Two parallel-plate capacitors, each having a capacitance of 2.0 uF are connected

Question

Two parallel-plate capacitors, each having a capacitance of 2.0 uF are connected in parallel across a 12-V battery. a) Find the charge on each capacitor. b) Find the total energy stored in the capacitors. The parallel combination is then disconnected from the battery and a dielectric slab of constant 2.5 is inserted between the plates of the second capacitor (C2), completely filling the gap. After the dielectric is inserted, find c) the 15.0uF 10.0uF 20.0uF 4.00uF 15.0V potential difference across each capacitor, d) the charge on each capacitor, and e) the total energy stored in the capacitors. Answer: a) 24 uC for each, b) 2.9 x10^-4 J c) 6.86 V each d) Q1 =13.7uC , Q1 = 34.3uC e) 1.65 x 10^-4 J )

Explanation / Answer

(a) Voltage across each capacitor=12 V

charge on C1=2*12=24 uC

charge on C2=24 uC

(b) Energy on each=0.5CV^2=1.44*10^(-4) J

total energy=2.88*10^(-4) J

(c) New C2=C2'=2*2.5=5 uF

potential difference across each=(4/7)*12=6.85 V

(d)Q1=C1*6.85=13.7 uC

Q2=C2'*6.85=34.3 uC

(e) Energy=0.5*(C1+C2')*6.85^2=1.65*10^(-4) J

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