Friday July 11 8:30 AM 9:0 AM Catch up Lecture 3 Slide 5 Electron projected in electric field problem Chapter 23 Problem 29 Cylindrical shell problem surrounding wire Show Faraday Ice Pail no chrage inside shell Lecture 4 Slide 34 Equipotential surfaces, Sharp points Total energy of charge distribution of charge 9:00 AM Lord Kelvin Water drop Generator explanation Lecture 5 Capacitance Reminder Put Microphone On 1
Example: An electron is projected perpendicularly to a downward electric field of E= 2000 N/C with a horizontal velocity v=10 6 m/s. How much is the electron deflected after traveling 1 cm. V e d E E 2
Chapter 23 Problem 29 3
Kelvin Water Drop Generator Electrons follow red path as the rings and the cans become increasingly polarized through induction 4
The figure belows shows a thin plastic rod of length L and uniform positive charge Q lying on an x axis. Find the electric field at point P 1 on the axis, at distance d from one end of the rod. Find the x and y components. L x < 0 d " V = k! $ x = k! ln( L # x ) x " # x x 0 E x =! dv dx = k" L (x)(l! x) = kq 1 (x)(l! x) x > 0 x 0 L d " V = k! $ x = k! ln( L + x ) x " # x x 0 E x =! dv dx =!k" L (x)(l + x) =!kq 1 (x)(l + x) 5
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Lecture 5 Capacitance Ch. 25 Cartoon - Capacitance definition and examples. Opening Demo - Discharge a capacitor Topics Capacitance - What is it? Parallel Plate Capacitor Dielectrics and induced dipoles Coaxial cable, Concentric spheres, Isolated sphere Two side by side spheres Energy density Graphical integration Combination of capacitance Demos Isolated sphere capacitor Circular parallel plate capacitor Cylindrical capacitor Dielectric Slab sliding into demo Elmo Problems Polling 8
What is Capacitance? Capacitance is a characteristic of a single isolated conducting object or a pair of conducting objects or even three objects. To keep it simple, suppose I have sphere and I put some charge on it say an amount q. Then it will have some voltage V. If now I double the q, the voltage will double. The point is that the ratio q/v is constant and it is called the capacitance. C= q/v. This will be true for any combination of objects. Even if there is no charge on the object it still has a capacitance as you will see. To help you understand capacitance, consider the isolated sphere again in more detail. If I put a chare q on the isolated sphere, the potential at any point r where r>r, the radius of the sphere, is given by the potential for a point charge. 9
q 0 + Isolated conducting sphere with total charge uniformly distributed on its surface. Recall we said that the potential for r > R is given by a point charge q at the center + q + R + + + At r > R V= kq r At r! R V= kq R C = q V = R k W = q 0 V + + + V = W q 0 10
Isolated conducting sphere Continued q R C = R 1 = 4!" 0R k k = 4!" 0 Units of Capacitance is 1F =1 Farad = 1 C/Volt q V = 4!"R! 0 = 8.85 " 10 #12 F m k = 8.99 X 10 9 Nm 2 /C 2 and R is in meters we have C! R 10 10 = 10"10 R(m) = 10 "12 R(cm) Earth: C = (6x10 8 cm)pf = 600 µf Conducting Marble: 1 pf You: 30 pf C = R( cm) pf 11
How does the capacitance of the charged conducting sphere change when another neutral sphere is nearby. q 0 W = q 0 V V = W q 0 q a d d a + + + + + + + Notice the redistribution of charge on both spheres. In effect the potential V of the original sphere is lowered for the same amount of charge. This means its effective capacitance has increased in the presence of the neutral sphere. Try to demonstrate it. 12
What is the capacitance of two charged conducting spheres? One of charge +q and one of charge - q. q d a a d C = q/v = 4πε 0 a (1/(1-m)) d =20 cm a =10 cm m = a/d = 0.5 C=10-10 (.1)(1/(1-m)) we do. C= 0.2 x 10-10 F C= 0.02 nf =20 pf If d gets very large, then C= 10 pf for the isolated sphere. The effective capacitance increases for the paired conductor case and that is why we make capacitors the way 13
Two isolated conducting objects such as two parallel plates Q i = C ij V j -Q! C = a 11 a 12 $ " # a 21 a 22 % & + Q 1 2 Q 1 = a 11 V 1 + a 12 V 2 Q 2 = a 21 V 1 + a 22 V 2! C = b 'b $ " # 'b b % & Q 1 =!b"v Q 2 = +b"v "V = V 2! V 1 14
Capacitance The most widely used capacitor consists of two parallel conducting plates of area A separated by a narrow air gap d. If charge +Q is placed on one plate, and -Q on the other, the potential difference between them is V, and then the capacitance is defined as. Applications Radio tuner circuit uses variable capacitor Blocks DC voltages in ac circuits Act as switches in computer circuits Triggers the flash bulb in a camera Converts AC to DC in a filter circuit 15
