(1)
the resultant capacitance will be
C=c1×c2/(c1+c2)
=0.33 microfarad
the resultant charge comes out to be
(q = CV) = 0.33 x 10^-6×100
q=0.33×10^-4 C
since the charge is same in series.
q1=33.3 micro-coulamb
q2=33.3 micro-coulamb
(2)
The capacitors are in parallel,
(a) equivalent capacitance
C=c1+c2
C=10+15
C=25 microfarad
(b) voltage is same in parallel,
q1=CV
q1=10×10^-6*6
q1=6×10^-5 C
q1=60 micro-coulamb
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