# Problem with solution Capacitor with and without dielectric in comparison Level 3 requires the basics of vector calculus, differential and integral calculus. Suitable for undergraduates and high school students.

Consider a plate capacitor with plate area $$A$$ and distance $$d$$ between electrodes. Inside the plate capacitor there is half a dielectric with relative permittivity $$\varepsilon_1$$ and the other half a dielectric with relative permittivity $$\varepsilon_2$$.

1. What is the capacitance of the capacitor with the two dielectrics?
2. By what factor does the voltage with dielectrics change compared to the voltage without dielectrics?
3. By what factor does the electrical energy with dielectrics change compared to the energy without dielectrics?
Solution tips

Use the formulas for capacitance, voltage, and electrical energy of the plate capacitor.

## Solution

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$$C ~=~ \varepsilon_0 \, \frac{A}{2d} \, ( \varepsilon_1 + \varepsilon_2 )$$

Solution for (b)

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$$\frac{ U_{\text v} }{ U_{\text d} } ~=~ \frac{2}{\varepsilon_1 ~+~ \varepsilon_2}$$

Solution for (c)

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$$\frac{ W_{\text v} }{ W_{\text d} } ~=~ \frac{\varepsilon_1 ~+~ \varepsilon_2}{2}$$