Question

Obtain an expression for (i) potential difference and (ii) capacitance of a parallel plate capacitor filled partly with dielectric material between plates.

Hint:

When dielectric is filled partially, treat it like two capacitors in series.

Answer 1

Step 1: Consider parallel plate capacitor.
Let distance between plates $= d = d_1 + d_2$, where $d_1$ is filled with air (or vacuum, $\varepsilon_0$), and $d_2$ is filled with dielectric of constant $K$.
Step 2: Electric field in each medium.
- In air gap: $E_1 = \dfrac{\sigma}{\varepsilon_0}$
- In dielectric: $E_2 = \dfrac{\sigma}{K\varepsilon_0}$
Step 3: Potential difference across capacitor.
\[ V = E_1 d_1 + E_2 d_2 = \frac{\sigma}{\varepsilon_0} d_1 + \frac{\sigma}{K\varepsilon_0} d_2 \] \[ V = \frac{\sigma}{\varepsilon_0}\left(d_1 + \frac{d_2}{K}\right) \]
Step 4: Capacitance formula.
Charge $Q = \sigma A$. \[ C = \frac{Q}{V} = \frac{\sigma A}{\dfrac{\sigma}{\varepsilon_0}\left(d_1 + \tfrac{d_2}{K}\right)} = \frac{\varepsilon_0 A}{d_1 + \tfrac{d_2}{K}} \]
Step 5: Conclusion.
Thus, $V = \dfrac{\sigma}{\varepsilon_0}(d_1 + d_2/K)$ and $C = \dfrac{\varepsilon_0 A}{d_1 + d_2/K}$.