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PhysicsTutor Capacitor Giambattista 17.54 Problem: • A parallel-plate capacitor with C=2.2F has a plate separation of 1.0 mm. • A) How much potential difference will the capacitor take before dielectric breakdown of air (critical field: Ebr=3106 V/m)? • B) What is the magnitude of the greatest charge the capacitor can store before breakdown? 2 Relevant ideas: 3 Relevant ideas: • Electric potential difference across cap. plates and E field inside related by separation d. 4 Relevant ideas: • Electric potential difference across cap. plates and E field inside related by separation d. • Maximum allowed field before breakdown then implies maximum voltage for given d. 5 Relevant ideas: • Electric potential difference across cap. plates and E field inside related by separation d. • Maximum allowed field before breakdown then implies maximum voltage for given d. • Charge on the plates and voltage across plates are related. Proportionality is controlled by the capacitance C, which is given. 6 Equations associated with ideas: 7 Strategy Strategy • Use the voltage (=potential difference) to field strength relation to obtain the breakdown voltage for this capacitor. Strategy • Use the voltage (=potential difference) to field strength relation to obtain the breakdown voltage for this capacitor. • Using the known capacitance C relate the breakdown voltage to charge Q on the plates. Strategy • Use the voltage (=potential difference) to field strength relation to obtain the breakdown voltage for this capacitor. • Using the known capacitance C relate the breakdown voltage to charge Q on the plates. • This is the maximum charge one can store on the plates (under breakdown the charge equilibrates). Solution Solution • _______________________ Solution • _______________________ • _______________________ Solution • _______________________ • _______________________ • _______________________ Solution • _______________________ • _______________________ • _______________________ • _______________________