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# Capacitance

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### Capacitance

1. 1. 25-2. Capacitance charge storage, energy storage Given As a Result Capacitor: Any two isolated conductors
2. 2. 25-2. Capacitance Units of capacitance: Farad (F) = Coulomb/Volt Potential (or Voltage) Capacitance
3. 3. 25-2. Capacitance +q -q E field between the plates: (Gauss’ Law) Relate E to potential difference V : What is the capacitance C  ? Area of each plate = A Separation = d charge/area =  = q/A
4. 4. 25-2. Capacitance Capacitance and Your iPod
5. 5. 25-2. Capacitance <ul><li>A huge parallel plate capacitor consists of two square metal plates of side 50 cm, separated by an air gap of 1 mm </li></ul><ul><li>What is the capacitance? </li></ul>C =  0 A/d = (8.85 x 10 –12 F/m)(0.25 m 2 )/(0.001 m) = 2.21 x 10 –9 F ( Very Small!! ) Units of capacitance: Farad (F) = Coulomb/Volt
6. 6. 25-2. Capacitance <ul><li>Does the capacitance of a capacitor increase, decrease, or remain the same? </li></ul><ul><li>When the charge q on it is doubled </li></ul><ul><li>When the potential difference across it tripled </li></ul>Determine what is NOT changed.
7. 7. 25-2. Capacitance <ul><li>A parallel plate capacitor of capacitance C is charged using a battery. </li></ul><ul><li>Charge = Q, potential difference = V. </li></ul><ul><li>Plate separation is INCREASED while battery remains connected. </li></ul><ul><li>V is fixed by battery! </li></ul><ul><li>C decreases (=  0 A/d) </li></ul><ul><li>Q=CV; Q decreases </li></ul><ul><li>E = Q/  0 A decreases </li></ul>Does the Electric Field Inside: (a) Increase? (b) Remain the Same? (c) Decrease? +Q – Q
8. 8. 25-3. Calculating the Capacitance Cylindrical Capacitor
9. 9. 25-3. Calculating the Capacitance Spherical Capacitor
10. 10. 25-3. Calculating the Capacitance Capacitance of a single conducting sphere Spherical Capacitor
11. 11. 25-4. C in Parallel and in Series Q: the same V: the same Parallel Series
12. 12. 25-4. C in Parallel and in Series <ul><li>A wire is an equipotential surface. </li></ul><ul><li>V AB = V CD = V </li></ul><ul><li>Q total = Q 1 + Q 2 </li></ul><ul><li>C eq V = C 1 V + C 2 V </li></ul><ul><li>Equivalent parallel capacitance = sum of capacitances </li></ul>V: the same Parallel A B C D C 1 C 2 Q 1 Q 2 C eq Q total
13. 13. 25-4. C in Parallel and in Series <ul><li>Q 1 = Q 2 = Q (WHY??) </li></ul><ul><li>V AC = V AB + V BC </li></ul>Q: the same Series A B C C 1 C 2 Q 1 Q 2 C eq Q
14. 14. 25-4. Example 1 <ul><li>Q = CV; V = 120 V </li></ul><ul><li>Q 1 = (10  F)(120V) = 1200  C </li></ul><ul><li>Q 2 = (20  F)(120V) = 2400  C </li></ul><ul><li>Q 3 = (30  F)(120V) = 3600  C </li></ul><ul><li>Note that: </li></ul><ul><li>Total charge (7200  C) is shared between the 3 capacitors in the ratio C 1 :C 2 :C 3 — i.e. 1:2:3 </li></ul>10  F 30  F 20  F 120V What is the charge on each capacitor?
15. 15. 25-4. Example 2 <ul><li>Q = CV; Q is same for all capacitors </li></ul><ul><li>Combined C is given by: </li></ul><ul><li>C eq = 5.46  F </li></ul><ul><li>Q = CV = (5.46  F)(120V) = 655  C </li></ul><ul><li>V 1 = Q/C 1 = (655  C)/(10  F) = 65.5 V </li></ul><ul><li>V 2 = Q/C 2 = (655  C)/(20  F) = 32.75 V </li></ul><ul><li>V 3 = Q/C 3 = (655  C)/(30  F) = 21.8 V </li></ul>10  F 30  F 20  F 120V What is the potential difference across each capacitor?
16. 16. 25-4. Example 3 <ul><li>Then, we have two 10  F capacitors in series </li></ul><ul><li>So, there is 5V across the 10  F capacitor of interest </li></ul><ul><li>Hence, Q = ( 10  F )(5V) = 50  C </li></ul>What is the charge on the 10  F capacitor? 10  F 5  F 5  F 10V 10  F 10  F 10V
17. 17. 25-4. Example 4 (N-1) Capacitors with A and d: Parallel Connection + –
18. 18. 25-4. Example 5 <ul><li>10  F capacitor is initially charged to 120V. </li></ul><ul><li>20  F capacitor is initially uncharged. </li></ul><ul><li>Switch is closed, equilibrium is reached. </li></ul>Initial charge on 10  F = (10  F)(120V)= 1200  C After switch is closed, let charges = Q 1 and Q 2 . Charge is conserved: Q 1 + Q 2 = 1200  C And, V final is same: <ul><li>Q 1 = 400  C </li></ul><ul><li>Q 2 = 800  C </li></ul><ul><li>V final = Q 1 /C 1 = 40 V </li></ul>What is the charge on each capacitor? 10  F (C 1 ) 20  F (C 2 )
19. 20. 25-5. Energy Stored <ul><li>Start out with uncharged capacitor </li></ul><ul><li>Transfer small amount of charge dq from one plate to the other until charge on each plate has magnitude Q </li></ul><ul><li>How much work was needed? </li></ul>dq
20. 21. 25-5. Energy Stored in an E-Field <ul><li>Energy stored in capacitor: U = Q 2 /(2C) = CV 2 /2 </li></ul><ul><li>View the energy as stored in ELECTRIC FIELD </li></ul><ul><li>For example, parallel plate capacitor: </li></ul><ul><li>Energy DENSITY = energy/volume = u </li></ul>General expression for any region with vacuum (or air)
21. 22. 25-6. Capacitor with a Dielectric Dielectric: Noncoducting material HW5-8 Breakdown Potential: Dielectric Strength (kV/mm) X Length (mm)
22. 23. 25-6. Capacitor with a Dielectric or ??
23. 24. 25-6. Capacitor with a Dielectric
24. 25. 25-6. Capacitor with a Dielectric
25. 26. 25-6. Example <ul><li>Capacitor has charge Q, voltage V </li></ul><ul><li>Battery remains connected while dielectric slab is inserted. </li></ul><ul><li>Do the following increase, decrease or stay the same: </li></ul><ul><ul><li>Potential difference? </li></ul></ul><ul><ul><li>Capacitance? </li></ul></ul><ul><ul><li>Charge? </li></ul></ul><ul><ul><li>Electric field? </li></ul></ul>dielectric slab
26. 27. 25-6. Example <ul><li>Initial values: capacitance = C ; charge = Q ; potential difference = V ; electric field = E </li></ul><ul><li>Battery remains connected </li></ul><ul><li>V is FIXED; V new = V ( same ) </li></ul><ul><li>C new =  C ( increases ) </li></ul><ul><li>Q new = (  C)V =  Q ( increases ). </li></ul><ul><li>Since V new = V, E new = E ( same ) </li></ul>dielectric slab