C qv sag 33 50 40 30 20 10 0 messingwcapacitorsact

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Unformatted text preview: (EMB) “The capacitance decreases in C2 due to the dielectric thus the voltage must increase because the charge is the same for both. C = Q/V” (SAG) 33 50 40 30 20 10 0 Messing w/ Capacitors ACT Messing w/ Capacitors ACT Two identical parallel plate capacitors are connected to identical batteries. Then a dielectric is inserted between the plates of capacitor C1. Compare the energy stored in the two capacitors. C0 V A) U1 < U0 V B) U0 = U1 Compare using U = 1/2CV2 U1/U0 = κ 35 Potential Energy goes UP Potential Energy goes C1 κ=2 C) U1 > U0 BB Preflight 10 Preflight 10 BB A B C “U = Q^2/(2C). In this situation, Q does not change, but Q^2/(2C). capacitance for C2 is larger than C1; therefore U1>U2.” (BLF) (BLF) “It should be the same due to the same amount of It charge.” (TRCK) charge.” (TRCK) “Higher V means higher energy” (MCA) “lady gaga” (KA); “duuuuhhhhh... i don know ” (JSD) 33 40 30 20 10 0 Preflight 12 Preflight 12 BB V must be the same ! Q: Q1 Q2 = C1 C2 C1 Q1 = Q2 C2 60 50 40 30 20 U1 = 1 C1V 2 2 U: U 2 = 1 C2V 2 2 C U1 = 1 U 2 C2 10 0 Calculation Calculation V C0 x0 V An air­gap capacitor, having κ x0/4 BB capacitance C and width x is 0 0 connected to a battery of voltage V. A dielectric (κ) of width x0/4 is inserted into the gap as shown. • Conceptual Analysis: Q C≡ V What is Qf, the final charge on the capacitor? What changes when the dielectric added? (A) Only C (B) only Q (C) only V (D) C and Q (E) V and Q Adding dielectric changes the physical capacitor V does not change and C changes 38 C changes Q changes...
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