Capacitance from charge and voltage — sizing a capacitor from Q and V A capacitor stores 2 coulombs of charge when 500 V is applied across its plates. What is its capacitance?

Introduction / Context:Capacitance quantifies how much charge a device stores per volt. Converting between coulombs, volts, farads, and microfarads is a core skill for power electronics, filtering, and energy-storage calculations.

Given Data / Assumptions:

• Stored charge Q = 2 C.
• Applied voltage V = 500 V.
• Ideal capacitor (no leakage or series resistance considered).

Concept / Approach:Use the defining relation: C = Q / V. Convert the result in farads to microfarads for comparison to options (1 F = 10^6 µF).

Step-by-Step Solution:C = Q / V = 2 / 500 F.C = 0.004 F.Convert to microfarads: 0.004 F * 10^6 µF/F = 4000 µF.

Verification / Alternative check:Back-calc charge: Q = C * V = 0.004 * 500 = 2 C. This matches the given, confirming the size.

Why Other Options Are Wrong:

• 4 µF and 250 µF: underestimates by factors of 1000 and 16 respectively; would only store millisecond-scale energy at this voltage.
• 250 F: unrealistically large here; would store Q = 250 * 500 = 125,000 C, not 2 C.

Common Pitfalls:

• Forgetting to convert F to µF (factor of 10^6).
• Misplacing decimals when dividing Q by V.

Final Answer:4,000 µF