# Problem with solution Capacitor with and without dielectric in comparison

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 \(\varepsilon_2\).

- What is the capacitance of the capacitor with the two dielectrics?
- By what factor does the voltage with dielectrics change compared to the voltage without dielectrics?
- 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

## Solution for (a)

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