physics question- anyone who study physics in F.6 or F.7...

Ok. This is a question about combination of capacitors and energy stored in capacitor....

FIRST QUESTION-

Step One:
Two capacitors are connected in series to the energy source. They are C1 and C2 respectively. The combined capacitance is Ce. When the capacitors are fully charged, 173uJ is stored in Ce and 12V is developed across Ce.

Step two:
Energy stored in Ce is given by: 0.5*(Ce)*(V^2)=173uJ
where V = 12V

Hence, Ce=2.403uF

Step three:
If C1=4uF, Ce=2.403uF,

Because (1/Ce)=(1/C1)+(1/C2),

C2 = 6.02uF

** Note that 173uJ originally is the chemical energy stored in the battery; through charging, this energy is convected into electrical energy stored in the electrical field across the capacitor

SECOND QUESTION:

Step one: the maximum capacitance obtained from 1.5uF, 2uF, and 3uF:

of course the max. value is given by the sum of them, i.e. 1.5+2+3 = 6.5uF. Hence, all capacitors are connected in parallel to give the max.

Step two: the minimum capaciatance obtained from 1.5uF, 2uF, and 3uF:
the min. value is given by:

all capacitors are connected in series to give:

1/[(1/1.5u)+(2u)+(3u)] = 0.67uF---minimum value