Determination of C1 and C2
Determination of Charge on each capacitor in parallel combination
Energy stored in a capacitor
E=‌

1

2

CV2
In series combination
0.045‌=‌

1

2

‌

C1C2

C1+C2

(100)2
⇒‌‌‌

C1C2

C1+C2

‌=0.09×10−4 ......(i)
In Parallel combination
0.25‌=‌

1

2

(C1+C2)(100)2
⇒‌‌C1+C2‌=0.5×10−4 .......(ii)
On simplifying (i) & (ii)
C1C2‌=0.045×10−8
(C1−C2)2‌=(C1+C2)2−4C1C2
‌=(0.5×10−4)2−4×0.045×10−8
‌=0.25×10−8−0.180×10−8
(C1−C2)2‌=0.07×10−8
(C1−C2)‌=2.6×10−5
‌=0.26×10−4 .......(iii)
From (ii) and (iii) we have
⇒C1=0.38×10−4F
‌ and ‌‌C2=0.12×10−4F
Charges on capacitor C1 and C2 in Parallel combination
Q1‌=C1V=(0.38×10−4×100)
‌=0.38×10−2C
Q2‌=C2V=(0.12×10−4×100)
‌=0.12×10−2C
[Note: If the student writes the relations/ equations
E=‌

1

2

CV2
and ‌‌0.045=‌

1

2

(‌

C1C2

C1+C2

)(100)2
0.25=‌

1

2

(C1+C2)(100)2
But is unable to calculate C1 and C2, award him/her full 2 marks.
Also if the student just writes
Q1=C1V=C1(100)‌ and ‌Q2=C2V=C2(100)
Award him/her one mark for this part of the question.]