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: Treat distance and dielectric as two separate scaling factors.
Since $C = \dfrac{K\epsilon_0 A}{d}$, doubling the gap alone would scale capacitance by $\dfrac{1}{2}$, while filling it with a dielectric of $K=4$ alone would scale it by $4$.
Step 2: Multiply the two factors together.
\[ \frac{C_{\text{new}}}{C_{\text{old}}} = K\times\frac{d_{\text{old}}}{d_{\text{new}}} = 4\times\frac12 = 2 \]
Step 3: Apply to the given value.
\[ C_{\text{new}} = 2\times12 = 24\ \mu\text{F} \]
\[ \boxed{24\ \mu\text{F}} \]