8.5 Molecular model of a dielectric  (Page 4/13)

Inserting a dielectric into a capacitor connected to a battery

When a battery of voltage $V_0$ is connected across an empty capacitor of capacitance $C_0$ , the charge on its plates is $Q_0$ , and the electrical field between its plates is $E_0$ . A dielectric of dielectric constant $\kappa$ is inserted between the plates while the battery remains in place , as shown in [link] . (a) Find the capacitance C , the voltage V across the capacitor, and the electrical field E between the plates after the dielectric is inserted. (b) Obtain an expression for the free charge Q on the plates of the filled capacitor and the induced charge $Q_{\text{i}}$ on the dielectric surface in terms of the original plate charge $Q_0$ .

Strategy

We identify the known values: $V_0$ , $C_0$ , $E_0$ , $\kappa$ , and $Q_0$ . Our task is to express the unknown values in terms of these known values.

Solution

(a) The capacitance of the filled capacitor is $C=\kappa C_0$ . Since the battery is always connected to the capacitor plates, the potential difference between them does not change; hence, $V=V_0$ . Because of that, the electrical field in the filled capacitor is the same as the field in the empty capacitor, so we can obtain directly that

(b) For the filled capacitor, the free charge on the plates is

The electrical field E in the filled capacitor is due to the effective charge $Q-Q_{\text{i}}$ ( [link] (b)). Since $E=E_0$ , we have

Solving this equation for $Q_{\text{i}}$ , we obtain for the induced charge

Significance

Notice that for materials with dielectric constants larger than 2 (see [link] ), the induced charge on the surface of dielectric is larger than the charge on the plates of a vacuum capacitor. The opposite is true for gasses like air whose dielectric constant is smaller than 2.

Got questions? Get instant answers now!