Physics 24100 Electricity & Optics Lecture 9 Chapter 24 sec. 3-5 Fall 2017 Semester Professor Koltick
Parallel Plate Capacitor Area, A C = ε 0A
Two Parallel Plate Capacitors Area, A 1 C 1 = ε 0A 1 Area, A 2 C 2 = ε 0A 2
One Big Parallel Plate Capacitor Area, A 1 Area, A 2 C = ε 0 A 1 + A 2
Two Parallel Plate Capacitors Area, A 1 Remember: conuctors in electrical contact have the same electric potential. A B We coul measure the potential ifference between points A an B if they were connecte to the plates with wires. Area, A 2 Q = C V C = ε 0 A 1 + A 2
Capacitors In Parallel Symbol for a capacitor: Capacitors in parallel: C 1 C C 2 Equivalent capacitance: C 1 C = C 1 + C 2 C 2
Capacitors in Parallel Consier two capacitors, C 1 an C 2, with the same potential ifference, V, across them. The charge on each one is Q 1 = C 1 V Q 2 = C 2 V If they are connecte in parallel then the potential ifference oesn t change but now we can write: Q = Q 1 + Q 2 = C 1 + C 2 V = C V where C = C 1 + C 2
Capacitors in Series Consier two capacitors, C, each with potential ifference, V, an charge Q = CV: C Q +Q Q +Q V If they are brought into electrical contact the potential ifference is 2V but the charge is still Q. The effective capacitance is C where C = Q 2V = C 2 Equivalent series capacitance is reuce. V C
Capacitors in Series Consier two parallel plate capacitors: A A 1 2 C 1 = ε 0A 1 C 2 = ε 0A 2
Capacitors in Series Consier two parallel plate capacitors: 1 2 C 1 = ε 0A 1 C 2 = ε 0A 2
Capacitors in Series Consier two parallel plate capacitors: A 1 2 C = ε 0 A 1 + 2
Capacitance in Series C = ε 0 A 1 + 2 1 C = 1 + 2 ε 0 A = 1 C 1 + 1 C 2 Equivalent series capacitance: C 1 C 2 C = 1 C 1 + 1 C 2 1
Question What is the value of the equivalent capacitance: C 1 C = 1 C + 1 C = C 2 C C C equiv = C + C 2 = 3 2 C (a) C equiv = 2 3 C (b) C equiv = 3 2 C (c) C equiv = 3C () C equiv = 1 3 C
Conuctor inserte between plates Two capacitors of area A in series A parallel-plate capacitor with conuctor inserte in the mile a b -Q +q +q E=0 outsie capacitors q E between plates A 0 -Q -q -q 0 0 0 E=0 in conuctor E q A 1 1 1 a b a b A 0 C C C A A A a -q -q +q b 0 +q C a b between plates
Another Question How woul the capacitance of a parallel plate capacitor change if we replace half the space with a conuctor? Conuctor (a) It woul ouble (b) It woul be half (c) No change
Aing Conucting Material Conuctor C C C = ε 0 A 4 = 4C C equiv = 1 C + 1 C 1 = 1 4C = 2C 2 The capacitance woul ouble.
Lecture 9-6 Capacitors with Dielectrics So far, we have iscusse capacitors with air or vacuum between the plates. However, most real-life capacitors have an insulating material, calle a ielectric, between the two plates. The ielectric serves several purposes: Provies a convenient way to maintain mechanical separation between the plates Provies electrical insulation between the plates Increases the capacitance of the capacitor
Lecture 9-9 Polar an Nonpolar Dielectrics Polar ielectrics: ielectric material whose molecules have permanent electric ipole moments, such as water. U p E E ext = 0: the orientations of the permanent electric ipoles are istribute ranomly, so the net ipole moment of the material is zero. E ext 0 : the molecular ipoles try to align themselves with the fiel against ranom thermal motion, resulting in a net ipole moment. Nonpolar ielectrics: molecules with no permanent electric ipole moments. E ext 0 : charges separate, inuce ipole moment emerges.
