Charge stored by a 1 µF capacitor when charged to 1 V: Express the stored charge in electrons or coulombs (assume ideal behavior and V = 1 V).

Introduction:The basic capacitor relation Q = C * V connects capacitance, voltage, and stored charge. For a given capacitance, the stored charge scales linearly with the voltage applied. Interpreting this in both coulombs and number of electrons builds intuition for charge magnitude.

Given Data / Assumptions:

• C = 1 µF
• V = 1 V
• Ideal capacitor, no leakage

Concept / Approach:

Compute charge using Q = C * V, then convert coulombs to electrons using 1 electron = 1.602e-19 C (approximately). Alternatively, remember that 1 µC ≈ 6.24 × 10^12 electrons.

Step-by-Step Solution:

Verification / Alternative check:

The known identity 1 µC ≈ 6.24 × 10^12 electrons confirms the computed value directly for V = 1 V and C = 1 µF.

Why Other Options Are Wrong:

• One coulomb of charge: This would require C * V = 1 F * 1 V, not 1 µF.
• One volt: Volt is potential, not stored charge.
• One microampere of current: Current is flow per second, not stored charge.
• 1 microcoulomb of charge: Correct in coulombs but the question’s selected answer asks for an electron count; 6.24 × 10^12 electrons is the electron-equivalent of 1 µC.

Common Pitfalls:

• Forgetting that the charge depends on the applied voltage; here V = 1 V is explicitly assumed.
• Confusing charge (C) with current (A).

Final Answer:

6.24 × 10^12 electrons