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3 years ago

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A parallel plate capacitor is created by placing two large square conducting plates of length and width 0.1m facing each other,

separated by a 1 cm gap with vaccum between the plates. We connect the capacitor to power supply, charge it to a potential difference VO 5 kV, and disconnect the power supply. We then insert a sheet of insulating plastic material between the plates, completely filling the space between them. We find that the potential difference decreases to 1 kV while the charge on each capacitor plate remains constant. Find the original capacitance CO. 885pf 17.7pf 8.85pf 1.77pf

1 answer:

borishaifa [10]3 years ago

5 0

Answer:

8.854 pF

Explanation:

side of plate = 0.1 m ,

d = 1 cm = 0.01 m,

V = 5 kV = 5000 V

V' = 1 kV = 1000 V

Let K be the dielectric constant.

So, V' = V / K

K = V / V' = 5000 / 1000 = 5

C = ε0 A / d = 8.854 x 10^-12 x 0.1 x 0.1 / 0.01 = 8.854 x 10^-12 F

C = 8.854 pF

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