A) When this capacitor/batery circuit reaches equilibrium,what is the charge on

Question

A) When this capacitor/batery circuit reaches equilibrium,what is the charge on each plate of the capacitor? B) When C1is fully charged, the battery is removedand C1 is put into a circuit with a switch and anothercapacitor C2 which has a capacitance of8.95F but it initially uncharged. When switch S isclosed,the charge from C1 is distributed between thecapacitors untill they reach the same potential. What is the finalpotential V?

Explanation / Answer

A) Use C = Q/V => Q = CV = (3.55F)(6.3V) = 22.4C(coulombs) B) Again C = Q/V but now you need to know the equivalentcapacitance. Since the potential V is the same for both C, theymust be in parallel so C = C1+C2 = 12.5F Now V = Q/C = 22.4C/12.5F = 1.79V

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