Step 1: Understanding the Concept:
The capacitance (\(C\)) of a parallel plate capacitor is a measure of its ability to store charge.
It is determined by physical factors: the area of overlap between the plates (\(A\)), the distance separating the plates (\(d\)), and the permittivity of the material between them.
When a dielectric material is introduced, it polarizes under the electric field, creating an internal field that opposes the external one. This effectively increases the charge-storing capacity.
Step 2: Key Formula or Approach:
The general formula for the capacitance is:
\[ C = \frac{K \epsilon_0 A}{d} \]
Where \(K\) is the dielectric constant (\(K=1\) for air), \(\epsilon_0\) is the permittivity of free space.
Initial state: \(C_0 = \frac{\epsilon_0 A}{d} = 6 \text{ pF}\).
Step 3: Detailed Explanation:
According to the problem, two specific changes are implemented:
1. The distance between the plates is reduced to half: \(d' = \frac{d}{2}\).
2. The dielectric constant is increased to \(K = 5\).
The new capacitance (\(C'\)) is:
\[ C' = \frac{K \epsilon_0 A}{d'} \]
Substitute the values of \(K\) and \(d'\):
\[ C' = \frac{5 \cdot \epsilon_0 A}{d/2} \]
Using the property of fractions:
\[ C' = 5 \cdot 2 \cdot \left( \frac{\epsilon_0 A}{d} \right) \]
We recognize that the term in the parenthesis is the original capacitance \(C_0\):
\[ C' = 10 \cdot C_0 \]
Given \(C_0 = 6 \text{ pF}\):
\[ C' = 10 \times 6 = 60 \text{ pF}. \]
This significant increase (10 times) occurs because reducing the distance doubles the capacity and the dielectric increases it by a factor of five.
Step 4: Final Answer:
The resulting capacitance is 60 pF.