Physics 1502: Lecture 5 Today’s Agenda Announcements: –Lectures posted on: –HW assignments, solutions etc. Homework #2:Homework #2: –On Masterphysics today: due next Friday –Go to masteringphysics.com and register –Course ID: MPCOTE33308 Labs: Begin next week

Today’s Topic : Chapter 22: Electric potential –Definition –How to compute it: E from V »Example: dipole –Equipotentials and Conductors –Electric Potential Energy » of Charge in External Electric Field

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

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 5, ACT 1 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

a b c uncharged conductor sphere with charge Q Calculating Electric Potentials Calculate the potential V(r) at the point shown (r<a) Where do we know the potential, and where do we need to know it? Determine E(r) for all regions in between these two points Determine  V for each region by integration Check the sign of each potential difference  V  V > 0 means we went “uphill”  V < 0 means we went “downhill” I II III IV r... and so on... (from the point of view of a positive charge) V=0 at r= ... we need r<a...

E from V? We can obtain the electric field E from the potential V by inverting our previous relation between E and V: Expressed as a vector, E is the negative gradient of V Cartesian coordinates: Spherical coordinates:

E from V: an Example What electric field does this describe?... expressing this as a vector: Something for you to try: Can you use the dipole potential to obtain the dipole field? Try it in spherical coordinates... you should get: Consider the following electric potential:

Electric Dipole z a a  +q -q r r 1 r 2 Last time, we derived for r>>a: Calculate E in spherical coordinates: the dipole moment 

The Bottom Line If we know the electric field E everywhere, allows us to calculate the potential function V everywhere (define V A = 0 above) If we know the potential function V everywhere, allows us to calculate the electric field E everywhere. Units for Potential! 1 Joule/Coulomb = 1 VOLT

Lecture 5, ACT 2 Consider the dipole shown at the right. –Fix r = r 0 >> a –Define  max such that the polar component of the electric field has its maximum value (for r=r 0 ). z a a  +q -q r r 1 r 2 Remember 1 What is  max ? (a)  max = 0 (b)  max = 45  (c)  max = 90 

Dipole Field z a a  +q -q r

Equipotentials GENERAL PROPERTY: –The Electric Field is always perpendicular to an Equipotential Surface. Why?? Dipole Equipotentials Defined as: The locus of points with the same potential. Example: for a point charge, the equipotentials are spheres centered on the charge. The gradient (  ) says E is in the direction of max rate of change. Along the surface, there is NO change in V (it’s an equipotential!) So, there is NO E component along the surface either… E must therefore be normal to surface

Claim The surface of a conductor is always an equipotential surface (in fact, the entire conductor is an equipotential) Why?? If surface were not equipotential, there would be an Electric Field component parallel to the surface and the charges would move!! Note Positive charges move from regions of higher potential to lower potential (move from high potential energy to lower PE). Equilibrium means charges rearrange so potentials equal. Conductors

Charge on Conductors? How is charge distributed on the surface of a conductor? –KEY: Must produce E=0 inside the conductor and E normal to the surface. Spherical example (with little off-center charge): E=0 inside conducting shell charge density induced on inner surface non-uniform charge density induced on outer surface uniform E outside has spherical symmetry centered on spherical conducting shell. +q

A Point Charge Near Conducting Plane + a q V=0

A Point Charge Near Conducting Plane + - a q The magnitude of the force is The test charge is attracted to a conducting plane Image Charge

Charge on Conductor How is the charge distributed on a non-spherical conductor?? Claim largest charge density at smallest radius of curvature. 2 spheres, connected by a wire, “far” apart Both at same potential Smaller sphere has the larger surface charge density ! But: 

Equipotential Example Field lines more closely spaced near end with most curvature. Field lines  to surface near the surface (since surface is equipotential). Equipotentials have similar shape as surface near the surface. Equipotentials will look more circular (spherical) at large r.

Electric Potential Energy The Coulomb force is a CONSERVATIVE force (i.e. the work done by it on a particle which moves around a closed path returning to its initial position is ZERO.) Therefore, a particle moving under the influence of the Coulomb force is said to have an electric potential energy defined by: The total energy (kinetic + electric potential) is then conserved for a charged particle moving under the influence of the Coulomb force. this “q” is the ‘test charge” in other examples...

St Elmo’s Fire

Lightning

Energy Units MKS: U = QV  1 coulomb-volt = 1 joule for particles (e, p,...) 1 eV = 1.6x joules Accelerators Electrostatic: VandeGraaff electrons  100 keV ( 10 5 eV) Electromagnetic: Fermilab protons  1TeV ( eV)