Respuesta :
Answer;
= 4.19 Joules
Solution;
Energy stored in capacitor = U = 8.38 J
U =(1/2)CV^2
C =(eo)A/d
C*d=(eo)A=constant
C2d2=C1d1
C2=C1d1/d2
Initial separation between the plates =d1= 2.30mm .
Final separation = d2 = 1.15 mm
But; Energy=U =(1/2)q^2/C
U2C2 = U1C1
U2 =U1C1 /C2
U2 =U1d2/d1
Final energy = Uf = initial energy × d2/d1
= 8.38 ×1.15/2.30
= 4.19 Joules
Thus; The final energy = 4.19 Joules
Energy stored by parallel-plate vacuum capacitor, when the separation is decreased to 1.15 mm for the constant plates on the charge is 4.19 joules.
What is electric potential energy of capacitor?
The electric potential energy of capacitor is the energy stored by it. It can be given as,
[tex]U=\dfrac{1}{2}CV^2[/tex]
Here, (C) is the capacitance and (V) is the potential difference.
As the capacitance is the ratio of area (A) of plate to the distance (d) between them, multiplied with permittivity of dielectric[tex](\varepsilon_0)[/tex].
Thus electric potential energy of capacitor can be given as,
[tex]U=\dfrac{1}{2}\times\dfrac{A}{d}\times\varepsilon_0V^2[/tex]
Given information-
The energy store in a parallel-plate vacuum capacitor is 8.38 j.
The initial distance between the two plates of the capacitor is 2.30 mm.
The final distance between the two plates of the capacitor is 1.15 mm.
The electric potential energy of capacitor when, the distance between the two plates of the capacitor is 2.30 mm.
[tex]8.38=\dfrac{1}{2}\times\dfrac{A}{0.0023}\times\varepsilon_0V^2[/tex]
Ans the electric potential energy of capacitor when, the distance between the two plates of the capacitor is 1.15 mm.
[tex]U=\dfrac{1}{2}\times\dfrac{A}{0.00115}\times\varepsilon_0V^2[/tex]
As the area and potential difference is the same. Thus on comparing the above two equation we get,
[tex]U=8.38\times\dfrac{0.00115}{0.0023}\\U=4.19 J[/tex]
Hence, the energy stored, when the separation is decreased to 1.15 mm for the constant plates on the charge is 4.19 joules.
Learn more about the electric potential energy of capacitor here;
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