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.

Respuesta :

Answer:[tex]179.79 \times 10^{-9} J[/tex]

Explanation:

Given

Separation (d)=4 mm

Area of cross-section[tex]=25 cm^2[/tex]

Potential difference=255 V

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

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

Energy stored[tex]=\frac{cv^2}{2}[/tex]

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