A 2.90 uF capacitor is charged to 470 V and a 3.55 uF capacitor is charged to 50

Question

A 2.90 uF capacitor is charged to 470 V and a 3.55 uF capacitor is charged to 505 V.

Part A
These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each capacitor?


Part B
What will be the charge on each capacitor?



Part C
What is the voltage for each capacitor if plates of opposite sign are connected?



Part D
What is the charge on each capacitor if plates of opposite sign are connected?

Explanation / Answer

C1 =2.90 uF , V1= 470 V C2 =3.55 uF ,V2= 505 V The energy is conserved and I agree with Kirchwey the charge is also conserved. No wonder since charge is related to energy just as mass is related to energy in mechanical system. E1+E2= Et ( before and after they are connected) E=0.5 *C*V^2 E1= 0.5*C1*V1^2= 1.97x10^-9 J E2= 0.5*C2*V2^2= 3.18x10^-9 J Et= 5.15x10^-9 J In parallel the total capacitance is Ct= C1+C2=6.45µF Then since Et=0.5 CtVt^2 Vt= sqrt ( 2Et/Ct) Vt=1.597 *10^-3 V B) Qt=Ct*Vt =1.03*10^-8 C) Opposite connected , which means capacitor is in series . Cs=(c1c2)/(c1+c2) =1.61 uF , Qt= sqrt(2*Et*c) V1 =Qt /C1 ,V2 = Qt/C2 D) when connected opposite terminal , the charge on both capacitor will remain same.i.e Qt

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