Since C2 and C3 are in series, so, C' = 100 pF
Since C1 and C' are parallel
so C" = C1 + C' = 100 + 100 or C" = 200 pF
Since C4 and C" are in series, so net capacitance of the network is

1

C

=

1

C"

+

1

C4

=

1

200

+

1

100

=

1+2

200

or C2 =

200

3

pF = 66.7 pF
Net charge stored on the combination is
Q = CV =

200

3

× 10−12 × 300 = 2 × 10−8 C
As C″ and C4 are in series, so Q" = Q4 = Q or Q" = Q4 = 2 × 10−8 C
and hence V" =

Q"

C"

=

2×10−8C

200×10−12F

= 100 V
and V4 =

Q4

C4

=

2×10−8C

100×10−12F

= 200 V
Since C1 and C' are in parallel, so
V1 = V' = V" or V1 = V' = 100 V
and hence Q1 = C1V1 = 100 × 10−12 × 100 = 1 × 10−8 C
and Q' = C1V1 = 100 × 10−12 × 100 = 1 × 10−8 C
and Q' = C'V' = 100 × 10−12 × 100 = 1 × 10−8 C
Since C2 and C3 are in parallel, so Q2 = Q3 = Q' or Q2 = Q3 = 1 × 10−8 C and hence V2 =

Q2

C2

=

1×10−8C

200×10−12F

= 50 V
and V3 =

Q3

C3

=

1×10−8C

200×10−12F

= 50 V