This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: SOLUTIONS, ASSIGNMENT 13 24.30. I DENTIFY : A C d = . The stored energy can be expressed either as 2 2 Q C or as 2 2 CV , whichever is more convenient for the calculation. S ET U P : Since d is halved, C doubles. E XECUTE : (a) If the separation distance is halved while the charge is kept fixed, then the capacitance increases and the stored energy, which was 8.38 J, decreases since 2 2 . U Q C = Therefore the new energy is 4.19 J. (b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance leads to a doubling of the stored energy to 16.8 J, using 2 2 U CV = , when V is held constant throughout. E VALUATE : When the capacitor is disconnected, the stored energy decreases because of the positive work done by the attractive force between the plates. When the capacitor remains connected to the battery, Q CV = tells us that the charge on the plates increases. The increased stored energy comes from the battery when it puts more charge onto the plates. 24.38. I DENTIFY : V Ed = and / C Q V = . With the dielectric present, C KC = . S ET U P : V Ed = holds both with and without the dielectric. E XECUTE : (a) 4 3 (3.00 10 V/m)(1.50 10 m) 45.0 V V Ed- = = = . 12 10 (5.00 10 F)(45.0 V) 2.25 10 C Q C V-- = = = . (b) With the dielectric, (2.70)(5.00 pF) 13.5 pF C KC = = = . V is still 45.0 V, so 12 10 (13.5 10 F)(45.0 V) 6.08 10 C Q CV-- = = = ....
View Full Document
This note was uploaded on 03/11/2010 for the course PHYSICS 1523 taught by Professor Dfs during the Fall '10 term at Dallas.
- Fall '10