A parallel-plate capacitor has capacitance 2.50 mu F. (a) How much energy is sto

Question

A parallel-plate capacitor has capacitance 2.50 mu F. (a) How much energy is stored in the capacitor if it is connected to a 19.00-V battery? (b) If the battery is disconnected and the distance between the charged plates doubled, what is the energy stored? mu j (c) The battery is subsequently reattached to the capacitor, but the plate separation remains as in part(b). How much energy is stored? mu J (a)Determine the capacitance that can be applied to a -filled parallel-plate capacitor having a plate area of 180 cm^2 and insulation thickness of 0.0420

Explanation / Answer

(a)

formula of energy
E = 1/2 (CV^2)
= 0.5(2.50*10^-6)(19^2)
= 451uJ

(b)E=Q^2/2C

C^1=C0/2

E^1=2E=902uJ

(C)E=1/2*C0/2*V^2

=225.5uJ

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