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To calculate the capacitance of a parallel-plate capacitor, the formula used is:
[C = frac{varepsilon_0 cdot A}{d}]
Where:
– (C) is the capacitance in Farads (F),
– (varepsilon_0) is the permittivity of free space ((8.85 times 10^{-12} , text{F/m})),
– (A) is the area of one of the plates in square meters ((text{m}^2)),
– (d) is the separation between the plates in meters (m).
Given values are (A = 20 , text{units}^2) and (d = 5 , text{m}). The units of area aren’t specified as square meters directly, but assuming the “units” meant (text{m}^2) for simplicity:
[C = frac{8.85 times 10^{-12} cdot 20}{5}]
[C = frac{8.85 times 10^{-12} cdot 20}{5} = frac{177 times 10^{-12}}{5}]
[C = 35.4 times 10^{-12} , text{F}]
[C = 35.4 , text{pF}]
The capacitance of the material in
a
Explanation: The capacitance of any material is given by, C = εA/d, where ε = εoεr is the
permittivity in air and the material respectively. Thus C = 1 X 8.854 X 10-12 X 20/5 =
35.36pF