Answer:
Explanation:
Given that,
We have two capacitors connected in series
C1=2.0-μF
C2=4.0-μF
Then the equivalent of their series connection
1/Ceq = ½ + ¼
1/Ceq= (2+1)/4
1/Ceq=¾
Taking the reciprocal
Ceq= 4/3 μF
The capacitors are connected to a battery of 1kv
V=1000Volts
We know that,
Q=CV
Where Q is charge
C is capacitance and
V is voltage
Then, Q=4/3 ×1000
Q=4000/3 -μC
Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF
After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,
For parallel connection, they have the same voltage but different charges.
When connected in parallel, there is a charge redistribution,
And the total charge will be 2•4000/3=8000/3 -μF
Then, Q1 +Q2= 8000/3 μF
Now the charge on each capacitor will be, let them have a common voltage V
Q=CV
Then, Q1=C1V
Q1= 2×V=2V
Q2= 4×V=4V
Then, Q1+Q2=8000/3
4V+2V=8000/3
6V=8000/3
V=8000/(3×6)
V=4000/9
V=444.44Volts
Now, Q1=2V
Q1=2×4000/9
Q1=8000/9 μF
Also, Q2=4V
Q2=4×4000/9
Q2=16000/9 μF
