Respuesta :
Answer:
[tex]U_c'=\frac{1}{2} \times U_c[/tex] the energy is half of the initial energy.
Explanation:
After the parallel plates capacitor becomes fully charged and it is disconnected from the battery and then the distance between the capacitor plates is is doubled such that there is no leakage of charge.
Energy of a capacitor is given as:
[tex]U_c=\frac{CV^2}{2}[/tex] ...................(1)
where:
C = capacitance of the capacitor
V = potential difference between the capacitor plates
And we also have:
[tex]C=\frac{\epsilon.A}{d}[/tex] ....................(2)
where:
A = area of the capacitor plate
d = distance between the capacitor plate
[tex]\epsilon=[/tex] permittivity of the medium between the plates
When the distance between the plates is doubled then the capacitance decreases to half:
[tex]C'=\frac{\epsilon.A}{2d}\\C' =\frac{1}{2}\times( \frac{\epsilon.A}{d})[/tex]
[tex]C'=\frac{1}{2}\times C[/tex]
Now form eq. (1) the energy of capacitor after the change:
[tex]U'_c=\frac{C'.V^2 }{2}[/tex]
[tex]U'_c=\frac{\frac{C}{2}.V^2 }{2}[/tex]
[tex]U'_c=\frac{1}{2}\times \frac{C.V^2}{2}[/tex]
[tex]U_c'=\frac{1}{2} \times U_c[/tex] the energy is half of the initial energy.
