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## Chapter 22

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**Flux**• Number of objects passing through a surface**Electric Flux, F**• is proportional to the number of electric field lines passing through a surface • Assumes that the surface is perpendicular to the lines • If not, then we use a cosine of the angle between them to get the components that are parallel • Mathematically:**Simple Cases**A A A q F=EAcosq F=EA E cosq F=0 E E E**From S to**• A S represents a sum over a large a collection of objects • Integration is also a sum over a collection of infinitesimally small objects, in our case, small areas, dA • So**Gauss’s Law**• The field lines emitted by a charge are proportional to the size of the charge. • Therefore, the electric field must be proportional to the size of the charge • In order to count the field lines, we must enclose the charges in some geometrical surface (one that we choose)**Mathematically**Charge enclosed within bounding limits of this closed surface integral**3 Shapes**• Sphere • Cylinder • Pillbox**Sphere**• When to use: around spherical objects (duh!) and point charges • Hey! What if an object is not one of these objects? • Closed surface integral yields: • r is the radius of the geometrical object that you are creating**Sphere Example**What if you had a sphere of radius, b, which contained a material whose charge density depend on the radius, for example, r=Ar2where A is a constant with appropriate units? At r=b, both of these expressions should be equal**Cylinder**• When to use: around cylindrical objects and line charges • Closed surface integral yields: • r is the radius of the geometrical object that you are creating and L is the length of the cylinder L**Cylinder Example**What if you had an infinitely long line of charge with a linear charge density, l?**Pillbox**• When to use: around flat surfaces and sheets of charge • Closed surface integral yields: • A is the area of the pillbox**Charge Isolated Conductor in Electrostatic Equilibrium**• If excess charge is placed on an isolated conductor, the charge resides on the surface. Why? • If there is an E-field inside the conductor then it would exert forces on the free electrons which would then be in motion. This is NOT electrostatic. • Therefore, if there is no E-field inside, then, by Gauss’s Law, the charge enclosed inside must be zero • If the charges are not on the outside, you are only left with the surface • A caveat to this is that E-field lines must be perpendicular to the surface else free charges would move.**A**E +++++++++++++++++++++++++++++++++++++++++++++ E Electric field on an infinitely large sheet of charge**Electric field on a conducting sheet**A E +++++++++++++++++++++++++++++++++++++++++++++ So a conductor has 2x the electric field strength as the infinite sheet of charge**A differential view of Gauss’s Law**• Recall the Divergence of a field of vectors How much the vector diverges around a given point Div=+large Div=0**Divergence Theorem (aka Gauss’s Thm or Green’s Thm)**Suspiciously like LHS of Gauss’s Law A place of high divergence is like a faucet Bounded surface of some region Sum of the faucets in a volume = Sum of the water going thru the surface**Div(E)**So how the E-field spreads out from a point depends on the amount of charge density at that point