Lessons · circuits

Energy in a capacitor: the area under the voltage–charge line

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A $C = 500\ \mu$ F capacitor is charged until it holds $Q = 0.1$ C, reaching $V = 200$ V. How much energy is stored — and why is it $\tfrac12 CV^2$ and not $CV^2$ ? As charge piles on, the voltage climbs as $V = q/C$ ; the energy is the area under that rising line. Drag the cursor and watch the stored energy grow as the triangle under $V(q)$ .

Energy stored

U = \tfrac12 C V^2 = Q^2/2C

(the shaded area)

10 J

Final voltage

V = Q/C

200 V

If the voltage were constant:

W = VQ

(the rectangle — twice the energy)

20 J

Charge delivered

Q = C V

0.1 C

Energy is the area under the voltage–charge line

U = \int_0^Q V\,dq

Pushing a charge $dq$ across the capacitor's voltage $V$ costs energy $V\,dq$ . The total stored energy is the integral of voltage over charge — the area under the $V$ – $q$ graph. Because $V = q/C$ rises from zero, that area is a triangle, not a rectangle.

check $\tfrac{dU}{dq} = V$ ; check $U = \int_0^q V\,dq'$ ; recover $\tfrac12 C V^2$ ; the constant-voltage case is the rectangle $VQ$ (twice the energy)

✓ The energy's slope is the voltage: $U'(q) = V(q)$ — the area's rate of growth is the curve's height. [structural]
✓ The stored energy is exactly the area: $U(q) = \int_0^q V\,dq'$ . [structural]
✓ The memorized $\tfrac12 CV^2$ is the area at full charge: $\tfrac12 C V^2 = U(Q)$ with $V = Q/C$ . [structural]
✓ A battery holding the voltage constant would do $W = VQ$ — a rectangle, twice the stored energy (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “Charging a capacitor to voltage

V

with a battery of EMF

V

stores all the energy the battery delivers.”

The battery delivers $W = VQ$ (it moves charge $Q$ across a fixed voltage $V$ — a rectangle). But the capacitor's voltage rises from zero as it fills, so it stores only the triangle under the $V$ – $q$ line, $\tfrac12 VQ = \tfrac12 CV^2$ . Exactly half the delivered energy is lost as heat in the charging resistance — and that fraction is independent of the resistance. The factor of $\tfrac12$ is the area of a triangle, not a fudge.

Modeling assumptions — author-asserted, disclosed not discharged

Ideal capacitor: fixed capacitance $C$ , no leakage, charge spread uniformly so $V = q/C$ holds throughout.
The 'where did half the energy go' step assumes the charge is delivered through some resistance (any real wire); the dissipated half does not depend on how small that resistance is.

The V–q graph, fully annotated

A static rendering (Matplotlib): the shaded area under V is the accumulated integral U, and the slope of U is V. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):