A large (0.3 radius) parallel plate capacitor is designed for operation outside

Question

A large (0.3 radius) parallel plate capacitor is designed for operation outside a submarine in a fresh water lake. It is created by a pair of opposing metal disks. In air they are permanently connected to a 100 V battery and become charged. The charge on the left side is + 8.854 times 10^-12 C and the charge on the right side is - 8.854 times 10^-12 C. The space between the plates is then filled with H_2O as the submarine dives into the lake. Assuming that the water is an insulator, what is the charge q on the capacitor after it dives? TABLE 21.3 Dielectric constants of some materials at 20 degree C

Explanation / Answer

charge in air=Qa=8.854x10^-12 C

volatge= 100V

Q= Ca V

Ca is the capacitance in air

Ca = Q/V= 8.854x10^-12/100

Ca= 8.854x10^-14 F

now when water is between the plates, the capacitance = Cw = K Ca = 80(8.854x10^-14)= 7.08x10^-12 F

voltage= 100 v

so q= cv

charge in water = Qw= Cw x V

Qw= (7.08x10^-12)(100)

Qw=7.08x10^-10 Coloumb

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