A parallel-plate capacitor has plates with an area of 410 cm2 and an air-filled

Question

A parallel-plate capacitor has plates with an area of 410 cm2 and an air-filled gap between the plates that is 2.10 mm thick. The capacitor is charged by a battery to 550 V and then is disconnected from the battery.

A. How much energy is stored in the capacitor? Units: J

B. The separation between the plates is now increased to 4.10 mm . How much energy is stored in the capacitor now? Units:  J

C. How much work is required to increase the separation of the plates from 2.10 mm to 4.10 mm ? Units:  J

Explanation / Answer

here,

1J = 10^6 uJ

plates area, a = 410 cm^2 = 0.0410 m^2
Volatge, v = 550 V
Distance between the plates, d = 2.10 mm = 0.0021 m

Part A:
Energy Stored in capacitor is given,
E1 = 0.5 * eo * v^2/d * A
E1 = 0.5 * 8.85*10^-12*550^2/(0.0021*0.0410)
E1 = 0.016 J
or
E1 = 16000 mJ

Part B:
d = 0.0041 mm, then

E2 = 0.5 * eo * v^2/d * A
E2 = 0.5 * 8.85*10^-12*550^2/(0.0041*0.0410)
E2 = 0.008 J
or
E2 = 8000 mJ

Then Energy will halfed.

Part C:
Work Done = E1 - E2
w = 16000 - 8000
w = 8000 mJ

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