Respuesta :
Answer:
U= [tex]0.112 x10^{-6}[/tex] J
u= 0.708 [tex]\frac{J}{m^{3} }[/tex]
Explanation:
diameter= 2 cm
[tex]r= 1 cm * \frac{1m}{100cm} = 0.01 m[/tex]
distance= 0.5 mm
[tex]d= 0.5 mm * \frac{1m}{1000m}= 0.5 x10^{-3}[/tex]
Area A= [tex]\pi *r^{2}= 0.314x10^{-3} m^{2}[/tex]
Volume v= [tex]A*d = 0.314 x10^{-3}m^{2}*0.5x10^{-3} m =0.157x10^{-6} m^{3}[/tex]
[tex]v= 0.157 x10^{-6} m^{3}[/tex]
Constant vacuum permittivity
[tex]E_{o}= 8.85x10^{-12}[/tex]
a).
[tex]C= \frac{A*E_{o} }{d} = \frac{0.314x10^{-3} *8.85x10^{-12} }{0.5x10^{-3} } \\C= 5.56 x10^{-12}[/tex]F
[tex]U= \frac{1}{2} *C *V^{2}\\ U=\frac{1}{2} * 5.56x10^{-12}*(200)^{2} \\U=0.112 x^{-6} J\\[/tex]
b).
[tex]u=\frac{U}{v}[/tex]
[tex]u=\frac{0.112 x10^{-6}J}{0.157x10^{-6}m^{3} } \\u=0.708 \frac{J}{m^{3} }[/tex]
The total energy stored in the electric field is U= 11.2×10⁻⁸ J
The energy density u= 0.713 J/m³
Electrical energy:
The total electrical energy (U) stored in the capacitor is given by:
[tex]U=\frac{1}{2}CV^2[/tex]
where, C is the capacitance,
and, V is the potential applied
Now, the capacitance (C):
[tex]C=\frac{\epsilon_o A}{d}[/tex]
here, A is the area of the plate,
and, d is the distance between the plates
according to the question:
d = 0.5mm = [tex]5\times10^{-4}m[/tex]
the radius of the plates is r = 2.0cm/2 = 1 cm = [tex]10^{-2}m[/tex]
so, [tex]A=\pi r^2=3.14\times10^{-4}m^2[/tex]
[tex]C=\frac{\epsilon_o A}{d} =\frac{8.85\times10^{-12}\times3.14\times10{-4}}{5\times10^{-4}}=5.56\times10^{-12}F[/tex]
now the energy,
[tex]U=\frac{1}{2}CV^2=\frac{1}{2}\times5.56\times10{-12}\times200\times200J \\ \\ U=11.2\times10^{-8}J[/tex]
The energy density u is the electrical energy stored per unit volume in the capacitor.
The volume of the capacitor,
[tex]v=A\times d\\\\v=3.14\times10^{-4}\times 5\times 10^{-4}m^3=1.57\times10^{-7}m^3[/tex]
The energy density:
[tex]u=\frac{U}{v}=\frac{11.2\times10^{-8}}{1.57\times10^{-7}}J\\ \\ u=0.713J/m^3[/tex]
Learn more about electrical energy:
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