Charge Equation - Electric Fields, Electrostatics & Capacitors Step by Step

A charge-equation scan exercises Q = C * V in both directions, holding voltage fixed while capacitance moves and then holding capacitance fixed while voltage moves across rows. The arithmetic is exact only for the stated capacitances, voltages, charges, and rows in the ideal capacitor, lumped-element model. Real capacitors add leakage, dielectric absorption, ESR, voltage-rating limits, plate/fringe geometry, meter/probe calibration limits, finite precision, and measurement uncertainty.

The relation Q = C V turns capacitance and voltage into stored charge. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Capacitance converts voltage to charge

Capacitance tells how many coulombs are stored per volt.

Q = C V

Substitute exact values

Use 2 farads and 6 volts.

Q = 2\ \text{F}\cdot 6\ \text{V}

Multiply

The capacitor stores 12 coulombs.

Q = 12\ \text{C}

Capacitance is charge per volt

Hold voltage at 6 volts. Larger capacitance stores more charge at that same voltage.

\begin{array}{c|c|c}C&V&Q\\1\ \text{F}&6\ \text{V}&6\ \text{C}\\2\ \text{F}&6\ \text{V}&12\ \text{C}\\3\ \text{F}&6\ \text{V}&18\ \text{C}\\\end{array}

Voltage is the other direct direction

Now hold capacitance at 2 farads. The charge changes in the same ratio as the voltage.

\begin{array}{c|c|c}C&V&Q\\2\ \text{F}&3\ \text{V}&6\ \text{C}\\2\ \text{F}&6\ \text{V}&12\ \text{C}\\2\ \text{F}&9\ \text{V}&18\ \text{C}\\\end{array}