free + Q charges - Q Dielectrics between Capacitor Plates Electric fiel E between plates can be calculate from Q q. ( Q q)/ A E, V E 0 neutral -q +q Polarization Charges ± q C Q Q V ( Q q ) / A 0 A 1 q 1 Q 0
Lecture 9-10 free + Q charges - Q neutral -q +q Dielectrics between Capacitor Plates Electric fiel E between plates can be calculate from Q q. ( Qq)/ A ( Qq) E, V E A 0 0 C Q Q V ( Q q ) / A 0 A 1 q 1 Q p// E0 E E0 '//( ) How about a conuctor? 0
Aing an Insulating Material (Dielectric) Empirical observation: inserting a ielectric material reuces the electric potential ifference. If Q is constant, the capacitance must increase. The ielectric constant, κ, of the material is the ratio of the capacitance when the space between the conuctors is fille with the ielectric material, to the original capacitance. Dielectric (insulator) C C κ = C C > 1 C = κε 0A
Permittivity The ratio κ is calle the ielectric constant. We can efine ε = κ ε 0 which is calle the permittivity of the material with ielectric constant κ. We can replace ε 0 with ε in formulas when the space between conuctors is not vacuum. For example, the parallel plate capacitor C = ε 0A C = ε A
Dielectric Constants material k ielectric strength (kv/mm) air (1 atm) 1.00054 3 paraffin 2.1 2.5 10 glass (Pyrex) 5.6 14 mica 5.4 10 100 polystyrene 2.55 24 H 2 O (20 0 C) 80? Strontium titanate 240 8 Sulfur hexafluorie 1.002026 9
Why? Remember the electric ipole? Insulators in an electric fiel are compose of electric ipoles: + + + + + + + + + + + + - - - - - - - - - - - - The ipoles line up with the electric fiel lines. The surface charge on the ielectric partially cancels the charge on the plates of the capacitor. The effective Q" is reuce We have to put more actual charge on the plates to get the same electric fiel an potential. C is increase.
Dielectric Breakown If the electric fiel is too strong, it can ionize the ielectric (rip electrons off its molecules). Free electrons = conuctor... (sparks). Can result in amage to the ielectric. DIELECTRIC STRENGTH: maximum electric fiel a ielectric can tolerate before breaking own. BREAKDOWN POTENTIAL: maximum electric potential a evice can tolerate before breaking own.
Inserting Dielectric Material with Battery Connecte 1. Charge a parallel plate capacitor fille with vacuum (air) to potential ifference V. 2. Keep the battery connecte Q CV Deposits charge 0 0 V remains fixe Q 0 Q 0 3. Insert a ielectric of ielectric constant κ C QCV Q 0 C 0 Q V So, Q increases from Q 0 an E remains fixe 1 2 2 U CV U0 an an E 0 V E Q Q 2 0 0 0 1 uu E 2 0 0
Dielectric Combinations κ 1 κ 2 A A /2 /2 A/2 κ 1 κ 2 A/2
Non-uniform Parallel-plate Capacitor 1 Equivalent to 2 capacitors in parallel 1 0A/2 2 0A/2 C C1C2 0A 1 2 2 Q 2Q V C A ( ) 0 1 2 Potential rop V in each is the same. 1 Q1 CV 1 Q, 12 Note Q1 Q if 2 1 2 even though 1 2 1 2 2 Q2 CV 2 Q 12 V V, E E
Lecture 9-12 Capacitor Examples C C C C 2C C/2 1 0A/2 2 0A/2 C C1C2 0A 1 2 2 C C C?C?=2/3 1 1 1 /4 3 /4 C C C A A 0 1 2 0 0 A 1 3 1 4 4
0 Non-uniform Parallel-plate Capacitor 2 Equivalent to 2 capacitors in series 1 1 1 /4 3 /4 C C C A A 1 2 0 0 1 3 1 A 4 4 Q Q1 3 V C A 4 4 0 (Free) charge in each Q is the same. Q ( A/ ) 1 V Q/ C V, 0 1 1 0 ( 0A/ ) C1 4 Q ( A/ ) 3 V Q/ C V, 0 2 2 0 ( 0A/ ) C2 4 Note V1/ V2 1/3 if 1