Question

An air-filled capacitor of capacitance \( C \) is filled with a dielectric (\( k = 3 \)) of width \( \frac{d}{3} \), where \( d \) is the separation between the plates. The new capacitance is:

• A. \( \dfrac{9}{5}C \)
• B. \( \dfrac{5}{4}C \)
• C. \( \dfrac{4}{3}C \)
• D. \( \dfrac{9}{7}C \)

Hint:

For capacitors partially filled with dielectric:
If layers are along thickness, treat them as series capacitors
Always compare thicknesses carefully
Use the original capacitance \( C = \frac{\varepsilon_0 A}{d} \) to simplify calculations

Answer 1

The Correct Option is D

Step 1: Original capacitance: \[ C = \frac{\varepsilon_0 A}{d} \]
Step 2: The capacitor now has two layers in series:
Dielectric slab of thickness \( \frac{d}{3} \) and dielectric constant \( k = 3 \)
Air slab of thickness \( \frac{2d}{3} \)
Step 3: Capacitance of dielectric part: \[ C_1 = \frac{k\varepsilon_0 A}{d/3} = \frac{3\varepsilon_0 A}{d/3} = \frac{9\varepsilon_0 A}{d} = 9C \]
Step 4: Capacitance of air part: \[ C_2 = \frac{\varepsilon_0 A}{2d/3} = \frac{3\varepsilon_0 A}{2d} = \frac{3}{2}C \]
Step 5: Since the slabs are in series: \[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{9C} + \frac{2}{3C} = \frac{7}{9C} \] \[ \Rightarrow C_{\text{eq}} = \frac{9}{7}C \]