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?

Answer 1

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