charge +Q. There is a hollow cavity inside the conductor that contains a point charge Q. In electrostatic equilibrium, is the charge on the outer surface of the conductor 2Q Q Q 2Q

Experimental Tests of Gauss’ Law Measuring charge on ball is equivalent to testing the inverse square law. The exponent 2 was found to be accurate to 1016 .

Field at a Conductor Surface At static equilibrium, E = 0 inside conductor, E = E at surface of conductor. Gauss’ law applied to pillbox surface:  The local character of E ~  is incidental. E always dependent on ALL the charges present.

Dilemma? E outside charged sheet of charge density  was found to be E just outside conductor of surface charge density  is What gives? Resolution: There’re 2 surfaces on the conductor plate. The surface charge density on either surface is . Each surface is a charge sheet giving E = /20. Fields inside the conductor cancel, while those outside reinforce. Hence, outside the conductor

Application: Shielding & Lightning Safety Coaxial cable Car hit by lightning, driver inside unharmed. Strictly speaking, Gauss law applies only to static E. However, e in metal can respond so quickly that high frequency EM field ( radio, TV, MW ) can also be blocked (skin effect).

Plate Capacitor inside outside Charge on inner surfaces only