Step 1: Understanding the Concept:
A capacitor is a passive electronic component consisting of two conducting plates separated by an insulating medium, used to store electrical energy in the form of an electrostatic field. The ability of a capacitor to store charge is called capacitance (\(C\)). This property is primarily determined by the physical geometry of the device, such as the area of the plates (\(A\)) and the distance between them (\(d\)), as well as the electrical properties of the material placed between the plates. When there is a vacuum or air between the plates, the capacitance is at its base level. However, when an insulating material, known as a dielectric, is inserted between the plates, the capacity of the system to store charge at a given potential difference significantly increases. The factor by which the capacitance increases is known as the dielectric constant (\(K\)) of the material.
Step 2: Key Formula or Approach:
The formula for the capacitance of a parallel plate capacitor in a vacuum is:
\[ C_{0} = \frac{\epsilon_{0}A}{d} \]
Where \(\epsilon_{0}\) is the permittivity of free space. When a dielectric material with a constant \(K\) is introduced to completely fill the space between the plates, the permittivity of the medium changes from \(\epsilon_{0}\) to \(\epsilon = K\epsilon_{0}\). The new capacitance \(C'\) is expressed as:
\[ C' = \frac{K\epsilon_{0}A}{d} = KC_{0} \]
Step 3: Detailed Explanation:
To understand why the capacitance increases, we must look at the molecular level behavior of the dielectric material. When a dielectric is placed in the electric field (\(E_{0}\)) generated by the charged plates of the capacitor, the molecules of the dielectric undergo polarization. Even if the molecules are non-polar, the external field causes a slight shift in the electron cloud relative to the nucleus, creating induced dipoles. If the molecules are polar, they tend to align themselves with the field.
This alignment or polarization creates a net "induced" surface charge on the faces of the dielectric that are in contact with the capacitor plates. These induced charges create an internal electric field (\(E_{ind}\)) that acts in the opposite direction to the external field (\(E_{0}\)). As a result, the net electric field (\(E_{net}\)) inside the dielectric is reduced:
\[ E_{net} = E_{0} - E_{ind} = \frac{E_{0}}{K} \]
Since the potential difference (\(V\)) between the plates is the product of the electric field and the separation distance (\(V = E \cdot d\)), the new potential difference \(V'\) becomes:
\[ V' = \frac{E_{0}}{K} \cdot d = \frac{V_{0}}{K} \]
By definition, capacitance is the ratio of stored charge to the potential difference (\(C = Q/V\)). If the capacitor is isolated (the charge \(Q\) remains constant), the decrease in potential difference leads to an increase in capacitance:
\[ C' = \frac{Q}{V'} = \frac{Q}{V_{0}/K} = K \left( \frac{Q}{V_{0}} \right) = KC_{0} \]
In this specific problem, we are given a dielectric constant of \(K = 3\). Substituting this value into our derivation, we find:
\[ C' = 3 \times C = 3C \]
This means the new capacitor can store three times as much charge for the same voltage compared to the original vacuum capacitor.
Step 4: Final Answer:
The new capacitance of the capacitor when completely filled with a dielectric constant of 3 is \(3C\).