Two parallel-plate capacitors, 6.0 mF each, are connected in parallel to a 10 V battery. One of the capacitors is then squeezed so that its plate separation is 50.0% of its initial value. Because of the squeezing, (a) how much additional charge is transferred to the ca- pacitors by the battery and (b) what is the increase in the total charge stored on the capacitors?

The additional charge transferred to the capacitors by the battery is 60 μC.

The increase in the total charge stored on the capacitors is 60 μC.

Explanation:

                        C = εоA / d

If the separation is halved, then the capacitance will be doubled and according to the equation q = CV, the charge will be doubled too.

                   q = CV

                      = (6e - 6) (10)

                      = 60 μC

      Final charge of the capacitor:

                   q = (2C)V

                      = (2 6e - 6) (10)

                      = 120 μC

       additional charge transmitted is:

                               q' = 120 - 60

                                   = 60 μC

                                = (C1 + C2) V

                                   = (6 + 6) (10)

                                   = 120 μF

        final total charge:

                                = (C1 + C2) V

                                   = (2 6 + 6) (10)

                                   = 180 μF

        Increase in the charge:

                               q' = 180 - 120

                                   = 60 μC

cant without the picture! ?

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