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Marina86 [1]
3 years ago
5

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 main

tained to a constant potential difference of 255 V. Determine the charge on the plates and the energy stored by the capacitor.
Physics
1 answer:
Nikolay [14]3 years ago
3 0

Answer:179.79 \times 10^{-9} J

Explanation:

Given

Separation (d)=4 mm

Area of cross-section=25 cm^2

Potential difference=255 V

C=\frac{\epsilon A}{d}=5.53 \times 10^{-12}F

charge (Q)=CV=5.53\times 225=1411.10\times 10^{-12} C

Energy stored=\frac{cv^2}{2}

E=\frac{5.53\times 10^{-12} 255^2}{2}=179.79 \times 10^{-9} J

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