Physics 1502: Lecture 6 Today’s Agenda Announcements: –Lectures posted on: –HW assignments, solutions etc. Homework #2:Homework #2: –On Masterphysics today: due Friday –Go to masteringphysics.com Labs: Begin this week

Today’s Topic : End of Chapter 22: Electric potential –Equipotentials and Conductors –Electric Potential Energy » of Charge in External Electric Field Chapter 23: Electrostatic energy –Definition and concept

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 We define the electric potential difference as: independent of q 0 independent of path We 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:

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 6, ACT 1 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 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

Van de Graaff Generator Schematic view of a classical Van de Graaff generator. 1) hollow metal sphere 2) upper electrode 3) upper roller (for example in acrylic glass) 4) side of the belt with positive charges 5) opposite side of the belt with negative charges 6) lower roller (metal) 7) lower electrode (ground) 8) spherical device with negative charges, used to discharge the main sphere 9) spark produced by the difference of potacrylic glass

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)

Definitions & Examples d A a b L a b ab 

Definition of Capacitance Example Calculations (1) Parallel Plate Capacitor (2) Cylindrical Capacitor (3) Isolated Sphere Energy stored in capacitors Dielectrics Capacitors in Circuits Overview Text Reference: Chapter 23

Capacitance A capacitor is a device whose purpose is to store electrical energy which can then be released in a controlled manner during a short period of time. A capacitor consists of 2 spatially separated conductors which can be charged to +Q and -Q respectively. The capacitance is defined as the ratio of the charge on one conductor of the capacitor to the potential difference between the conductors. Is this a "good" definition? Does the capacitance belong only to the capacitor, independent of the charge and voltage ? + -

Calculate the capacitance. We assume + , -  charge densities on each plate with potential difference V: Example 1: Parallel Plate Capacitor d A Need Q: Need V:from defn: –Use Gauss’ Law to find E

Gaussian surface encloses non-zero net charge Recall: Two Infinite Sheets (into screen) E=0 E   Field outside the sheets is zero Field inside sheets is not zero: Gaussian surface encloses zero net charge A A

Example 1: Parallel Plate Capacitor d A Calculate the capacitance: Assume +Q,-Q on plates with potential difference V. As hoped for, the capacitance of this capacitor depends only on its geometry (A,d). 

Dimensions of capacitance C = Q/V => [C] = F(arad) = C/V = [Q/V] Example: Two plates, A = 10cm x 10cm d = 1cm apart => C = A  0 /d = = 0.01m 2 /0.01m * 8.852e-12 C 2 /Jm = 8.852e-12 F

Lecture 6 - ACT 2 d A Suppose the capacitor shown here is charged to Q and then the battery disconnected. Now suppose I pull the plates further apart so that the final separation is d 1. d 1 > d d1d1 A If the initial capacitance is C 0 and the final capacitance is C 1, is … A) C 1 > C 0 B) C 1 = C 0 C) C 1 < C 0