Charge and potential difference?

Question

Hi,

Capacitors C1 = 10 µF and C2 = 24 µF are each charged to 10 V, then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C1 connected to the negative plate of C2 and vice versa. Afterward, what are the charge on and the potential difference across each capacitor?

I know that the charge on the plates will be effected when hooked up initially and then it will go to the standard parallel configuration. But I am still confused as to what to do. Any ideas?


Thanks!

Answer 1

Safest bet the charge in the system is preserved and so is the total energy (even if one of them becomes negative)

E=0.5QV=0.5QV^2
E=0.5CV^2
E(total)=E1+E2 but since we reverse polarity
E(total)=E2-E1 (since C2 is larger then C1)
E(total)=0.5(C1+C2)V^2

Vf=sqrt[2E(total)/C(equivalent)]

C(equivalent)=C1+C2

We have

Vf=sqrt[2(0.5(C2 – C1)V^2/(C1+C2)]
Vf=V sqrt[(C1 - C2)/C1+C2)]
Vf=10 sqr[(24-10)/10+24)]
Vf=6.42V

Answer 2

imagine we've a parallel plate capacitor and we commence including electrons to between the plates through connecting both plates to a battery. What occurs is that the detrimental fee of the electrons gathering on the first plate begins to repel electrons in the different plate, so what you come across is that the 2d plate seems to amass efficient expenditures. As further and extra electrons assemble on the first plate, further and extra efficient expenditures assemble on the 2d. meaning the flair distinction between the plates grows and, if I save going, it is going to ultimately grow to be sufficient for the electrons to leave the first plate, be pulled through the textile keeping apart the plates, and arrive on the 2d. that is fairly a lot what occurs in a lightning strike ... expenditures assemble on the floor and cloud until eventually the flair distinction is so tremendous that the air can no longer end them from combining and ZAP! Now, capacitance tells you thoughts including a fee to a capacitor impacts the flair distinction between the plates. A intense capacitance means that you could upload more than a number of expenditures to the plates with a small enhance in ability distinction so that you could save more beneficial fee previously the textile between the plates breaks down and helps the prices to pass. A low capacitance means that including a small fee dramatically will enhance the flair distinction. So capacitance is defined because the ratio of fee to ability or, in different words, how replacing the fee impacts the voltage. once you seem at your graph, you're searching at how replacing the fee impacts the flair distinction. The capacitance is the slope of that graph. So Q and V replace, notwithstanding the relationship between them is what capacitance measures!