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Announcements I need paperwork/notification for any special needs by the end of lectures today.  If possible, Test Preparation Homework for Exam 1 will be available on the Physics 2135 web site (under Handouts) late Friday (otherwise Sunday). It will be handed out in lecture next Monday. This is next Tuesday’s homework. Do not ignore it!  Test 1 Room Assignments (next slide) are available on the Physics 2135 web site (under Course Information). This information will be included with Test Preparation Homework 1.  Exam 1 is Tuesday, February 17, 5:00-6:00 pm.

Know the exam time! Find your room ahead of time! If at 5:00 on test day you are lost, go to 104 Physics and check the exam room schedule, then go to the appropriate room and take the exam there. Exam is from 5:00-6:00 pm!  Physics 2135 Test Rooms, Spring 2015: InstructorSectionsRoom Dr. KurterF, H104 Physics Dr. MadisonE, G199 Toomey Dr. ParrisK, M125 Butler-Carlton (Civil Eng.) Mr. UpshawA, C, J, LG-3 Schrenk Dr. WaddillB, DG-31 EECH (Electrical Eng.) Special AccommodationsTesting Center

More Announcements  Exam 1 special arrangements: Test Center students. The Test Center s you confirming your appointment. The Testing Center will tell you in that when your exam starts, and no one will be admitted after 5:15 pm, unless you have made other arrangements. If you have not received a confirming , you are NOT on the Test Center list! Other special cases: you should already have been in correspondence with me or your recitation instructor. Anybody else must let me know by the end of the 1:00 lecture today about special needs for the exam.

Today’s agenda: Capacitance. You must be able to apply the equation C=Q/V. Capacitors: parallel plate, cylindrical, spherical. You must be able to calculate the capacitance of capacitors having these geometries, and you must be able to use the equation C=Q/V to calculate parameters of capacitors. Circuits containing capacitors in series and parallel. You must understand the differences between, and be able to calculate the “equivalent capacitance” of, capacitors connected in series and parallel.

Capacitors and Dielectrics Capacitance A capacitor is basically two parallel conducting plates with air or insulating material in between. V0V0 V1V1 E L A capacitor doesn’t have to look like metal plates. Capacitor for use in high-performance audio systems.

When a capacitor is connected to an external potential, charges flow onto the plates and create a potential difference between the plates. Capacitor plates build up charge. The battery in this circuit has some voltage V. We haven’t discussed what that means yet. The symbol representing a capacitor in an electric circuit looks like parallel plates. Here’s the symbol for a battery, or an external potential V

If the external potential is disconnected, charges remain on the plates, so capacitors are good for storing charge (and energy). Capacitors are also very good at releasing their stored charge all at once. The capacitors in your tube-type TV are so good at storing energy that touching the two terminals at the same time can be fatal, even though the TV may not have been used for months. High-voltage TV capacitors are supposed to have “bleeder resistors” that drain the charge away after the circuit is turned off. I wouldn’t bet my life on it. Graphic from V conducting wires On-line “toy” here.here

assortment of capacitors

The magnitude of charge acquired by each plate of a capacitor is Q=CV where C is the capacitance of the capacitor. The unit of C is the farad but most capacitors have values of C ranging from picofarads to microfarads (pF to  F). micro  10 -6, nano  10 -9, pico  (Know for exam!) C is always positive. +Q + -Q - V C Here’s this V thing again. It is the potential difference provided by the “external potential.” For example, the voltage of a battery. V is really a  V. V is really  V.

Today’s agenda: Capacitance. You must be able to apply the equation C=Q/V. Capacitors: parallel plate, cylindrical, spherical. You must be able to calculate the capacitance of capacitors having these geometries, and you must be able to use the equation C=Q/V to calculate parameters of capacitors. Circuits containing capacitors in series and parallel. You must understand the differences between, and be able to calculate the “equivalent capacitance” of, capacitors connected in series and parallel.

