A parallel plate air capacitor has a capacitance \(C\). When it is half filled as shown in the figure with a dielectric constant \(K=5\), the percentage increase in the capacitance is:
Step 1: Understanding the Concept:
When a capacitor is partially filled with a dielectric slab of thickness \(d'\) parallel to the plates, it can be treated as a series combination of two capacitors: one with the dielectric and one with air. Step 2: Key Formula or Approach:
1. Initial capacitance: \(C = \frac{\epsilon_0 A}{d}\)
2. Series capacitance: \(\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}\) Step 3: Detailed Explanation:
Let the plate area be \(A\) and total separation be \(d\).
Initial capacitance \(C_{initial} = \frac{\epsilon_0 A}{d}\).
After filling half the distance (\(d/2\)) with dielectric \(K=5\):
Capacitor 1 (Dielectric part): \(C_1 = \frac{K \epsilon_0 A}{d/2} = \frac{5 \epsilon_0 A}{d/2} = \frac{10 \epsilon_0 A}{d} = 10C\).
Capacitor 2 (Air part): \(C_2 = \frac{\epsilon_0 A}{d/2} = \frac{2 \epsilon_0 A}{d} = 2C\).
The equivalent capacitance \(C_{eq}\) is:
\[ C_{eq} = \frac{C_1 C_2}{C_1 + C_2} = \frac{10C \cdot 2C}{10C + 2C} = \frac{20C^2}{12C} = \frac{5}{3}C \]
The increase in capacitance is \(\Delta C = C_{eq} - C_{initial} = \frac{5}{3}C - C = \frac{2}{3}C\).
Percentage increase:
\[ % \text{ Increase} = \frac{\Delta C}{C_{initial}} \times 100 = \frac{2/3 C}{C} \times 100 = \frac{200}{3} \approx 66.67% \] Step 4: Final Answer:
The percentage increase in capacitance is 66.67%.
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