Definitions & Examples d A a b L C 1 C 2 a b C 3 C ab 

Today… Energy stored in the electric field –as distinguished from electric potential energy of a charge located in an electric field. Definition of Capacitance Example Calculations Parallel Plate Capacitor Cylindrical Capacitor Combinations of Capacitors Capacitors in Parallel Capacitors in Series Text Reference: Chapter 25.2, 4 Examples: 25.2,3,5,6 and 7

Last time… Two ways to say same thing: E ( x, y, z ) and V ( x, y, z ) Potential energy = potential x test charge ( q ) Conductors in E -fields “become” equipotential surfaces/volumes – E -field always normal to surface of conductor

Electric Potential Energy In addition to discussing the energy of a "test" charge in a Coulomb field, we can speak of the electric potential energy of the field itself! Reasons? – Work had to be done to assemble the charges (from infinity) into their final positions. – This work is the potential energy of the field. – The potential energy of a system of N charges is defined to be the algebraic sum of the potential energy for every pair of charges. This theme continues in the course with E in capacitors and B in inductors. Let’s start with a couple of example calculations.

Electric Potential Energy Example 1: Nuclear Fission What is the potential energy of the two nuclei? = 200 MeV 235 U nucleus + n (92 protons) BaKr (56 p) (36 p) d = m Compare this to the typical energy released in a chemical reaction, ~10eV. By allowing the two fragments to fly apart, this potential energy  kinetic energy  heat  drives a turbine  generate electricity. (What holds the protons together in the nucleus to begin with? Gluons! The “strong force”  very short range, very very strong!) (The electrons are all “far” away)

Electric Potential Energy Example 2: What is the potential energy of this collection of charges? Step 1:Bring in +2 q from infinity. This costs nothing. Step 3:Bring in 2nd -q charge. It is attracted to the +2 q, but repelled from the other -q charge. The total work (all 3 charges) is -q-q -q-q +2 q d d Step 2:Bring in one -q charge. This is attracted! Work required is negative 

Electrical Potential Energy Example 2: What is the potential energy of this collection of charges? -q-q -q-q +2 q d d What is the significance of the minus sign? A negative amount of work was required to bring these charges from infinity to where they are now. (i.e., the attractive forces between the charges are larger than the repulsive ones).

Lecture 7, ACT 1 Consider the 3 collections of point charges shown below. –Which collection has the smallest potential energy? (a) (b) (c) d d d -Q-Q -Q-Q -Q-Q d d d -Q-Q +Q+Q +Q+Q d d -Q-Q +Q+Q +Q+Q We have to do positive work to assemble the charges in (a) since they all have the same charge and will naturally repel each other. In (b) and (c), it’s not clear whether we have to do positive or negative work since there are 2 attractive pairs and one repulsive pair. (a) (b) (c) U 0 (a) (b) (c)

Electric potential and potential energy Consider three charges as our “sources”: We want to find the electric potential that these charges give rise to or “source” at some arbitrary point, P Use superposition of V ( r ) from before –now all three charges “source ” V ( r ) throughout all space –evaluate this anywhere and at P in particular: Q1Q1 Q2Q2 Q3Q3 P r 1p r 2p r 3p

Electric potential and potential energy What about electric potential energy of an added fourth charge, q test ? Those three charges source V ( r ) The potential energy of an added test charge is just q test V (its position) Use V ( r ) from previous slide –multiply by q test q test Q1Q1 Q2Q2 Q3Q3 r 1p r 2p r 3p

Potential energy, potential and potential energy Potential energy stored in a static charge distribution –work we do to assemble the charges Electrostatic potential at any point in space –sources give rise to V ( r ) Electric potential energy of a charge in the presence of a set of source charges –potential energy of the test charge equals the potential from the sources times the test charge

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. The capacitance belongs only to the capacitor, independent of the charge and voltage.

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

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

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 ). Note that C ~ length; this will always be the case! 

Practical Application: Microphone (“condenser”) Sound waves incident  pressure oscillations  oscillating plate separation d  oscillating capacitance ( )  oscillating charge on plate  oscillating current in wire ( )  oscillating electrical signal d Moveable plateFixed plate Current sensor Battery See this in action at !

