应该是和国内大2的差不多A4.5Fcapacitorischargedtoapotentialdifferenceof13.0V.Thewiresconnectingthecapacitortothebatteryarethendisconnectedfromthebatteryandconnectedacrossasecond,initiallyuncharged,capacitor.T

Question

应该是和国内大2的差不多
A 4.5 F capacitor is charged to a potential difference of 13.0 V.The wires connecting the capacitor to the battery are then disconnected from the battery and connected across a second,initially uncharged,capacitor.The potential difference across the 4.5 F capacitor then drops to 5 V.What is the capacitance of the second capacitor?

Answer 1

其实,这一题应该是国内高二的水平吧,（有可能难在英语上吧）
From above,you can see that two capacitor is connected.From the law of charge conservation ,the charge of all is 4.5*13.0C,and it is the total of the charge of two capacitors.
The charge remained of the 4.5F capacitor fell to 4.5*5C.Therefore,the charge of the second capacitor is 4.5*8C.
Two capacitors are equipotential.So the potential difference of the second capacitor is equally 5V.
The answer is 7.2F.
我英语不好,不知你能不能看懂……