Fixed Charge Energy Inverse Scan - Electric Fields, Electrostatics & Capacitors Step by Step

At fixed charge, increasing capacitance lowers the voltage needed to hold that charge, so the stored energy also falls in the inverse ledger. The exact rows assume an ideal capacitor and fixed charge source; real measurements depend on dielectric absorption, leakage, voltage probes, capacitance tolerance, charge injection, and uncertainty in the stated inputs.

At fixed stored charge, larger capacitance lowers voltage and stored energy through exact inverse rows. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Hold charge fixed and change capacitance

Keep stored charge fixed at 12 coulombs. A larger capacitance needs less voltage for that same charge, so the stored energy falls.

V=\frac{Q}{C}\qquad U=\tfrac{1}{2}QV

More capacitance lowers energy at fixed charge

Each row first solves voltage from Q divided by C, then computes one-half Q V. This makes the inverse dependence on capacitance visible without decimals.

\begin{array}{c|c|c|c}Q&C&V&U\\12\ \text{C}&2\ \text{F}&6\ \text{V}&36\ \text{J}\\12\ \text{C}&3\ \text{F}&4\ \text{V}&24\ \text{J}\\12\ \text{C}&6\ \text{F}&2\ \text{V}&12\ \text{J}\\\end{array}

The same result is Q squared over two C

Substituting voltage equals Q over C into one-half Q V gives the fixed-charge route. The capacitance sits in the denominator.

U=\frac{Q^{2}}{2C}

The middle row closes both energy routes

For the checked middle row, the charge, capacitance, voltage, and energy all agree before rendering.

Q=12\ \text{C}\qquad C=3\ \text{F}\qquad U=24\ \text{J}