Example: Cylindrical Capacitor Calculate the capacitance: Assume + Q, - Q on surface of cylinders with potential difference V. a b L r

Recall: Cylindrical Symmetry Gaussian surface is cylinder of radius r and length L Cylinder has charge Q  Apply Gauss' Law: + + ErEr L ErEr Q

Example: Cylindrical Capacitor Calculate the capacitance: Assume + Q, - Q on surface of cylinders with potential difference V. If we assume that inner cylinder has + Q, then the potential V is positive if we take the zero of potential to be defined at r = b :  a b L r +Q+Q - Q- Q

Lecture 7, ACT 2 In each case below, a charge of + Q is placed on a solid spherical conductor and a charge of - Q is placed on a concentric conducting spherical shell. –Let V 1 be the potential difference between the spheres with ( a 1, b ). –Let V 2 be the potential difference between the spheres with ( a 2, b ). –What is the relationship between V 1 and V 2 ? (Hint – think about parallel plate capacitors.) What we have here are two spherical capacitors. Intuition: for parallel plate capacitors: V = ( Q / C ) = ( Qd )/( A  0 ). Therefore you might expect that V 1 > V 2 since ( b-a 1 ) > ( b-a 2 ). In fact this is the case as we can show directly from the definition of V ! (a) V 1 < V 2 (b) V 1 = V 2 (c) V 1 > V 2 a2a2 b +Q+Q -Q-Q a1a1 b +Q+Q -Q-Q

Capacitors in Parallel Find “equivalent” capacitance C in the sense that no measurement at a, b could distinguish the above two situations. Aha! The voltage across the two is the same…. Parallel Combination: Equivalent Capacitor:   C 1 C 2 V a b Q2Q2 Q1Q1 -Q1-Q1 -Q2-Q2 C  V a b Q -Q-Q

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 –assume there is no net charge on node between C 1 and C 2 C ab  +Q+Q -Q-Q C 1 C 2 ab +Q+Q -Q-Q RHS: LHS:  -Q-Q +Q+Q

Examples: Combinations of Capacitors C 1 C 2 C 3 C 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., 

Lecture 7, ACT 3 What is the equivalent capacitance, C eq, of the combination shown? (a) C eq = (3/2) C (b) C eq = (2/3) C (c) C eq = 3 C o o C C C C eq 3A What is the relationship between V 0 and V in the systems shown below? conductor (Area A ) V +Q+Q -Q-Q d /3 (a) V = (2/3) V 0 (b) V = V 0 (c) V = (3/2) V 0 d (Area A ) V0V0 +Q+Q -Q-Q 3B

Lecture 7, ACT 3 What is the equivalent capacitance, C eq, of the combination shown? (a) C eq = (3/2) C (b) C eq = (2/3) C (c) C eq = 3 C o o C C C C eq 3A C C C C C1C1

Lecture 7, ACT 3 The electric field in the conductor = 0. The electric field everywhere else is : E = Q /( A  0 ) To find the potential difference, integrate the electric field: (a) V = (2/3) V 0 (b) V = V 0 (c) V = (3/2) V 0 What is the relationship between V 0 and V in the systems shown below? d (Area A ) V0V0 +Q+Q -Q-Q conductor (Area A ) V +Q+Q -Q-Q d /3 3B

Lecture 7, ACT 3, Another Way (a) V = (2/3) V 0 (b) V = V 0 (c) V = (3/2) V 0 What is the relationship between V 0 and V in the systems shown below? d (Area A ) V0V0 +Q+Q -Q-Q conductor (Area A ) V +Q+Q -Q-Q d /3 3B The arrangement on the right is equivalent to capacitors (each with separation = d /3) in SERIES!! conductor d /3 (Area A ) V +Q+Q -Q-Q d /3 +Q+Q -Q-Q 

Summary A Capacitor is an object with two spatially separated conducting surfaces. The definition of the capacitance of such an object is: The capacitance depends on the geometry : d A Parallel Plates a b L r +Q+Q -Q-Q Cylindrical a b +Q+Q -Q-Q Spherical

Next time: Energy stored in a capacitor Dielectrics Reading assignment: Ch Examples: 25.4,5,6,7,8,9 and 10

Appendix: Another example Suppose we have 4 concentric cylinders of radii a, b, c, d and charges + Q, - Q, + Q, - Q Question: What is the capacitance between a and d ? Note: E -field between b and c is zero! WHY?? A cylinder of radius r 1 : b < r 1 < c encloses zero charge! -Q-Q +Q+Q -Q-Q +Q+Q a b c d Note: This is just the result for 2 cylindrical capacitors in series! 

Conductors versus Insulators Charges move to Charges cannot cancel out electricmove at all in the conductor E=0  equipotentialCharge distribution surfaceon insulator unaffected by external fields All charge on surface

What does grounding do? 1. Acts as an “infinite” source or sink of charge. 2. The charges arrange themselves in such a way as to minimize the global energy (e.g., E  0 at infinity, V  0 at infinity). 3. Typically we assign V = 0 to ground.