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Chapter 17. Capacitance and Capacitors! C = q / V V= voltage q = charge Capacitors in parallel Capacitors in series. Capacitors, Batteries, & Circuits. Single Capacitor Example. Voltage = 12 volts Capacitance = 20 microfarads Charge = ?
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Chapter 17 Capacitance and Capacitors! • C = q / V V= voltage q = charge • Capacitors in parallel • Capacitors in series
Single Capacitor Example Voltage = 12 volts Capacitance = 20 microfarads Charge = ? q = C * V = 20 e-6 f * 12 V = 240 microcoulombs
Example: Capacitors in Parallel A circuit consists of a 6.0 mFcapacitor, a 12.0 mF capacitor, a 12 Vbattery, and a switch. Initially, theswitch is open and the capacitors discharged.The switch is then closed, and thecapacitors charge. When the capacitors are fully chargedand the open-circuit terminal voltage of thebattery is restored: (a) What is potential of each conductor inthe circuit?(b) What is the charge on each capacitor plate? (c) What total charge has passed through the battery?
Capacitors in Parallel The connected plates are at the same potential, so they form an effective capacitor with a plate area that is the sum of the two plate areas. Since C~A, Ceff = C1+C2, i.e., capacitors in parallel add capacitances.
Example: Capacitors in Series A circuit consists of a 6.0 mF capacitor, a 12.0 mFcapacitor, a 12 V battery, and a switch. Initially, theswitch is open and the capacitors discharged. Theswitch is then closed, and the capacitors charge. When the capacitors are fully charged and theopen-circuit terminal voltage of the battery is restored: (a) What is potential of each conductor in the circuit?(b) What is the charge on each capacitor plate? (c) What total charge has passed through the battery?
Capacitors in Series The connected string of capacitors all have the same charge Q on their plates, and the potential drops across the component capacitors add.
Example:Using the C Equivalence Formula A 6.0 mF capacitor and a 12.0 mF capacitor, each initially uncharged, are connected in series across a 12 V battery. (a) Find the charge Q on each capacitor. (b) Find the potential difference across each capacitor.
Dielectric Materials There is a class of polarizable dielectric materials that have an important application in the construction of capacitors. In an electric field their tiny dipoles line up, reducing the E field and potential difference and increasing the capacitance: E off E on
Electric Fields and Dielectrics In an external field EO, neutral molecules can polarize. The induced electric field E’ produced by the dipoles will be in the opposite direction from the external field EO. Therefore, in the interior of the slab the resulting field is E = EO-E’. The polarization of the material has the net effect of producing a sheet of positive charge on the right surface and a sheet of negative charge on the left surface, with E’ being the field made by these sheets of charge.
Capacitors and Dielectrics If a capacitor connected to a battery, so that it has a charge q, and then a dielectric material of dielectric constant ke is placed in the gap, the potential is unchanged but the charge becomes keq. If a capacitor is given a charge q, and then a dielectric material of dielectric constant ke is placed in the gap, the charge q is unchanged, but the potential drops to V/ke.
Example: A Dielectric in a Parallel-Plate Capacitor A parallel plate capacitor has square plates of edge-length 10 cm and a gap of 4.0 mm. A dielectric slab with k = 2.0 has dimensions 10 cm x 10 cm x 4 mm. (a) What is the capacitance without the dielectric? (b) What is the capacitance with the dielectric in the gap? (c) What is the capacitance if a similar dielectric with a thickness of 3.0 mm is placed in the gap?
A d Energy Storage with Dielectrics Electric Field Mechanical Stress on Dielectric