Parallel Plate Capacitor 16
Electric Field of Parallel Plate Capacitor Gaussian surface d + + + + + + + + + + + E - - - - - - - - - - - + q - q A Area of plate =A EA = q! 0 E =! 0 q A V = Ed = qd! 0 A C = q V = q C =! 0A d qd!0 A Coulomb/Volt = Farad V f! V i =! f # i! E " d! r Integrate from - charge to + charge so that V f! V i = + " Edr = Ed V = Ed +!! E! d! r = "Edr 17
Show Demo Model, calculate its capacitance, and show how to charge it up with a battery. Circular parallel plate capacitor r d C =! 0A d r r = 10 cm = 0.1m A = πr 2 = π(.1m) 2 A =.03 m 2 d = 1 mm =.001 m! 0 = 8.85 " 10 #12 F m C 2! 11 2 C.03m Coulomb = (10 2 ) }Farad Nm.001m Volt C = 3 " 10!10 F C = 300pF p = pico = 10-12 18
Demo Continued Connect Battery to aluminum plate to charge it up with plates close together. Connect voltmeter to plates. Disconnect battery and move plates apart and watch voltage increase Recall C =! 0A d = q V 1. As d increases, C decreases and V increases because q is constant. 2. If you leave battery connected, V is constant and q decreases. Demonstrate that a piece of wire has capacitance by touching electroscope. 19
Dielectrics A dielectric is any material that is not a conductor, but polarizes well. Even though they don t conduct they are electrically active Examples. Stressed plastic or piezo-electric crystal will produce a spark. When you put a dielectric in a uniform electric field (like in between the plates of a capacitor), a dipole moment is induced on the molecules throughout the volume. This produces a volume polarization that is just the sum of the effects of all the dipole moments. If we put it in between the plates of a capacitor, the surface charge densities due to the dipoles act to reduce the electric field in the capacitor. 20
Permanent dipoles Induced dipoles _ ++ _ E 0 = the applied field E = the field due to induced dipoles E = E 0 - E 21
Dielectrics The amount that the field is reduced defines the dielectric constant κ from the formula the dielectric. E = E 0!, where E is the new field and E 0 is the old field without Since the electric field is reduce, the voltage difference is reduced, and the capacitance is increased. V = V 0! C = Q V = Q V 0! =!C 0 where κ is typically between 2 6 with water equal to 80. See pg 669 for table of dielectric constants Repeat demo with dielectric slab sliding in between plates. Watch how capacitance and voltage change. Also show aluminum slab. 22
d C = q V V = E 0 d " E 0 =! 0! = q A E 0 q =! 0A V = qd! 0A C =! 0A d E = E 0! V = E 0! d C = q V C =!q V 0 V = V 0! C =!C 0 23
Gauss s Law for Dielectrics Induced charge! 0! E " d! A "# = q $ q'! 0 "! E # d! A "$ = q Induced charge factored into kappa 24
Model of coaxial cable for calculation of capacitance Outer metal braid Signal wire - to + 25
Find the capacitance of a ordinary piece of coaxial cable (TV cable) V f! V i =! f # i! E " d! r! E. dˆr = Edr cos180 =!Edr Integrate from b to a or - to + Er = 2k! r rˆ a V a! V b =!" E.! d r! a a = " Edr = 2k# " dr = +2k# ln r r b b b b a V a! V b = 2k" ln b a! = Q L k = 1 4!"0 air V a is higher than V b 26
capacitance of a coaxial cable cont. So, V C C Q b = ln 2!"0L a = = Q V = 2!" ln 0 b a C 2!" = L ln C L = b a 0 Q2!" Qln L Now if a=0.5mm and b=2.0mm, then C 43 L 6 " 10 ln4 pf m 0 b a L 6 " 10 1.38!11!11 = And if κ = 2, then C pf = 86 L m = ε 0 (for air) a = 0.5 mm b = 2.0 mm κ 2 For κ = 2 27
Capacitance of two concentric spherical shells dr -q +q Integration path b a V = Va! Vb =!# E! " d r! a = + # Edr as b! E. d! r = Eds cos180 =!E(!dr) b E a a Va! Vb =!" Edr =!" kq dr =!kq" dr a V = kq 1 r b b a b r 2 = kq( 1 a! 1 b ) = kq(b! a ab ) ab C = q / V = 4!" 0 b # a a b r 2 Let b get very large. Then C = 4!" 0 a for an isolated sphere 28
Electric Potential Energy of Capacitor As we begin charging a capacitor, there is initially no potential difference between the plates. As we remove charge from one plate and put it on the other, there is almost no energy cost. As it charges up, this changes. At some point during the charging, we have a charge q on the positive plate. The potential difference between the plates is q V = C As we transfer an amount dq of positive charge from the negative plate to the positive one, its potential energy increases by an amount du. du = Vdq = q C dq. The total potential energy increase is Q 2 2 q q Q U =! dq = = C 2C 2C Also U 0 = = 2 = 2 1 using 1 QV 2 1 CV 2 Q 2 C C = 29 Q V
