Question

Given a 2.00 μF capacitor, a 5.00 μF capacitor, and a 6.50 V battery, find the charge on each capacitor if you connect them in the following ways.

(a) in series across the battery

2.00 μF capacitor	...μC
5.00 μF capacitor	... μC

(b) in parallel across the battery

2.00 μF capacitor	...μC
5.00 μF capacitor	...μC

Answer 1

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)

             = 9.2857*10^-6 C

            = 9.2857 μC

b)

The charge across capacitor 2.75 µF is

         q1 =C1V = (2.00*10^-6F)(6.50 V)

                         = 13*10^-6 C

                          = 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 !!