Question

What must be the potential difference across the plates of the capacitor filled with a dielectric so that it stores the same amount of electrical energy as the empty capacitor?

Answer 1

Answer:

$V_2=5.66V$

Explanation:

The electrical energy stored in the empty capacitor is defined as:

$U_0=\frac{C_0V_1^2}{2}$

Where $V_1$ is the potential difference across the plates of the capacitor and $C_0$ is its capacitance.

The capacitance of the capacitor with a dielectric is given by:

$C_2=kC_0(1)$

The electrical energy stored in the capacitor filled with a dielectric is:

$U_2=\frac{C_2V_2^2}{2}$

We have $U_0=U_2$. Thus:

$\frac{C_0V_1^2}{2}=\frac{C_2V_2^2}{2}$

Replacing (1) and solving for $V_2$:

$V_2^2=\frac{C_0V_1^2}{C_2}\\V_2^2=\frac{C_0V_1^2}{(kC_0)}\\V_2^2=\frac{V_1^2}{k}\\V_2=\frac{V_1}{\sqrt{k}}\\V_2=\frac{12V}{\sqrt{4.5}}\\V_2=5.66V$