Thus, the equivalent capacitance C of the
network is
C = C¢ + C4 = 10
3
+10


  mF =13 3mF
(b) Clearly, from the figure, the charge on each of the capacitors, C1,
C2 and C3 is the same, say Q Let the charge on C4 be Q¢ Now, since
the potential difference across AB is  Q/C1, across BC is Q/C2, across
CD is  Q/C3 , we have
1
2
3
500 V
Q
Q
Q
C
C
C
+
+
=