FCC Exam Question: 6A199

If the dielectric constants of a capacitor dielectric material between the capacitor plates were changedfrom 1 to 1:

A. The capacitance would be twice its original value
B. The capacitance would be half its original value
C. The capacitance would decay rapidly
D. None of the above
Correct Answer: A

Explanation: The capacitance of a capacitor is directly proportional to the dielectric constant (also known as relative permittivity, κ) of the material between its plates. The formula for capacitance (C) is C = (κ * ε₀ * A) / d, where ε₀ is the permittivity of free space, A is the plate area, and d is the distance between plates. If the dielectric constant changes from 1 to 2, it means the value of κ has doubled. Consequently, with all other factors (ε₀, A, d) remaining constant, the capacitance (C) will also double. Therefore, the capacitance would be twice its original value. Option B is incorrect because doubling the dielectric constant increases capacitance, not halves it. Option C is incorrect as a change in dielectric constant primarily affects the steady-state value of capacitance, not causing it to decay rapidly.

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Includes Elements 1, 3, 6, 7R, 8, and 9.