Capacitance increases with ________

Question

Examoogle · Accepted Answer

Increase in plate area

Explanation: Capacitance is defined as the ability of a system to store an electric charge per unit potential difference. For a parallel plate capacitor, it is directly proportional to the surface area of the conductive plates.

Working Principle: The capacitance $C$ of a parallel plate capacitor is determined by the geometry and the dielectric medium. When the surface area $A$ of the plates is increased, there is more space to accommodate charge carriers at a given electric field intensity. Consequently, for a constant voltage, a larger area allows more charge $Q$ to be stored, resulting in higher capacitance.

Key Formulas: $C = \frac{\varepsilon A}{d}$ — where $C$ is capacitance, $\varepsilon$ is permittivity, $A$ is plate area, and $d$ is distance between plates.; $Q = CV$ — charge stored as a function of capacitance and potential difference.

Key Points: Capacitance is directly proportional to the area of overlap between plates ($C \propto A$).; Capacitance is inversely proportional to the separation distance between plates ($C \propto \frac{1}{d}$).; Increasing the dielectric constant ($\varepsilon_r$) of the material between the plates also increases capacitance.

Advantages: Higher energy density with larger plate area.; Easier control of capacitance values during manufacturing by adjusting physical dimensions.

Disadvantages: Increased plate area leads to larger physical dimensions of the component.; Higher capacitance can sometimes lead to slower charging/discharging time constants ($RC$ circuit).

Applications: Filtering and smoothing circuits in power supplies.; Energy storage in pulse power systems.; Tuning circuits in radio frequency applications.

Additional Info: The SI unit for capacitance is the Farad (F).. Option B is incorrect because a decrease in area results in a decrease in charge storage capacity.. Option C is incorrect because increasing distance $d$ increases the denominator, thereby decreasing $C$.. Option D is irrelevant as capacitance depends on dielectric constant (permittivity), not simply density.

Answer: A is correct — Capacitance is directly proportional to the area of the capacitor plates, so increasing the area increases the capacitance.

Exam Tip: Always remember the mnemonic 'Area is Up, Capacitance is Up; Distance is Up, Capacitance is Down'.

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