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##### 21. Gauss’s Law

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**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. 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**21.2. Electric Flux & Field**8 lines out of surfaces 1, 2, & 3. But 88 = 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. 88 = 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**Outer sphere has 4 times area. But E is 4 times weaker. So is the same For a point charge: E r2 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. 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 = qq = 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 +2C 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 • 0 • 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 1016 .**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 = /20. 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