Physics 1202: Lecture 5 Today’s Agenda Announcements: –Lectures posted on: –HW assignments, solutions etc. Homework #2:Homework #2: –On Masterphysics today: due Friday next week –Go to masteringphysics.com Labs: Begin next week

Today’s Topic : End of Chapter 16: Electric potential –Equipotentials and Conductors –Electric Potential Energy » of Charge in External Electric Field Capacitors:Electrostatic energy –Definition and concept –Capacitors in parallel and in series –Energy stored –Dielectrics

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 By analogy with the electric field Defined using a test charge q 0 We define a potential V due to a charge q –Using potential energy of a charge q and a test charge q 0 

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...

E from V? We can obtain the electric field E from the potential V by inverting our previous relation between E and V: We found F In general true for all direction

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. 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

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

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.

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

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: Parallel Plate Capacitor d A Need Q: Need V:recall where  x = d or so

Recall:Two infinite planes Same charge but opposite Fields of both planes cancel out outside They add up inside Perfect to store energy !

Example: 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 5 – ACT 1 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

Can we define the capacitance of a single isolated sphere ? The sphere has the ability to store a certain amount of charge at a given voltage (versus V=0 at infinity) Example : Isolated Sphere Need  V:V  = 0 V R = k e Q/R So, C = R/k e

Capacitors in Parallel Find “equivalent” capacitance C in the sense that no measurement at a,b could distinguish the above two situations. V a b Q2Q2 Q1Q1  V a b Q Parallel Combination: Equivalent Capacitor:  Total charge: Q = Q 1 + Q 2  C = C 1 + C 2

Capacitors in Series Find “equivalent” capacitance C in the sense that no measurement at a,b could distinguish the above two situations. The charge on C 1 must be the same as the charge on C 2 since applying a potential difference across ab cannot produce a net charge on the inner plates of C 1 and C 2. ab  +Q-Q ab +Q-Q RHS: LHS: 

Examples: Combinations of Capacitors a b ab  How do we start?? Recognize C 3 is in series with the parallel combination on C 1 and C 2. i.e. 

Energy of a Capacitor How much energy is stored in a charged capacitor? –Calculate the work provided (usually by a battery) to charge a capacitor to +/- Q: Calculate incremental work  W needed to add charge  q to capacitor at voltage V: - + But  W is also the change in potential energy  U Q q qq VqVq V q =q/C V The total U to charge to Q is shaded triangle: In terms of the voltage V:

Lecture 5 – ACT 2 d A d1d1 A The same capacitor as last time. The capacitor is charged to Q and then the battery disconnected. Then I pull the plates further apart so that the final separation is d 1. d 1 > d If the initial energy is U 0 and the final capacitance is U 1, is … A) U 1 > U 0 B) U 1 = U 0 C) U 1 < U 0

Summary 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. How do the quantities Q, W, C, V, E change? How much do these quantities change?.. exercise for student!! Q: W: C: V: E: remains the same.. no way for charge to leave. increases.. add energy to system by separating decreases.. since energy , but Q remains same increases.. since C , but Q remains same remains the same.. depends only on chg density answers:

Where is the Energy Stored? Claim: energy is stored in the Electric field itself. Think of the energy needed to charge the capacitor as being the energy needed to create the field. The Electric field is given by: The energy density u in the field is given by: To calculate the energy density in the field, first consider the constant field generated by a parallel plate capacitor: Units: J/m 3 

Dielectrics Empirical observation: Inserting a non-conducting material between the plates of a capacitor changes the VALUE of the capacitance. Definition: The dielectric constant of a material is the ratio of the capacitance when filled with the dielectric to that without it. i.e. –  values are always > 1 (e.g., glass = 5.6; water = 78) –They INCREASE the capacitance of a capacitor (generally good, since it is hard to make “big” capacitors –They permit more energy to be stored on a given capacitor than otherwise with vacuum (i.e., air)

Parallel Plate Example Charge a parallel plate capacitor filled with vacuum (air) to potential difference V 0. An amount of charge Q = C V 0 is deposited on each plate Now insert material with dielectric constant . –Charge Q remains constant –So…, C =  C 0 Voltage decreases from V 0 to Electric field decreases also: