Parallel Plate CapacitorCylindrical Capacitor (Cable)Spherical CapacitorCapacitors in ParallelCapacitors in SeriesEnergy Stored in a CapacitorEnergy Stored in an Electric FieldDielectricsPHY2061 Enriched Physics 2 Lecture Notes Capacitance Capacitance Disclaimer: These lecture notes are not meant to replace the course textbook. The content may be incomplete. Some topics may be unclear. These notes are only meant to be a study aid and a supplement to your own notes. Please report any inaccuracies to the professor. +q −q A set of conductors can store electric charge. The net charge Q=0=q−q, but the magnitude of charge on each conductor is |q|. This charge q is proportional to the potential difference between the conductors: qCV VV V VqCV+−=Δ Δ= − ≡= The constant of proportionality between charge and potential difference is C≡capacitance. Unit is Farad (F) ≡ Coulomb/Volt. 6121F = 10 F1pF = 10 Fμ−− To set up a potential difference between 2 conductors requires an electric “pump”, such as a battery (see next chapter). C ΔV +−+q −q A larger capacitance implies that a large charge q is stored for the same potential difference V. Capacitance depends only on the geometry of the conductors, not the charge q or voltage V. We can see this through examples. D. Acosta Page 1 9/26/2006PHY2061 Enriched Physics 2 Lecture Notes Capacitance Parallel Plate Capacitor ΔV +−d A Consider the top view of the 2 plates: E Create a Gaussian surface (box) that extends inside and outside one of the conductor surfaces. Gauss’ Law enc0Sqdε⇒Φ= ⋅ =∫EAv 0=E inside a conductor 0d⋅=EA on left/right edges 0≠E on front outside face only enc00SqdEAqEAεε⇒Φ= ⋅ = =⇒=∫EAv The electric potential difference between the 2 plates is given by: + − E ds opposite directionsVVV d d dsVdVd++−−Δ= − =− ⋅ ⋅ =−Δ=Δ⇒=∫Es Es EEE So for parallel plates: 00 0212002C 8.85 10 8.85 pF/mNmVAΔqEA A VddqCVACdεε εεε−⎛⎞== = Δ⎜⎟⎝⎠=Δ⇒= = × = D. Acosta Page 2 9/26/2006PHY2061 Enriched Physics 2 Lecture Notes Capacitance Cylindrical Capacitor (Cable) Let inner conductor have radius a, and outer radius b. linder between conductors (E=0 inside conductors). Take Gaussian surface as cyenc02 0 on cylindeSErL q dεπε=⋅=EA0r endsqdΦ= ⋅ =∫EAv ⇒02 Lrπε1qE⇒= 0000 opposite directions, but opposite again2ln22ln /2ln /bbaaVVV d d ds ds drqdrVdrdrLrqbVLaLqVbaLCbaπεπεπεπε++−−+−Δ= − =− ⋅ ⋅ =− =−Δ=− = =Δ=⎛⎞⇒= Δ⎜⎟⎝⎠⇒=∫∫∫∫Es Es EEE Spherical Capacitor a, and outer radius b. Take Gaussian surface as sphere between conductors (E=0 inside conductors). Let inner sphere have radius 2 qKarb⇒= <<E Gauss’ Law r VVV d2aaVdrdrKqr opposite directions, but opposite again11111bad ds ds drKqababqVKb aabCbbdrVKqrKba+−−⋅ =− =−⎛⎞ ⎛ ⎞= −⎟ ⎜ ⎟⎠ ⎝ ⎠⎛⎞⇒= Δ⎜⎟−⎝⎠⇒=−∫E += − =− ⋅Es EsΔ+−Δ=− = =∫∫∫EEΔ= −⎜⎝D. Acosta Page 3 9/26/2006PHY2061 Enriched Physics 2 Lecture Notes Capacitance Capacitors in Parallel Consider N capacitors all connected in parallel to the same source of potential difference V. Across each capacitor i the charge on one of the plates is: iiqCV= The total charge on all the plates with the same electric potential is: 1Nii iiCVVC==∑ So we can write the equivalent capacitance Cequiv as: NC V= arallel is the sum of the pacitors, adding nce the capacitance proportional to the area, it increases in direct proportion. For N capacitors in series, the magnitude of the charge q on each plate must be the same. Consider the electric conductor connecting any 2 capacitors, and suppose that a charge +q is on the plate of one of the capacitors the conductor is connected to. Since the conductor as originally uncharged, a charge –q must exist on the plate of the second capacitor. 11NNiiQq====∑∑ Qequivequiv1iiCC==∑ other words, the equivalent capacitance of N capacitors in pInindividual capacitances. Considering the example of parallel plate caseveral in parallel is equivalent to extending the area of the plates. SiisCapacitors in Series wNow a capacitor has the same charge magnitude on each plate, so by inference we can determine that the magnitude of charge on each plate in the series of capacitor must bethe same. he potential difference across any capacitor is given by iiqVTC= ectric potential supplied by the battery or The total potential difference must add up to elpower supply: 1equivNiiqqVCC===∑ So the equivalent capacitance of capacitors connected in series is given by: 1equiv11NiC==∑ iC D. Acosta Page 4 9/26/2006PHY2061 Enriched Physics 2 Lecture Notes Capacitance The potential difference across any capacitor can be deter mined by: 11NjVq==∑jiCqVC= nergy Stored in a Capacitor Let’s calculate the work required of a battery or power supply to move an infinitesimal harge onto the plate of a capacitor already containing a charge iE dq′q′c. This is the same Recall that the electric potential difference across a device is equal to the potential energy difference per unit charge: as finding the change in the potential energy of the capacitor. UVqΔΔ= egative of the work done by the electric ds equal to the work done by the power pply or battery to move the charge (the charge must move against the direction of the The potential energy difference is equal to the nfield to set up the configuration, or in other worsuelectric field): appWUqV=Δ = Δ So the work done to move an infinitesimal charge dq′ onto the plate of a capacitor is iven by: g appdW dq V′=Δ the capacitor already has a charge , then qV′q′IfΔC= So appqdW dqC′′= So to charge up a capacitor initially uncharged to a total charge q will require integrating over the above expression: 2app app02CCU=Δ∫∫ 2app112qqdW qdqqC′′== ==WWD. Acosta Page 5 9/26/2006PHY2061 Enriched Physics 2 Lecture Notes Capacitance Since e electric potential energy stored in a capacitor can be qCV=Δ for a capacitor, thexpressed in 2 ways: ()22122qUΔ ΔCVC= = nnected to an electrical circuit. For example, a flash bulb on a camera works in this way. Using both forms of the relation for the energy in a capacitor, we can see which capacitor has a greater energy when two are connected in series or parallel. When two capacitors This potential energy can be
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