When the plate area of a capacitor increases?

Analysis

The question asks about the relationship between the physical dimensions of a parallel-plate capacitor (specifically the plate area) and its capacitance.

Concept Explanation

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

Where:

• $C$ is the capacitance (measured in Farads).
• $\varepsilon_0$ is the permittivity of free space (a constant).
• $\varepsilon_r$ is the relative permittivity (dielectric constant) of the material between the plates.
• $A$ is the area of the overlapping plates.
• $d$ is the distance between the plates.

From this formula, we can see that the capacitance ($C$) is directly proportional to the area ($A$) of the plates. This means that if you increase the area of the plates, you are providing more space for charge to be stored at a given voltage, thereby increasing the capacitance.

Step-by-Step Solution

1. Identify the variables: We are looking at the effect of changing the area ($A$) on the capacitance ($C$).
2. Apply the relationship: Since $C \propto A$, an increase in $A$ must result in an increase in $C$.
3. Evaluate the options: • A is incorrect because voltage rating is determined by the dielectric strength, not just the area. • B is incorrect because it suggests an inverse relationship. • C is incorrect because capacitance is physically dependent on the area. • D is correct because it aligns with the mathematical relationship $C \propto A$.

Conclusion

When the plate area of a capacitor increases, the capacitance increases because there is more surface area available to store electric charge.

Exam Tip: Remember the mnemonic "C-A-D": Capacitance is proportional to Area and inversely proportional to Distance. If you remember this, you will never miss a question about how physical changes affect a capacitor!