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

Today… Calculate E from V Definition of Capacitance Example Calculations-Parallel Plate Capacitor Combinations of Capacitors Capacitors in Parallel Capacitors in Series Lightning! Appendices A. Calculate electric field of dipole from potential B. Cylindrical capacitor examples

Last time… If we know E ( x, y, z ) we can calculate V ( x, y, z ): Potential energy of a test charge in external field: potential x test charge ( q ) Conductors in E -fields “become” equipotential surfaces/volumes – E -field always normal to surface of conductor

We can obtain the electric field E from the potential V by inverting our previous relation between E and V : V V +d V Expressed as a vector, E is the negative gradient of V Cartesian coordinates: Spherical coordinates: E from V ? Note: The units of E [= N/C] can be expressed [V/m].

This graph shows the electric potential at various points along the x-axis. 2) At which point(s) is the electric field zero? A B C D Preflight 7:

Consider the following electric potential: 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 (see Appendix A): E from V : an Example

The electric potential in a region of space is given by The x -component of the electric field E x at x = 2 is (a)E x = 0 (b)E x > 0(c)E x < 0 1 Lecture 7, ACT 1

The electric potential in a region of space is given by The x -component of the electric field E x at x = 2 is (a)E x = 0 (b)E x > 0(c)E x < 0 We know V (x) “everywhere” To obtain E x “everywhere”, use 1 Lecture 7, ACT 1

allows us to calculate the potential function V everywhere (keep in mind, we often define V A = 0 at some convenient place)  If we know the electric field E everywhere, allows us to calculate the electric field E everywhere If we know the potential function V everywhere, Units for Potential! 1 Joule/Coul = 1 VOLT The Bottom Line

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. [The unit of capacitance is the Farad: 1 F = 1C/V]

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 (see Appendix B for other examples) 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 !

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.) (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

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

Two identical parallel plate capacitors are shown in an end- view in A) of the figure. Each has a capacitance of C. 4) If the two are joined together as shown in B), forming a single capacitor, what will be the final capacitance? a) C/2 b) C c) 2C Preflight 7:

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., 

6) Which of these configurations has the lowest overall capacitance? Three configurations are constructed using identical capacitors a) Configuration A b) Configuration B c) Configuration C C C C C C Configuration A Configuration B Configuration C Preflight 7:

A circuit consists of three unequal capacitors C 1, C 2, and C 3 which are connected to a battery of emf  The capacitors obtain charges Q 1 Q 2, Q 3, and have voltages across their plates V 1, V 2, and V 3. C eq is the equivalent capacitance of the circuit. 8) Check all of the following that apply: a) Q 1 = Q 2 b) Q 2 = Q 3 c) V 2 = V 3 d) E = V 1 e) V 1 C 1 Preflight 7:

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

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 C C C C C1C1

Lightning (a.k.a. the atmosphere is a BIG capacitor!!) Factoids: _ + _ _ + + Collisions produce charged particles. The heavier particles (-) sit near the bottom of the cloud; the lighter particles (+) near the top. Stepped Leader Negatively charged electrons begin zigzagging downward. Attraction As the stepped leader nears the ground, it draws a streamer of positive charge upward. Flowing Charge As the leader and the streamer come together, powerful electric current begins flowing Contact! Intense wave of positive charge, a “return stroke,” travels upward at 10 8 m/s

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

Appendix A: Electric Dipole Now use (in spherical coordinates) to find the electric field everywhere.  the dipole moment z a a  +q+q -q-q r r 1 r 2 In Lecture 5 (appendix), we found the potential of an electric dipole to be:

Appendix A: Dipole Field z a a  +q+q -q-q r

Appendix A: Sample Problem 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 -q-q r r 1 r 2 (a)  max = 0 (b)  max = 45  (c)  max = 90  What is  max ? The expression for the electric field of a dipole ( r >> a ) is: The polar component of E is maximum when sin  is maximum. Therefore, E  has its maximum value when  = 90 .

Appendix B: Cylindrical Capacitor Examples Calculate the capacitance: Assume + Q, - Q on surface of cylinders with potential difference V. a b L r Gaussian surface is cylinder of radius r (a < r < b) and length L  Apply Gauss' Law: 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 : 

Appendix B: 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! 