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Find the capacitance of a parallel-plate capacitor made of two 4.0 cm x 4.0 cm metal plates if the plates are separated by 5.0 mm.Group of answer choices2.8x10-10 F1.0x10-6 F2.8x10-12 F2.0x10-5 F

Sagot :

Given the area is:

A=4.0 cmx4.0 cm

The seperation distance is d=5.0 mm

The capacitance is :

[tex]\begin{gathered} C=\frac{\epsilon_oA}{d} \\ \Rightarrow C=8.85\times10^{-12}\frac{4.0\times10^{-2}\times4.0\times10^{-2}}{5\times10^{-3}} \\ \Rightarrow C=2.83\times10^{-12}F \end{gathered}[/tex]

Thus the answer is

[tex]2.8\times10^{-12}F[/tex]

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