Respuesta :
The capacitance of a parallel-plate capacitor in vacuum is given by
[tex]C= \frac{A \epsilon_0}{d} [/tex]
where
A is the area of each plate
d is the separation between the plates
[tex]\epsilon_0 = 8.85 \cdot 10^{-12} F/m[/tex] is the electric permittivity in vacuum
For the capacitor in our problem, the plates are circular, with a diameter of d=4.0 cm, so their radius is r=2.0 cm=0.02 m and their area is
[tex]A=\pi r^2 =\pi (0.02 m)^2 = 1.26 \cdot 10^{-3} m^2[/tex]
while the separation between the plates is
[tex]d=0.50 mm=5 \cdot 10^{-4} m[/tex]
Therefore the capacitance is
[tex]C= \frac{A\epsilon_0 }{d}= \frac{(1.26 \cdot 10^{-3} m^2)(8.85 \cdot 10^{-12} F/m)}{5 \cdot 10^{-4}m}=2.23 \cdot 10^{-11}F [/tex]
[tex]C= \frac{A \epsilon_0}{d} [/tex]
where
A is the area of each plate
d is the separation between the plates
[tex]\epsilon_0 = 8.85 \cdot 10^{-12} F/m[/tex] is the electric permittivity in vacuum
For the capacitor in our problem, the plates are circular, with a diameter of d=4.0 cm, so their radius is r=2.0 cm=0.02 m and their area is
[tex]A=\pi r^2 =\pi (0.02 m)^2 = 1.26 \cdot 10^{-3} m^2[/tex]
while the separation between the plates is
[tex]d=0.50 mm=5 \cdot 10^{-4} m[/tex]
Therefore the capacitance is
[tex]C= \frac{A\epsilon_0 }{d}= \frac{(1.26 \cdot 10^{-3} m^2)(8.85 \cdot 10^{-12} F/m)}{5 \cdot 10^{-4}m}=2.23 \cdot 10^{-11}F [/tex]
Capacitance, is the ability of a system to store the electric charge. the capacitance of two 4.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart is [tex]\bold {2.23\times 10^-^1^1 F}[/tex].
The capacitance of a parallel-plate capacitor in vacuum,
Where,
A - Area of plate = [tex]\bold {1.26 \times 10^-^3 m^2}[/tex]
d - distance between plate = 4 cm
[tex]\bold {\epsilon_ 0}[/tex] - electric permittivity - [tex]\bold {8.85 \times 10^-^1^2 Fm^-^1}[/tex]
Put the values in the formula,
[tex]\bold {C = \dfrac {\bold {(1.26 \times 10^-^3 m^2)} \times \bold {(8.85 \times 10^-^1^2 Fm^-^1)} }{5 \times 10^-^4m}}\\\\\bold {C = 2.23\times 10^-^1^1 F}[/tex]
Therefore the capacitance of two 4.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart is [tex]\bold {2.23\times 10^-^1^1 F}[/tex].
To know more about capacitance,
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