The capacitance of the capacitor is [tex]1.8\cdot 10^{-9}F[/tex]
Explanation:
The capacitance of a parallel-plate capacitor is given by the equation
[tex]C=\frac{k\epsilon_0 A}{d}[/tex]
where
k is the dielectric constant of the medium
[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
For the capacitor in this problem, we have:
k = 2.1 is the dielectric constant
[tex]d=3.5\cdot 10^{-5}m[/tex] is the separation between the plates
[tex]A=0.063m \cdot 0.054m = 0.0034 m^2[/tex] (I assumed that 5.4 m is a typo, since it is not a realistic size for the side of the plate)
Therefore, the capacitance of the capacitor is
[tex]C=\frac{(2.1)(8.85\cdot 10^{-12})(0.0034)}{3.5\cdot 10^{-5}}=1.8\cdot 10^{-9}F[/tex]
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