Graphical interpretation of integration V q/c 1 U = C N! i= 1 dq " qi qi q Q V = q/c = Area under the triangle U Q =! Vdq Q 0 du = Vdq where V = Q q q2! dq = = Q2 C 2C 0 2 0 q C Area under the triangle is the value of the integral Area of the triangle is also = 1 b! h 2 Area = 1 2 (b)(h) = 1 2 (Q)(Q C ) = 1 2 Q 2 C Q! 0 q dq C 30
Where is the energy stored in a capacitor? Find energy density for parallel plate capacitor. When we charge a capacitor we are creating an electric field. We can think of the work done as the energy needed to create that electric field. For the parallel plate capacitor the field is constant throughout, so we can evaluate it in terms of electric field E easily. U = 1 2 QV = 1 2 (!AE)(Ed) = 1 2!E2 (Ad) U Ad = 1 2!E2 = " volume occupied by E We are now including dielectric effects: ε Electrostatic energy density general result for all 1 " =! E 2 geometries. 2 To get total energy you need to integrate over volume. 31
How much energy is stored in the Earth s atmospheric electric field? (Order of magnitude estimate) atmosphere h Earth R 20 km E = m V 100 =10 1 U = " 2 0E! Volume 2 Volume = 4! R Volume = "! 2 h 2 6 2 4 18 3 4 (6! 10 ) (2! 10 ) = 8.6 10 m R = 6x10 6 m U U = 1 2 (10!11 = 4.3! 10 C2 )(10 2 V Nm 2 m )2 (8.6 " 10 18 m 3 ) 11 J The total solar influx is 200 Watts/m 2 Usun 6 2 16 = 200 # 3.14(6! 10 ) " 2! 10 J s = 2! 10 21 J day Only an infinitesimal fraction gets converted to electricity. 32
Parallel Combination of Capacitors Typical electric circuits have several capacitors in them. How do they combine for simple arrangements? Let us consider two in parallel. +Q 1 +Q 2 C eq C eq = Q V -Q 1 -Q 2 We wish to find one equivalent capacitor to replace C 1 and C 2. Let s call it C. The important thing to note is that the voltage across each is the same and equivalent to V. Also note what is the total charge stored by the capacitors? Q. Q = Q1 + Q2 = C1V + C 2V = ( C1 + C 2) V Q V = C1 + C 2! C eq = C1 + C 2 33
Series Combination of Capacitors +Q 1 2 eq -Q +Q V 1 V 2 -Q C eq = Q V V = Q C eq What is the equivalent capacitor C eq? Voltage across each capacitor does not have to be the same. The charges on each plate have to be equal and opposite in sign by charge conservation. The total voltage across each pair is: V = V 1 + V 2 = Q C1 So 1 Ceq = 1 C1 + Q C 2 + 1 C 2 = Q( 1 C1 + 1 ) = Q( 1 ) C 2 C eq Therefore, C eq = C1C 2 C1 + C 2 34
Sample problem C 1 = 10 µf C 2 = 5.0 µf C 3 = 4.0 µf a) Find the equivalent capacitance of the three capacitors C 1 and C 2 are in series. 1 1 1 C C 2 C 1 = +! C12 = C 12 C1 C 2 C1 + 10! 5 50 C 12 = = = 3.3µF 10 + 5 15 2 C 12 and C 3 are in parallel. Ceq = C12 + C3 = 3.3 + 4.0 = 7.3µ F 35
Sample problem (continued) C 1 = 10 µf C 2 = 5.0 µf C 3 = 4.0 µf b) If V = 100 volts, what is the charge Q 3 on C 3? C = Q/V # 6 Q3 = C3V = 4.0" 10! 100 Q3 = 4.0" 10! 4 Coulombs c) What is the total energy stored in the circuit? U = 1 2 1! 6 4 2 2 eq! C 2 V = " 7.3 " 10 2! 2 U = 3.6" 10 J F " 10 V = 3.6 " 10 J 36
Chapter 25 Problem 28 In Fig. 25-28, a potential difference V = 100 V is applied across a capacitor arrangement with capacitances C1 = 10.0 µf, C2 = 5.00 µf, and C3 = 15.0 µf. Find the following values. 5 uf (a) the charge on capacitor 3 100 V 10 uf 15 uf (b) the potential difference across capacitor 3 (c) the stored energy for capacitor 3 37
5 uf (d) the charge on capacitor 1 (e) the potential difference across capacitor 1 100 V 10 uf (f) the stored energy for capacitor 1 15 uf (g) the charge on capacitor 2 (h) the potential difference across capacitor 2 (i) the stored energy for capacitor 2 38
Chapter 25 Problem 34 An air-filled parallel-plate capacitor has a capacitance of 1.4 pf. The separation of the plates is doubled and wax is inserted between them. The new capacitance is 2.3 pf. Find the dielectric constant of the wax. 39
Chapter 25 Problem 40 You are asked to construct a capacitor having a capacitance near 1 nf and a breakdown potential in excess of 10,000 V. You think of using the sides of a tall Pyrex drinking glass as a dielectric, lining the inside and outside curved surfaces with aluminum foil to act as the plates. The glass is 17 cm tall with an inner radius of 3.1 cm and an outer radius of 3.5 cm. The dielectric strength is 14 kv/mm. (a) What is the capacitance? (b) What is the breakdown potential of this capacitor? 40