Respuesta :
Answer:
A. [tex]U_0 = \dfrac{\epsilon_0 A V^2}{2d}[/tex]
B. [tex]U_1 = \dfrac{\epsilon_0 A V^2}{6d}[/tex]
C. [tex]U_2 = \dfrac{K\epsilon_0 A V^2}{2d}[/tex]
Explanation:
The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is
[tex]C=\dfrac{\epsilon A}{d}[/tex]
[tex]C[/tex] is the capacitance, [tex]A[/tex] is the common plate area, [tex]d[/tex] is the plate separation and [tex]\epsilon[/tex] is the permittivity of the material between the plates.
For air or free space, [tex]\epsilon[/tex] is [tex]\epsilon_0[/tex] called the permittivity of free space. In general, [tex]\epsilon=\epsilon_r \epsilon_0[/tex] where [tex]\epsilon_r[/tex] is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, [tex]\epsilon_r=1[/tex].
The energy stored in a capacitor is the average of the product of its charge and voltage.
[tex]U = \dfrac{QV}{2}[/tex]
Its charge, [tex]Q[/tex], is related to its capacitance by [tex]Q=CV[/tex] (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for [tex]U[/tex],
[tex]U = \dfrac{CV^2}{2}[/tex]
A. Substituting for [tex]C[/tex] in [tex]U[/tex],
[tex]U_0 = \dfrac{\epsilon_0 A V^2}{2d}[/tex]
B. When the distance is [tex]3d[/tex],
[tex]U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}[/tex]
[tex]U_1 = \dfrac{\epsilon_0 A V^2}{6d}[/tex]
C. When the distance is restored but with a dielectric material of dielectric constant, [tex]K[/tex], inserted, we have
[tex]U_2 = \dfrac{K\epsilon_0 A V^2}{2d}[/tex]