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

Question

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?

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Sapo 5 years 2021-08-22T12:34:06+00:00 1 Answers 473 views 2

Answers ( )

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    2021-08-22T12:35:35+00:00

    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.

    • Initial charge of the capacitor is:

                       q = CV

                          = (6e – 6) \times (10)

                          = 60 μC

          Final charge of the capacitor:

                       q = (2C)V

                          = (2 \times 6e – 6) \times (10)

                          = 120 μC

           additional charge transmitted is:

                                   q’ = 120 – 60

                                       = 60 μC

    • initial total charge:

                                    q_{i} = (C1 + C2) V

                                       = (6 + 6) \times (10)

                                       = 120 μF

            final total charge:

                                   q_{f} = (C1 + C2) V

                                       = (2 \times 6 + 6) \times (10)

                                       = 180 μF

            Increase in the charge:

                                   q’ = 180 – 120

                                       = 60 μC

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