Parallel Plate Capacitance V0V0 V1V1 E d We previously calculated the electric field between two parallel charged plates: This is valid when the separation is small compared with the plate dimensions. We also showed that E and  V are related: +Q-Q A This lets us calculate C for a parallel plate capacitor.

Reminders: Q is the magnitude of the charge on either plate. V is actually the magnitude of the potential difference between the plates. V is really |  V|. Your book calls it V ab. C is always positive.

V0V0 V1V1 E d +Q-Q A Parallel plate capacitance depends “only” on geometry. This expression is approximate, and must be modified if the plates are small, or separated by a medium other than a vacuum (lecture 9). Greek letter Kappa. For today’s lecture (and for exam 1), use Kappa=1. Do not use  =9x10 9 ! Because it isn’t!

We can also calculate the capacitance of a cylindrical capacitor (made of coaxial cylinders). L Coaxial Cylinder Capacitance The next slide shows a cross-section view of the cylinders.

Q -Q b r a E d Gaussian surface Lowercase c is capacitance per unit length: This derivation is sometimes needed for homework problems! (Hint: 4.10, 11, 12.) Some necessary details are not shown on this slide! See lectures 4 and 6.

Isolated Sphere Capacitance An isolated sphere can be thought of as concentric spheres with the outer sphere at an infinite distance and zero potential. We already know the potential outside a conducting sphere: The potential at the surface of a charged sphere of radius R is so the capacitance at the surface of an isolated sphere is

Capacitance of Concentric Spheres If you have to calculate the capacitance of a concentric spherical capacitor of charge Q… In between the spheres (Gauss’ Law) You need to do this derivation if you have a problem on spherical capacitors! (not this semester) +Q -Q b a If there is spherical capacitor homework, details will be provided in lecture!

Example: calculate the capacitance of a capacitor whose plates are 20 cm x 3 cm and are separated by a 1.0 mm air gap. d = area = 0.2 x 0.03 If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Example: what is the charge on each plate if the capacitor is connected to a 12 volt* battery? 0 V +12 V  V= 12V *Remember, it’s the potential difference that matters. If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Example: what is the electric field between the plates? 0 V +12 V  V= 12V d = E If you keep everything in SI (mks) units, the result is “automatically” in SI units.

Demo: Professor Tries to Avoid Spot-Welding His Fingers to the Terminals of a Capacitor While Demonstrating Energy Storage Asynchronous lecture students: we’ll try to make a video of this.

Today’s agenda: Capacitance. You must be able to apply the equation C=Q/V. Capacitors: parallel plate, cylindrical, spherical. You must be able to calculate the capacitance of capacitors having these geometries, and you must be able to use the equation C=Q/V to calculate parameters of capacitors. Circuits containing capacitors in series and parallel. You must understand the differences between, and be able to calculate the “equivalent capacitance” of, capacitors connected in series and parallel.

Capacitors in Circuits Recall: this is the symbol representing a capacitor in an electric circuit. And this is the symbol for a battery… + - …or this… …or this.

Capacitors connected in parallel: C1C1 C2C2 C3C3 + - V The potential difference (voltage drop) from a to b must equal V. a b V ab = V = voltage drop across each individual capacitor. V ab Circuits Containing Capacitors in Parallel Note how I have introduced the idea that when circuit components are connected in parallel, then the voltage drops across the components are all the same. You may use this fact in homework solutions. C2C2 C3C3 + -

C1C1 C2C2 C3C3 + - V a Q = C V  Q 1 = C 1 V & Q 2 = C 2 V & Q 3 = C 3 V Now imagine replacing the parallel combination of capacitors by a single equivalent capacitor. By “equivalent,” we mean “stores the same total charge if the voltage is the same.” C eq + - V a Q total = C eq V = Q 1 + Q 2 + Q 3 Q3Q3 Q2Q2 Q1Q1 + - Q Important!

