Today s agenda: Capacitors and Capacitance. You must be able to apply the equation C=Q/V.

Today s agenda: Capacitors and 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: the basics What is a capacitor? device for storing charge simplest example: two parallel conducting plates separated by air E A V 0 d V 1 assortment of capacitors

Capacitors in circuits symbol for capacitor (think parallel plates) symbol for battery, or external potential battery voltage V is actually potential difference between the terminals + - V when capacitor is connected to battery, charges flow onto the plates - Capacitor plates build up charges +Q and -Q conducting wires +- V - when battery is disconnected, charge remains on plates

C Capacitance How much charge can a capacitor store? Better question: How much charge can a capacitor store per voltage? Capacitance: Q V V is really V, the potential difference across the capacitor capacitance C is a device property, it is always positive unit of C: farad (F) 1 F is a large unit, most capacitors have values of C ranging from picofarads to microfarads (pf to F). micro 10-6, nano 10-9, pico 10-12 (Know for exam!)

Capacitance of parallel plate capacitor electric field between two parallel charged plates: Q E. A 0 0 -Q E +Q Q is magnitude of charge on either plate. potential difference: d d 1 0 0 0 V 0 d V 1 V V V E d E dx Ed A capacitance: C Q Q Q 0A V Ed Q d d 0A

Parallel plate capacitance depends only on geometry. C 0 A This expression is approximate, and must be modified if the plates are small, or separated by a medium other than a vacuum (lecture 9). d -Q +Q E V 0 d V 1 A C 0 A d Greek letter Kappa. For today s lecture (and for exam 1), use =1. is NOT the same as k=9x10 9!

Capacitance of coaxial cylinder capacitors do not have to consist of parallel plates, other geometries are possible capacitor made of two coaxial cylinders: L

b b ΔV= V -V = - Ed =- E dr b a r a a from Gauss law: (see lectures 4 and 6) 2kλ E= r ΔV=-2k λ b a dr r = - 2k λ ln b a Gaussian surface Q λ L λ L C= = = ΔV ΔV b 2k λ ln a b r Q a L 2πε0L C= = b b 2k ln ln a a d E -Q capacitance per unit length: C = L 2πε 0 b ln a

Example application: coaxial cables, capacitance per length is a critical part of the specifications. L

Isolated Sphere Capacitance 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: Q V. 4 r The potential at the surface of a charged sphere of radius R is V 0 0 Q 4 R so the capacitance at the surface of an isolated sphere is Q C 40R. V

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) E Q 4 r 0 2 a b Q b dr Q 1 1 a 2 0 0 V 4 r 4 a b +Q -Q C Q 40 V 1 1 a b You need to do this derivation if you have a problem on spherical capacitors!

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. C 0 A d C 12 8.8510 0.20.03 12 C5310 F C 53 pf 0.001 F d = 0.001m area = 0.2m x 0.03m 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? Q CV 0 V 12 Q 53 10 12 C V= 12V 10 Q 6.410 C +12 V *Remember, it s the potential difference that matters.

Example: what is the electric field between the plates? E V d 0 V E 12V 0.001 m E V= 12V V E 12000,"up." m d = 0.001 +12 V

Circuits Containing Capacitors in Parallel V ab Capacitors connected in parallel: C 1 a C 2 b C 3 + - all three capacitors must have the same potential difference (voltage drop) V ab = V V General concept: When circuit components are connected in parallel, then the voltage drops across these components are all the same.

Q 1 = C 1 V & Q 2 = C 2 V & Q 3 = C 3 V a C 1 Q 1 C 2 Q 2 C 3 Q 3 + - + - V Imagine replacing the parallel combination of capacitors by a single equivalent capacitor a Q C eq equivalent means stores the same total charge if the voltage is the same. + - V Q total = C eq V = Q 1 + Q 2 + Q 3 Important!

Summarizing the equations on the last slide: C 1 Q 1 = C 1 V Q 2 = C 2 V Q 3 = C 3 V Q 1 + Q 2 + Q 3 = C eq V a C 2 C 3 b Using Q 1 = C 1 V, etc., gives + - C 1 V + C 2 V + C 3 V = C eq V V C 1 + C 2 + C 3 = C eq (after dividing both sides by V) Generalizing: C eq = i C i (capacitances in parallel add up)

Circuits Containing Capacitors in Series Capacitors connected in series: C 1 C 2 C 3 + - +Q V -Q charge +Q flows from the battery to the left plate of C 1 charge -Q flows from the battery to the right plate of C 3 (+Q and Q: the same in magnitude but of opposite sign)

Charges +Q and Q attract equal and opposite charges to the other plates of their respective capacitors: +Q C 1 C A 2 B -Q +Q -Q +Q C 3 -Q + - These equal and opposite charges came from the originally neutral circuit regions A and B. V 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.

