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Difficulty: MediumCapacitors

In basic electronics, the capacitance of a parallel-plate capacitor is determined by its physical construction and the dielectric material between its plates. Which of the following modifications will increase the capacitance of a parallel-plate capacitor?

  1. A
    Increasing the distance separating the two conductive plates
  2. Increasing the surface area of the conductive platesAnswer
  3. C
    Winding the conductive material into tight coils to store energy in a magnetic field
  4. D
    Operating the component at higher AC frequencies to increase its Farad rating

Answer

Increasing the surface area of the conductive plates increases the capacitance of a parallel-plate capacitor.
Capacitance (CC) of a parallel-plate capacitor is given by C=εAdC = \frac{\varepsilon A}{d}, where ε\varepsilon represents the permittivity of the dielectric material, AA is the surface area of the plates, and dd is the distance between the plates. Because capacitance is directly proportional to plate area, increasing the plate surface area allows more charge to be stored at any given voltage, directly increasing total capacitance.

Step-by-Step Solution

1
Recall the fundamental formula defining the capacitance of a parallel-plate capacitor.
The formula is C=εAdC = \frac{\varepsilon A}{d}, where CC is capacitance, ε\varepsilon is dielectric permittivity, AA is the plate surface area, and dd is the distance between plates.
Identifying the relationship between physical parameters and capacitance establishes which changes increase or decrease the value.
2
Analyze the proportional relationship between plate area and capacitance.
Capacitance is directly proportional to plate surface area (C∝AC \propto A) and inversely proportional to separation distance (C∝1dC \propto \frac{1}{d}).
Because plate area AA appears in the numerator of the formula, expanding AA causes capacitance CC to increase.
3
Identify the correct modification that increases capacitance.
Increasing the surface area of the conductive plates increases the total capacitance.
A larger plate area accommodates a greater quantity of charge for a given electric potential.

Key Concept

Physical factors affecting parallel-plate capacitance (C=εAdC = \frac{\varepsilon A}{d})
Estimated Time:1m 0s
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Topics

Capacitors

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