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A parallel-plate capacitor in air has a plate separation d = 4 mm and square plates with area A = 25.0 cm^2. The plates are maintained to a constant potential difference of 255 V. Determine the charge on the plates and the energy stored by the capacitor.

User Aromore
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1 Answer

2 votes

Answer:
179.79 * 10^(-9) J

Step-by-step explanation:

Given

Separation (d)=4 mm

Area of cross-section
=25 cm^2

Potential difference=255 V


C=(\epsilon A)/(d)=5.53 * 10^(-12)F

charge
(Q)=CV=5.53* 225=1411.10* 10^(-12) C

Energy stored
=(cv^2)/(2)


E=(5.53* 10^(-12) 255^2)/(2)=179.79 * 10^(-9) J

User Alecto
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