Answer :
A) [tex]U_0 = \frac{\epsilon_0 A V^2}{2d}[/tex]
B) [tex]U_1=\frac{3\epsilon_0 A V^2}{2d}[/tex]
C) [tex]U_2=\frac{k\epsilon_0 A V^2}{2d}[/tex]
Explanation:
A)
First of all, the capacitance of a parallel-plate capacitor filled with air is given by
[tex]C=\frac{\epsilon_0 A}{d}[/tex] (1)
where
[tex]\epsilon_0[/tex] is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
The energy stored by a capacitor is given by
[tex]U_0 = \frac{1}{2}CV^2[/tex] (2)
where
C is the capacitance
V is the potential difference across the plates
Substituting (1) into (2), we find an expression of the tenergy stored:
[tex]U_0 = \frac{\epsilon_0 A V^2}{2d}[/tex]
B)
In this part of the problem, the capacitor is disconnected from the battery.
This means that now the charge on the capacitor remains constant. The charge can be written as
[tex]Q=CV[/tex]
Since the charge is the same as in part A), we can write it explicitely:
[tex]Q=CV=\frac{\epsilon_0 A V}{d}[/tex] (1)
We can write the energy stored in the capacitor using another equation:
[tex]U=\frac{Q^2}{2C}[/tex] (3)
In this case, the distance between the plates is increased to 3d, so the new capacitance is
[tex]C=\frac{\epsilon_0 A}{3d}[/tex] (2)
Substituting (1) and (2) into (3), we find the new energy stored:
[tex]U_1 = \frac{(\frac{\epsilon_0 A V}{d})^2}{2(\frac{\epsilon_0 A}{3d})}=\frac{3\epsilon_0 A V^2}{2d}[/tex]
3)
In this case, the capacitor is reconnected to the battery, so the potential difference is now equal to the initial potiential difference V.
In this case, however, a dielectric plate is moved inside the space between the plates. Therefore, the capacitance becomes
[tex]C=\frac{k\epsilon_0 A}{d}[/tex]
where
k is the dielectric constant of the dielectric material
To calculate the energy stored, we can use again the original formula
[tex]U=\frac{1}{2}CV^2[/tex]
And substituting C and V, we find
[tex]U_2=\frac{k\epsilon_0 A V^2}{2d}[/tex]