A 2.5 μF capacitor consists of 2 large, thin metal plates separated by a plexiglas sheet. The plexiglas sheet has a dielectric constant of 3.4. The capacitor is connected to a 60 V battery and fully charged. Then, whilst still connected to the battery, the plexiglas sheet is removed (assume that the spacing between the metal plates does not change). Calculate the amount of charge that flows to (or from) the capacitor plates when the sheet is removed. Additional charge flowing onto the plates is represented with a + sign and charge leaving the plates is represented with a - sign.

Q / Q₀ = 3.4 ,      ΔQ = 31.14 10⁻⁶ C

Explanation:

For this exercise we will calculate the charge of the capacitor with and without dielectric

the capacitance is given by

          C = Q /ΔV

we also use the ratio of capacitances

          C = k C₀

           Q /ΔV = kQ₀ /ΔV

           Q = k Q₀

            Q / Q₀ = k

            Q / Q₀ = 3.4

If we want a numerical value

           C = k Q / DV

           Q = C ΔV / k

            Q = 2.5 10⁻⁶ 60 / 3.4

            Q = 44.12 10⁻⁶ C

whereby the charge without dielectric is

          Q₀ = Q / 3.4

          Q₀ = 44.12 / 3.4 10⁻⁶

          Q₀ = 12.98 10⁻⁶ C

therefore the charge when removing the dielectric is reduced, the amount of charge that flows is

          ΔQ = Q - Q₀

          ΔQ = (44.12 - 12.98) 10⁻⁶

          ΔQ = 31.14 10⁻⁶ C

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explanation: