a) [tex]3.27\cdot 10^{-3} J[/tex]
b) [tex]11.60\cdot 10^{-3} J[/tex]
c) [tex]8.33\cdot 10^{-3} J[/tex]
Explanation:
a)
The energy stored in a capacitor is given by
[tex]U=\frac{1}{2}CV^2[/tex]
where
C is the capacitance of the capacitor
V is the potential difference across the plates of the capacitor
For the capacitor in this problem, before insering the dielectric, we have:
[tex]C=13.5 \mu F = 13.5\cdot 10^{-6}F[/tex] is its capacitance
V = 22.0 V is the potential difference across it
Therefore, the initial energy stored in the capacitor is:
[tex]U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J[/tex]
b)
After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:
[tex]C'=kC[/tex]
where
k = 3.55 is the dielectric constant of the material
C is the initial capacitance of the capacitor
Therefore, the energy stored now in the capacitor is:
[tex]U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2[/tex]
where:
[tex]C=13.5\cdot 10^{-6}F[/tex] is the initial capacitance
V = 22.0 V is the potential difference across the plate
Substituting, we find:
[tex]U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J[/tex]
C)
The initial energy stored in the capacitor, before the dielectric is inserted, is
[tex]U=3.27\cdot 10^{-3} J[/tex]
The final energy stored in the capacitor, after the dielectric is inserted, is
[tex]U'=11.60\cdot 10^{-3} J[/tex]
Therefore, the change in energy of the capacitor during the insertion is:
[tex]\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J[/tex]
So, the energy of the capacitor has increased by [tex]8.33\cdot 10^{-3} J[/tex]
