Problem with solution: Capacitor with and without dielectric in comparison

Since one half of the plate capacitor is filled with a dielectric and the other half with another dielectric, the problem can be considered as a parallel connection of two capacitors, each having a plate area \(A/2\) (because the dielectric fills only half of the capacitor).

In a parallel circuit of capacitors, the total capacitance is the sum of individual capacitances: 1 $$ C ~=~ C_1 + C_2 $$ Here \(C_1\) is the capacitance of one capacitor and \(C_2\) is the capacitance of the other capacitor, which still have to be determined. The capacitance of the first plate capacitor with relative permittivity \(\varepsilon_1\) and plate area \(\frac{A}{2}\) is: 2 $$ C_1 ~=~ \varepsilon_0 \, \varepsilon_1 \, \frac{A}{2d} $$

Similarly, the capacitance of the other capacitor with relative permittivity is \(\varepsilon_2\): 3 $$ C_2 ~=~ \varepsilon_0 \, \varepsilon_2 \, \frac{A}{2d} $$

The total capacitance is therefore the sum of Eq. 2 and Eq. 3: 4 $$ C ~=~ \varepsilon_0 \, \frac{A}{2d} \, ( \varepsilon_1 + \varepsilon_2 ) $$

The voltage and charge are proportional: 5 $$ Q ~=~ C \, U $$

Assuming that the charge \(Q\) is kept constant, the voltage \( U_{\text v} \) on the capacitor with plate area \(A\) and with vacuum in between is given by: 6 \begin{align} U_{\text v} &~=~ \frac{Q}{C} \\\\ &~=~ \frac{d \, Q}{\varepsilon_0 \, A} \end{align}

We only used the formula for the capacitance of the plate capacitor with the relative permittivity = 1 and the area \(A\).

And the voltage \( U_{\text d} \) with the two dielectrics results by substituting the total capacitance 4 into Eq. 5: 7 \begin{align} U_{\text d} &~=~ \frac{Q}{C} \\\\ &~=~ \frac{Q \, d}{\varepsilon_0 \, A} \, \frac{2}{\varepsilon_1 ~+~ \varepsilon_2} \end{align}

The comparison of q>6 and 7 shows that the voltage at the plate capacitor with the two dielectrics differs by the following factor: 8 $$ \frac{ U_{\text v} }{ U_{\text d} } ~=~ \frac{2}{\varepsilon_1 ~+~ \varepsilon_2} $$

The electrical energy \(W\) stored in the plate capacitor is given by: 9 $$ W ~=~ \frac{1}{2} \, C \, U^2 $$

Inserting the capacitance of the plate capacitor with vacuum between the plates (like in exercise b) gives the following energy: 10 $$ W_{\text v} ~=~ \frac{1}{2} \, \varepsilon_0 \, \frac{A}{d} \, U^2 $$

Substitute the capacitance 4 for a capacitor with the two dielectrics into equation 9: 11 $$ W_{\text d} ~=~ \frac{1}{2} \, \varepsilon_0 \, \frac{A}{d} \, \frac{\varepsilon_1 ~+~ \varepsilon_2}{2} \, U^2 $$

The comparison of 10 and 11 shows that the electric energy at the plate capacitor with the two dielectrics differs by the following factor: 12 $$ \frac{ W_{\text v} }{ W_{\text d} } ~=~ \frac{\varepsilon_1 ~+~ \varepsilon_2}{2} $$