Question

A parallel plate capacitor with air between the plates has capacitance \(12 \, \mu \text{F}\). If the distance between the plates is doubled and the space between the plates filled with a dielectric constant 4, find the capacitance of the capacitor.

• A. 24 \(\mu\text{F}\)
• B. 72 \(\mu\text{F}\)
• C. 6 \(\mu\text{F}\)
• D. 12 \(\mu\text{F}\)

Hint:

For parallel plate capacitors with dielectric and distance change: \(C_{\text{new}} = K C_0 (d_0/d)\).

Answer 1

The Correct Option is A

Step 1: Recall parallel plate capacitor formula.
\[ C = \frac{\epsilon_0 A}{d} \quad \text{for air}, \quad C = \frac{K \epsilon_0 A}{d} \quad \text{with dielectric} \]

Step 2: Identify given values.
Initial capacitance \(C_0 = 12 \, \mu\text{F}\), initial distance \(d_0\). After modification: distance doubled \(d = 2 d_0\), dielectric constant \(K = 4\).

Step 3: Apply formula with new values.
\[ C = \frac{K \epsilon_0 A}{d} = \frac{4 \cdot \epsilon_0 A}{2 d_0} = 2 \cdot \frac{\epsilon_0 A}{d_0} = 2 C_0 \]

Step 4: Compute new capacitance.
\[ C = 2 \cdot 12 = 24 \, \mu\text{F} \]

Step 5: Verify reasoning.
Increasing distance decreases capacitance by factor 2, dielectric increases it by factor 4, net factor 2. Result consistent.

Step 6: Final conclusion.
Hence, the new capacitance is:
\[ \boxed{24 \, \mu\text{F}} \]