Effect of decreasing dielectric constant (relative permittivity k) on a capacitor’s capacitance, with geometry unchanged

Introduction:
Capacitance depends on the physical geometry and the dielectric material between the plates. This question examines how changing the dielectric constant alone affects capacitance when plate area and separation remain fixed.

Given Data / Assumptions:

• Parallel-plate model is applicable.
• Plate area A and separation d are unchanged.
• Dielectric constant (relative permittivity) k is decreased.

Concept / Approach:
The basic relation for a parallel-plate capacitor is C = k * epsilon_0 * A / d, where epsilon_0 is the permittivity of free space and k is the relative permittivity of the dielectric. Capacitance is directly proportional to k; therefore, reducing k reduces C linearly when A and d are fixed.

Step-by-Step Solution:

Verification / Alternative check:
Practical observation: inserting a material with higher k (e.g., ceramic with high-k dielectric) increases capacitance for the same size, while switching to a lower-k material decreases it. Measurements align with the formula above.

Why Other Options Are Wrong:

• Increase: Opposite to the formula's direct proportionality to k.
• Remain the same: Would require k to be unchanged or A/d to counterbalance.
• Be destroyed: Changing k does not inherently damage the capacitor.

Common Pitfalls:
Confusing dielectric constant with dielectric strength, or assuming geometry changes when only material changes. Also, mixing AC breakdown behavior with static capacitance value.

Final Answer:
Decrease.