If a sheet of an insulating material is inserted between the plates of an air capacitor, the capacitance will

Analysis

The question asks how the capacitance of an air-filled capacitor changes when an insulating material (a dielectric) is placed between its plates.

Concept Explanation

The capacitance ($C$) of a parallel-plate capacitor is given by the formula:

Where:

• $\varepsilon_0$ is the permittivity of free space (air).
• $A$ is the area of the plates.
• $d$ is the distance between the plates.

When you insert an insulating material (a dielectric) with a dielectric constant ($\varepsilon_r$) greater than 1, the formula becomes:

Since the dielectric constant ($\varepsilon_r$) of any insulating material is always greater than that of air (which is approximately 1), multiplying the original capacitance by $\varepsilon_r$ will result in a larger value.

Step-by-Step Solution

1. Identify the initial state: The capacitor is filled with air, so its capacitance is $C_{air} = \frac{\varepsilon_0 A}{d}$.
2. Introduce the dielectric: When an insulator is inserted, the permittivity of the medium between the plates changes from $\varepsilon_0$ to $\varepsilon = \varepsilon_r \varepsilon_0$.
3. Calculate the new capacitance: The new capacitance $C_{new} = \frac{\varepsilon_r \varepsilon_0 A}{d}$.
4. Compare: Since $\varepsilon_r > 1$, it follows that $C_{new} > C_{air}$. Therefore, the capacitance increases.

Conclusion

Inserting a dielectric material reduces the electric field strength between the plates for a given charge, which allows the capacitor to store more charge at the same potential difference. Thus, the capacitance increases.

Exam Tip: Always remember the mnemonic "D-I-I": Dielectric Insertion Increases capacitance. This will help you quickly recall the relationship in the exam hall!