Step 1: Understanding the Concept:
The capacitance of a parallel-plate capacitor is a measure of its ability to store electric charge.
It depends on the surface area of the plates, the distance between them, and the permittivity of the medium separating the plates.
When an insulating material (dielectric) is placed between the plates, the electric field polarizes the atoms of the dielectric, which reduces the net electric field and potential difference for a given charge, thereby increasing the capacitance.
Step 2: Key Formula or Approach:
The initial capacitance in air (\(C_0\)) is:
\[ C_0 = \frac{\epsilon_0 A}{d} \]
When filled with a dielectric of constant \(K\), the permittivity \(\epsilon_0\) is replaced by \(K\epsilon_0\):
\[ C_{\text{new}} = \frac{K\epsilon_0 A}{d} = K \cdot C_0 \]
Step 3: Detailed Explanation:
According to the problem:
- Initial capacitance, \(C_0 = 8\mu\text{F}\).
- Dielectric constant, \(K = 5\).
Substitute these values into the proportionality relationship:
\[ C_{\text{new}} = 5 \times 8\mu\text{F} \]
\[ C_{\text{new}} = 40\mu\text{F} \]
The dielectric constant \(K\) acts as a scaling factor. Since \(K>1\) for all dielectrics, inserting such a material will always increase the capacitance of the component.
In this specific case, the ability of the capacitor to hold charge at a specific voltage increases five-fold.
Step 4: Final Answer:
The new capacitance value is \(40\mu\text{F}\).