Given that
C1 = 2.00 * 10^-6 F
C2 = 5.00 * 10^-6 F
voltage V = 6.50 V
The equivalent capacitance is
       1/C = 1/C1 + 1/C2
        1/C = 1/(
2.00*10^-6 F) + 1/(5.0*10^-6 F)
          C
=1.42857*10^-6 F
So the charge across each capacitance (2.00 µF and 5.00 µF)
is
        Q = CV
             =
(1.42857*10^-6 F)(6.50 V)
           Â
= 9.2857*10^-6 C
          Â
= 9.2857 μC
b)
The charge across capacitor 2.75 µF is
        q1 =C1V =
(2.00*10^-6F)(6.50 V)
                       Â
= 13*10^-6 C
                        Â
= 13.0 μC
The charge across capacitor 5.00 µF is
       q2 = C2V = (5.00 *
10^-6 F)(6.50 V)
                         =
32.510^-6 C
                         =
32.5 μC
I hope help you !!
