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Unformatted text preview: µ C. The voltage across capacitor 1 is therefore (18 µ C)/( 6.0 µ F) = 3.0 V. From the discussion in section 25-4, we conclude that the voltage across capacitor 2 must be 6.0 V – 3.0 V = 3.0 V. Consequently, the charge on capacitor 2 is (4.0 µ F)( 3.0 V) = 12 µ C....
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.
- Spring '10