Question

Derive the formula for the capacitance of a parallel plate capacitor, partially filled with a dielectric.

Hint:

When a dielectric partially fills a capacitor, treat the regions as separate capacitors in series, and apply the series formula to compute net capacitance.

Answer 1

Consider a parallel plate capacitor of plate area \( A \) and separation \( d \). Let the space between the plates be partially filled with a dielectric of dielectric constant \( K \) up to a thickness \( t \), and the remaining space \( (d - t) \) is air (or vacuum).
We can treat the system as two capacitors in series: 1. Capacitor with dielectric of thickness \( t \) and dielectric constant \( K \), 2. Capacitor with air gap of thickness \( d - t \) and dielectric constant 1.
Capacitance of each section:
- With dielectric: \[ C_1 = \frac{K \varepsilon_0 A}{t} \] - With air: \[ C_2 = \frac{\varepsilon_0 A}{d - t} \] Since they are in series, the equivalent capacitance \( C \) is given by:
\[ \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{t}{K \varepsilon_0 A} + \frac{d - t}{\varepsilon_0 A} = \frac{1}{\varepsilon_0 A} \left( \frac{t}{K} + (d - t) \right) \] \[ C = \frac{\varepsilon_0 A}{\left( \frac{t}{K} + d - t \right)} \] Final Result:
\[ \boxed{C = \frac{\varepsilon_0 A}{d - t + \frac{t}{K}}} \]