V ab a C 1 C 2 C A B 3 +Q -Q +Q -Q +Q -Q V 1 V 2 + - V V 3 b The charges on C 1, C 2, and C 3 are the same, and are Q = C 1 V 1 Q = C 2 V 2 Q = C 3 V 3 The voltage drops across C 1, C 2, and C 3 V ab = V 1 + V 2 + V 3. add up General concept: When circuit components are connected in series, then the voltage drops across these components add up to the total voltage drop.

replace the three capacitors by a single equivalent capacitor C eq +Q -Q V + - equivalent means it has the same charge Q and the same voltage drop V as the three capacitors Q = C eq V V

Collecting equations: Q = C 1 V 1 Q = C 2 V 2 Q = C 3 V 3 Important! 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: Q Q Q V = + + C C C 1 2 3 Q Q Q Q = + + C C C C eq 1 2 3 1 1 1 1 = + + C C C C eq 1 2 3

Generalizing: OSE: 1 1 = C C eq i i (capacitors in series)

Summary (know for exam!): Parallel Series C 1 C 1 C 2 C 3 C 2 C 3 equivalent capacitance C eq C i i 1 1 C C eq i i charge Q s add V s add voltage same V same Q

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 2 C 3 C1 Start by combining parallel combination of C 2 and C 3 C 23 = C 2 + C 3 = C + C = 2C

Now I see a series combination. eq 1 1 1 = + C C C eq 1 23 1 1 1 2 1 3 = + = + = C C 2C 2C 2C 2C C 23 = 2C C 1 = C C = eq 2 C 3

Example: for the capacitor circuit shown, C 1 = 3F, C 2 = 6F, C 3 = 2F, and C 4 =4F. (a) Find the equivalent capacitance. (b) if V=12 V, find the potential difference across C 4. C 1 C 2 C 4 C 3 V I ll work this at the blackboard. 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.

(a) Find C eq. (b) if V=12 V, find V 4. C 1 =3F C 2 =6F C 4 =4F C 1 and C 3 are not in parallel. Make sure you understand why! C 3 =2F C 2 and C 4 are not in series. Make sure you understand why! V=12 V C 1 and C 2 are in series. Make sure you use the correct equation! 1 1 1 1 1 2 1 3 1 = + = + = + = = C C C 3 6 6 6 6 2 12 1 2 Don t forget to invert: C 12 = 2 F.

(a) Find C eq. (b) if V=12 V, find V 4. C 12 =2F C 4 =4F C 12 and C 4 are not in series. Make sure you understand why! C 3 =2F V=12 V C 12 and C 3 are in parallel. Make sure you use the correct equation! C 123 = C 12 + C 3 = 2 + 2 = 4μF

(a) Find C eq. (b) if V=12 V, find V 4. C 123 =4F C 4 =4F C 123 and C 4 are in series. Make sure you understand why! Combined, they make give C eq. V=12 V Make sure you use the correct equation! 1 1 1 1 1 2 1 = + = + = = C C C 4 4 4 2 eq 123 24 Don t forget to invert: C eq = 2 F.

(a) Find C eq. (b) if V=12 V, find V 4. C eq =2F C eq = 2 F. V=12 V If you see a capacitor circuit on the test, read the problem first. Don t go rushing off to calculate C eq. Sometimes you are asked to do other things. Truth in advertising: there s a high probability you will need to calculate C eq at some point in the problem.

(a) Find C eq. (b) if V=12 V, find V 4. Q 1 =? C 1 =3F V 1 =? Q 3 =? C 3 =2F V 3 =? V=12 V Q 2 =? C 2 =6F V 2 =? Q 4 =? C 4 =4F V 4 =? 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. We know C 4 and want to find V 4. If we know Q 4 we can calculate V 4. Maybe that is a good way to proceed.

(a) Find C eq. (b) if V=12 V, find V 4. Q 123 =? C 123 =4F V 123 =? Q 4 =? C 4 =4F V 4 =? C 4 is in series with C 123 and together they form C eq. V=12 V Therefore Q 4 = Q 123 = Q eq. Q = C V = 2 12 = 24μC= Q eq eq 4 Q Q Q4 24 C = V = V 4 = = = 6V V C C 4 4

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? Our text s convention is V ab = V a V b. This is explained on page 759. This is in contrast to Physics 1135 notation, where V ab = V b V a. a C 1 C 2 C 4 b C 3 V 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 ab = -100 V; there is a 100 volt drop on going from a to b.

A toy to play with http://phet.colorado.edu/en/simulation/capacitor-lab (You might even learn something.) For now, select multiple capacitors. Pick a circuit.