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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.

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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