An electric field of 8.30 x 10^5 V/m is desired between two parallel plates, each of area 31.5 cm^2 and separated by 2.45 mm. There's no dielectric. What charge must be on each plate?

Respuesta :

Answer:

Charge on each plate = 2.31 x 10⁻⁸ C

Explanation:

We have the equations

           [tex]E=\frac{V}{d}\texttt{ and }V=\frac{Q}{C}[/tex]

Combining both equations

           [tex]E=\frac{\left (\frac{Q}{C}\right )}{d}=\frac{Q}{Cd}[/tex]

We also have the equation for capacitance

           [tex]C=\frac{\epsilon A}{d}[/tex]

That is

          [tex]E=\frac{Q}{\frac{\epsilon A}{d}\times d}=\frac{Q}{\epsilon A}\\\\Q=\epsilon AE[/tex]

Substituting

           [tex]Q=8.85\times 10^{-12}\times 31.5\times 10^{-4}\times 8.30\times 10^5=2.31\times 10^{-8}C[/tex]

Charge on each plate = 2.31 x 10⁻⁸ C

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