Physics 1502: Lecture 4 Today’s Agenda Announcements: –Lectures posted on: –HW assignments, solutions etc. Homework #1:Homework #1: –On Masterphysics today: due next Friday –Go to masteringphysics.com and register –Course ID: MPCOTE33308 Labs: Begin next week

Today’s Topic : End of Chapter 21: Gauss’s Law –Motivation & Definition –Coulomb's Law as a consequence of Gauss' Law –Charges on Insulators: »Where are they? Chapter 22: Electric potential –Definition –How to compute it

Karl Friedrich Gauss ( )

x y Infinite Line of Charge Symmetry  E field must be  to line and can only depend on distance from line NOTE: we have obtained here the same result as we did last lecture using Coulomb’s Law. The symmetry makes today’s derivation easier! ErEr Therefore, CHOOSE Gaussian surface to be a cylinder of radius r and length h aligned with the x-axis. Apply Gauss' Law: AND q = h h ErEr  SdE On the ends, EdSrhE    2  On the barrel, 

Gauss’ Law: Help for the Problems How to do practically all of the homework problems What Can You Do With This?? If you have (a) spherical, (b) cylindrical, or (c) planar symmetry AND: If you know the charge (RHS), you can calculate the electric field (LHS) If you know the field (LHS: usually because E=0 inside conductor), you can calculate the charge (RHS). Gauss’ Law is ALWAYS VALID!!

Application of Gauss’ Law: RHS: q = ALL charge inside cylinder=  A Planar Symmetry: Gaussian surface = Cylinder of area A RHS: q = ALL charge inside radius r Spherical Symmetry: Gaussian surface = Sphere of radius r RHS: q = ALL charge inside radius r, length L Cylindrical Symmetry: Gaussian surface = Cylinder of radius r LHS:

Insulators vs. Conductors Insulators – wood, rubber, styrofoam, most ceramics, etc. Conductors – copper, gold, exotic ceramics, etc. Sometimes just called metals Insulators – charges cannot move. –Will usually be evenly spread throughout object Conductors – charges free to move. –on isolated conductors all charges move to surface.

E E Conductors vs. Insulators E E E in = 0 E in < E

Hollow conductors

Conductors & Insulators How do the charges move in a conductor?? Hollow conducting sphere Charge the inside, all of this charge moves to the outside. Consider how charge is carried on macroscopic objects. We will make the simplifying assumption that there are only two kinds of objects in the world: Insulators.. In these materials, once they are charged, the charges ARE NOT FREE TO MOVE. Plastics, glass, and other “bad conductors of electricity” are good examples of insulators. Conductors.. In these materials, the charges ARE FREE TO MOVE. Metals are good examples of conductors.

Conductors vs. Insulators

Charges on a Conductor Why do the charges always move to the surface of a conductor ? –Gauss’ Law tells us!! –E = 0 inside a conductor when in equilibrium (electrostatics) ! »Why? If E  0, then charges would have forces on them and they would move ! Therefore from Gauss' Law, the charge on a conductor must only reside on the surface(s) ! Infinite conducting plane Conducting sphere

Lecture 4, ACT 1 Consider the following two topologies: 1A Compare the electric field at point X in cases A and B: (a)  1 <  (b)  1 =  (c)  1 >  (a) E A < E B (b) E A = E B (c) E A > E B 1B What is the surface charge density  1 on the inner surface of the conducting shell in case A? E 22 A) A solid non-conducting sphere carries a total charge Q = -3  C distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. B) Same as (A) but conducting shell removed. 11 -Q

Lecture 4, ACT 2 A line charge (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius r i = a, and outer radius r o = b as shown. –What is the value of the charge density  o (C/m 2 ) on the outer surface of the cylinder? (a) (b) (c) a b   

Electric Potential q A C B r A B r path independence equipotentials R R Rr V Q 4   r Q 4   R

Overview Introduce Concept of Electric Potential –Is it well-defined? i.e. is Electric Potential a property of the space as is the Electric Field? Calculating Electric Potentials –Charged Spherical Shell –N point charges –Electric Dipole Can we determine the Electric Field if we know the Electric Potential? Text Reference: Chapter 22

Electric Potential Suppose charge q 0 is moved from pt A to pt B through a region of space described by electric field E. Since there will be a force on the charge due to E, a certain amount of work W AB will have to be done to accomplish this task. We define the electric potential difference as: Is this a good definition? Is V B - V A independent of q 0 ? Is V B - V A independent of path? A B q 0 E

Independent of Charge? To move a charge in an E field, we must supply a force just equal and opposite to that experienced by the charge due to the E field. A B q 0 E F elec F we supply = -F elec 

Lecture 4, ACT 3 A single charge ( Q = -1  C) is fixed at the origin. Define point A at x = + 5m and point B at x = +2m. –What is the sign of the potential difference between A and B? (V AB  V B - V A ) (a) V AB <  (b) V AB =  (c) V AB >  x -1  C   A B

Independent of Path? A B q 0 E F elec -F elec This equation also serves as the definition for the potential difference V B - V A. The integral is the sum of the tangential (to the path) component of the electric field along a path from A to B. The question now is: Does this integral depend upon the exact path chosen to move from A to B? If it does, we have a lousy definition. Hopefully, it doesn’t. It doesn’t. But, don’t take our word, see appendix and following example.

Does it really work? Consider case of constant field: –Direct: A - B Long way round: A - C - B So here we have at least one example of a case in which the integral is the same for BOTH paths. A C B Eh r  dl

Electric Potential Define the electric potential of a point in space as the potential difference between that point and a reference point. a good reference point is infinity... we typically set V  = 0 the electric potential is then defined as: for a point charge, the formula is:

Potential from charged spherical shell E Fields (from Gauss' Law) Potentials r > R: r < R: R R Q 4   R Rr V Q 4   r r < R: r > R:

Potential from N charges The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately. x r1r1 r2r2 r3r3 q1q1 q3q3 q2q2 

Electric Dipole z a a  +q -q r r 1 r 2 The potential is much easier to calculate than the field since it is an algebraic sum of 2 scalar terms. Rewrite this for special case r>>a:  Can we use this potential somehow to calculate the E field of a dipole? (remember how messy the direct calculation was?)  r 2 -r 1

Appendix: Independent of Path? q A C B r A B r E We want to evaluate potential difference from A to B q A B r A B r E What path should we choose to evaluate the integral?. If we choose straight line, the integral is difficult to evaluate. Magnitude different at each pt along line. Angle between E and path is different at each pt along line. If we choose path ACB as shown, our calculation is much easier! From A to C, E is perpendicular to the path. ie

Appendix: Independent of Path? Evaluate potential difference from A to B along path ACB. q A C B r A B r E Evaluate the integral: by definition:

Appendix: Independent of Path? How general is this result? Consider the approximation to the straight path from A->B (white arrow) = 2 arcs (radii = r 1 and r 2 ) plus the 3 connecting radial pieces. q A B This is the same result as above!! The straight line path is better approximated by Increasing the number of arcs and radial pieces. q A B r A B r C r1r1 r2r2 For the 2 arcs + 3 radials path:

Appendix: Independent of Path? Consider any path from A to B as being made up of a succession of arc plus radial parts as above. The work along the arcs will always be 0, leaving just the sum of the radial parts. All inner sums will cancel, leaving just the initial and final radii as above.. Therefore it's general! q A B r1r1 r2r2