Capacitor charge–voltage relation: when the voltage across a capacitor is tripled, how does the stored charge change?

Introduction / Context:
The fundamental relationship for capacitors links charge, capacitance, and voltage. This is central to energy storage, ADC sample-and-hold design, and timing circuits. Understanding proportional scaling avoids sizing and stress errors in practice.

Given Data / Assumptions:

• Capacitance C is constant (same physical device).
• Voltage V across the capacitor is increased by a factor of 3.
• Ideal capacitor (no leakage or dielectric absorption considered).

Concept / Approach:
The governing relation is Q = C * V, where Q is stored charge. If C is fixed and V is multiplied by 3, the stored charge scales by the same factor. Energy (E = 0.5 * C * V^2) would scale by the square of the factor, but the question asks specifically about charge, not energy.

Step-by-Step Solution:

Verification / Alternative check:
Pick numbers: let C = 1 µF, V = 2 V → Q = 2 µC. Tripling V to 6 V → Q = 6 µC, exactly 3× larger.

Why Other Options Are Wrong:

• Is cut to one-third / doubles / quadruples: incorrect scaling for Q = C * V.
• Stays the same: only true if V or C do not change; V is tripled here.

Common Pitfalls:

• Confusing charge scaling (linear with V) with energy scaling (quadratic with V).
• Assuming non-ideal effects change this first-order proportionality.

Final Answer:
triples