In basic capacitor theory, select the correct relationship between charge (Q), capacitance (C), and voltage (V).

Introduction:
The most fundamental relation for a capacitor ties stored charge to the applied voltage through the capacitance. Mastering this identity is essential for voltage, energy, and transient calculations in analog and power electronics.

Given Data / Assumptions:

• Ideal capacitor with constant capacitance C over the operating range.
• Steady-state or instantaneous relation (no explicit time dependency needed).
• SI units: Q in coulombs, C in farads, V in volts.

Concept / Approach:
By definition, capacitance is the ratio of stored charge to the voltage across the device: C = Q / V. Rearranging yields Q = C * V and V = Q / C. These forms are interchangeable and are the starting point for many capacitor problems, including energy E = 0.5 * C * V^2 and dynamic current i(t) = C * dv/dt in time-varying signals.

Step-by-Step Solution:
1) Start from the definition: C = Q / V.2) Multiply both sides by V to isolate Q: Q = C * V.3) Recognize the inverse form for voltage: V = Q / C (useful when Q and C are known).4) Apply as needed in design or analysis to compute stored charge or voltage.

Verification / Alternative check:
Dimensional analysis: farad equals coulomb per volt. Therefore C * V has units (C/V) * V = C, matching charge in coulombs. This confirms the correctness of Q = C * V.

Why Other Options Are Wrong:
C = Q * V: Incorrect algebra; this would give units of C = C*V, dimensionally inconsistent.

V = I * R: Ohm law for resistors, unrelated to capacitors.

Q = V / C: Inverted; correct is V = Q / C, not Q = V / C.

Common Pitfalls:
Mixing up Q = C * V and V = Q / C, or trying to apply Ohm law to capacitive behavior in place of the capacitor definition. Always start from C = Q / V to avoid errors.

Final Answer:
Q = C * V