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