Q 1 = C 1 VQ 2 = C 2 VQ 3 = C 3 V Q 1 + Q 2 + Q 3 = C eq V Summarizing the equations on the last slide: Using Q 1 = C 1 V, etc., gives C 1 V + C 2 V + C 3 V = C eq V C 1 + C 2 + C 3 = C eq (after dividing both sides by V) Generalizing: C eq =  C i (capacitors in parallel) C1C1 C2C2 C3C3 + - V a b

Capacitors connected in series: C1C1 C2C2 + - V C3C3 An amount of charge +Q flows from the battery to the left plate of C 1. (Of course, the charge doesn’t all flow at once). +Q-Q An amount of charge -Q flows from the battery to the right plate of C 3. Note that +Q and –Q must be the same in magnitude but of opposite sign. Circuits Containing Capacitors in Series

C1C1 C2C2 + - V C3C3 +Q A -Q B The charges +Q and –Q attract equal and opposite charges to the other plates of their respective capacitors: -Q +Q These equal and opposite charges came from the originally neutral circuit regions A and B. Because region A must be neutral, there must be a charge +Q on the left plate of C 2. Because region B must be neutral, there must be a charge -Q on the right plate of C 2. +Q -Q

C1C1 C2C2 + - V C3C3 A -Q B +Q -Q Q = C 1 V 1 Q = C 2 V 2 Q = C 3 V 3 The charges on C 1, C 2, and C 3 are the same, and are But we don’t know V 1, V 2, and V 3 yet. a b We do know that V ab = V and also V ab = V 1 + V 2 + V 3. V3V3 V2V2 V1V1 V ab Note how I have introduced the idea that when circuit components are connected in series, then the voltage drop across all the components is the sum of the voltage drops across the individual components. This is actually a consequence of the conservation of energy. You may use this fact in homework solutions.

C eq + - V +Q-Q V Let’s replace the three capacitors by a single equivalent capacitor. By “equivalent” we mean V is the same as the total voltage drop across the three capacitors, and the amount of charge Q that flowed out of the battery is the same as when there were three capacitors. Q = C eq V

Collecting equations: Q = C 1 V 1 Q = C 2 V 2 Q = C 3 V 3 V ab = V = V 1 + V 2 + V 3. Q = C eq V Substituting for V 1, V 2, and V 3 : Substituting for V: Dividing both sides by Q: Important!

Generalizing: OSE:(capacitors in series)

Summary (know for exam!): Series same Q, V’s add Parallel same V, Q’s add C1C1 C2C2 C3C3 C1C1 C2C2 C3C3

C3C3 C2C2 C1C1 I don’t see a series combination of capacitors, but I do see a parallel combination. C 23 = C 2 + C 3 = C + C = 2C Example: determine the capacitance of a single capacitor that will have the same effect as the combination shown. Use C 1 = C 2 = C 3 = C.

C 1 = C C 23 = 2C Now I see a series combination.

Example: for the capacitor circuit shown, C 1 = 3  F, C 2 = 6  F, C 3 = 2  F, and C 4 =4  F. (a) Find the equivalent capacitance. (b) if  V=12 V, find the potential difference across C 4. I’ll work this at the blackboard. C3C3 C2C2 C1C1 C4C4 VV Homework Hint: each capacitor has associated with it a Q, C, and V. If you don’t know what to do next, near each capacitor, write down Q=, C=, and V=. Next to the = sign record the known value or a “?” if you don’t know the value. As soon as you know any two of Q, C, and V, you can determine the third. This technique often provides visual clues about what to do next.

You really need to know this: Capacitors in series… all have the same charge add the voltages to get the total voltage Capacitors in parallel… all have the same voltage add the charges to get the total charge (and it would be nice if you could explain why)

Homework Hint! What does our text mean by V ab ? C3C3 C2C2 C1C1 C4C4 VV ab Our text’s convention is V ab = V a – V b. This is explained on page 84. This is in contrast to Physics 23 notation, where V a  b = V b – V a. In the figure on this slide, if V ab = 100 V then point a is at a potential 100 volts higher than point b, and V a  b = -100 V; there is a 100 volt drop on going from a to b.

A “toy” to play with… (You might even learn something.) For now, select “multiple capacitors.” Pick a circuit.