Answer:
[tex]V_2=5.66V[/tex]
Explanation:
The electrical energy stored in the empty capacitor is defined as:
[tex]U_0=\frac{C_0V_1^2}{2}[/tex]
Where [tex]V_1[/tex] is the potential difference across the plates of the capacitor and [tex]C_0[/tex] is its capacitance.
The capacitance of the capacitor with a dielectric is given by:
[tex]C_2=kC_0(1)[/tex]
The electrical energy stored in the capacitor filled with a dielectric is:
[tex]U_2=\frac{C_2V_2^2}{2}[/tex]
We have [tex]U_0=U_2[/tex]. Thus:
[tex]\frac{C_0V_1^2}{2}=\frac{C_2V_2^2}{2}[/tex]
Replacing (1) and solving for [tex]V_2[/tex]:
[tex]V_2^2=\frac{C_0V_1^2}{C_2}\\V_2^2=\frac{C_0V_1^2}{(kC_0)}\\V_2^2=\frac{V_1^2}{k}\\V_2=\frac{V_1}{\sqrt{k}}\\V_2=\frac{12V}{\sqrt{4.5}}\\V_2=5.66V[/tex]
