A 1290.-nF parallel plate capacitor is connected to a 19.8-V battery and charged

Question

A 1290.-nF parallel plate capacitor is connected to a 19.8-V battery and charged.

a) What is the charge Q on the positive plate of the capacitor?
2.554×10-5 C

b) What is the electric potential energy stored in the capacitor?
2.529×10-4 J

The 1290.-nF capacitor is then disconnected from the 19.8-V battery and used to charge three uncharged capacitors, a 130.-nF capacitor, a 210.-nF capacitor, and a 335.-nF capacitor, connected in series.

c) After charging, what is the potential difference across each of the four capacitors?
[Enter the potential differences across the 1290.-nF capacitor, the130.-nF capacitor, the 210.-nF capacitor, and the 335.-nF capacitor, separated by commas, in units of volts. Example: if you think that the potential difference across the 1290.-nF capacitor is 7 V and across each of the others is 9 V, then you should enter 7,9,9,9].

d) How much of the electrical energy stored in the 1290.-nF capacitor was transferred to the other three capacitors?

!!!Please answer only c and d!!!

Explanation / Answer

CiVi = CeqVf + Ci*Vf

Ceq = C1*C2*c3/(C1*C2 + C2*C3 + C3*C1) = 130*210*335/(210*335+130*335+130*210) = 64nF

Vf = Ci*Vi/(Ceq+Ci) = 1290*19.8/(64+1290) = 18.864 V Answer

Q_total =Q_accross_Ceq = Ceq*Vf = 64n*18.864

V_130 = Q_total / C_130 = 64n*18.864 /130n = 9.29 V Answer

V_210 = 64n*18.864 /210n = 5.75 V Answer

V_335 = 64n*18.864 /335n = 3.6 V Answer

D) Energy = Ceq*V*V/2 = 64n*18.864*18.864/2 = 1.138*10-5 Answer

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