Electrostatics > The electric field in the presence of a dielectric > Problem 6.6.26

Statement

$6.6.26.$ A capacitor is filled with a dielectric and charged to a potential difference $V$. The plates are connected to each other for a very short time. When the potential difference has decreased by a factor of three, the plates are disconnected. After that, the potential difference slowly increases to $2/3$ of its initial value. How can this effect be explained? Find the dielectric permittivity of the substance filling the capacitor.

Solution

The operation of the capacitor is based on two effects. In an "empty" capacitor (without a dielectric), the field inside is created by the accumulation of charge on the plates. The capacitance of such a capacitor is
$$
C_1 = \frac{\varepsilon_0 S}{d} = C_0.
$$
If a dielectric is inserted between the plates, part of the field is due to its polarisation – the aligned orientation of elementary dipoles in the medium. The capacitance of such a capacitor is
$$
C_2 = \varepsilon C_0.
$$
The idea of the problem is that when the capacitor is short‑circuited, the free charges flow almost instantaneously, while the reorientation of the dipoles in the dielectric requires more time.

The initial field strength in the capacitor:
$$
E_0 = \frac{V}{d} = \frac{\sigma_0}{\varepsilon \varepsilon_0}.
$$
The initial free charge density on the plates:
$$
\sigma_0 = \varepsilon \varepsilon_0 E_0.
$$
The density of bound (polarisation) charge on the surface of the dielectric at the initial moment:
$$
\sigma_{p} = P = (\varepsilon - 1)\varepsilon_0 E_0.
$$

When short‑circuited, the free charges redistribute almost instantaneously, while the bound charges do not have time to change. Let the free charge density after this be $\sigma_1$. According to the condition, the potential difference has decreased by a factor of three, so the field strength becomes
$$
E_1 = \frac{V/3}{d} = \frac{E_0}{3}.
$$
The field inside the dielectric is determined as the sum of the fields of the free and bound charges. Remembering that these fields are oppositely directed:
$$
E_1 = \frac{\sigma_1 - \sigma_{p}}{\varepsilon_0}.
$$
From this, the free charge remaining on the plates is
$$
\sigma_1 = \varepsilon_0 E_0 \left(\varepsilon - \frac{2}{3}\right).
$$

After the plates are disconnected, the free charge $\sigma_1$ remains constant. However, the dipoles of the dielectric gradually begin to disorient, the bound charge $\sigma_{p}$ decreases, and the field inside the capacitor increases. According to the condition, when the polarisation has fully adjusted to the new field, the potential difference reaches $2V/3$. Then the final field strength is
$$
E_2 = \frac{2E_0}{3}.
$$
In the equilibrium state, the following holds:
$$
\sigma_1 = \varepsilon \varepsilon_0 E_2 = \varepsilon \varepsilon_0 \frac{2E_0}{3}.
$$

Equating the two expressions for $\sigma_1$:
$$
\varepsilon_0 E_0 \left(\varepsilon - \frac{2}{3}\right) = \varepsilon \varepsilon_0 \frac{2E_0}{3},
$$
$$
\varepsilon - \frac{2}{3} = \frac{2}{3}\varepsilon \quad\Rightarrow\quad \varepsilon = 2.
$$

Answer:

The dipole moments in the dielectric align in the electric field with a delay; $\varepsilon = 2$.