Lessons · circuits

Electric potential energy: why separating charges takes finite energy

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 3 modeling assumptions (author-asserted)

Two opposite point charges, of magnitudes $q_1 = q_2 = 2\ \mu$ C, are held a separation $R = 2$ cm apart and attract with $F = kq_1q_2/r^2$ . How much energy does it take to pull them fully apart — and why is that energy finite? Drag the cursor outward and watch the potential energy climb toward a ceiling as the area under $1/r^2$ converges. This is the electric twin of escape energy: the same inverse-square area, charge in place of mass.

PE gained reaching

r=2R

0.899 J

Separation energy

kq_1q_2/R

(area to

\infty

)

1.798 J

Force at the start

F(R)=kq_1q_2/R^2

89.9 N

Electric PE is the area under the force–distance curve

\Delta U = \int_R^r F\,dr' = \int_R^r \frac{kq_1q_2}{r'^2}\,dr'

The energy to pull two attracting charges apart is the integral of the Coulomb force over distance — the area under the $F$ – $r$ curve. For a constant force (a uniform field) this is a rectangle ( $qEd$ ); for the inverse-square force of point charges it is the area under a falling curve.

check $\tfrac{d}{dr}\Delta U = F$ ; check $\Delta U = \int_R^r F\,dr'$ ; check $\lim_{r\to\infty}\Delta U = kq_1q_2/R$ ; collapse the uniform-field case to $qEd$

✓ The PE's slope is the force: $U'(r) = F(r)$ — the area's rate of growth is the curve's height. [structural]
✓ The accumulated PE is the area: $\Delta U(r) = \int_R^r F\,dr'$ . [structural]
✓ The area to infinity converges: the energy to fully separate the pair $\to kq_1q_2/R$ is finite as $r\to\infty$ . [structural]
✓ In a uniform field $F\approx qE$ is constant, so $U \approx qEd$ — the area is a rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “It takes infinite energy to completely separate two bound charges, since the electric force reaches out forever.”

The force does reach out forever, but it weakens as $1/r^2$ — and the area under $1/r^2$ from $R$ to infinity is finite. That finite area is the binding energy $kq_1q_2/R$ : pull the charges to infinity and the work you must supply converges to it, exactly as a rocket needs only finite energy $GMm/R$ to escape a planet. (Bringing point charges all the way together, $r\to 0$ , is the end that diverges — not the separation.) Drag the cursor out: the potential energy rises toward a ceiling, not without bound.

Modeling assumptions — author-asserted, disclosed not discharged

Point charges (or spherical charge distributions), so outside them the force is $kq_1q_2/r^2$ .
Opposite charges (attraction); $q_1, q_2$ are magnitudes, and the energy shown is the work supplied to separate them. The potential energy is taken zero at infinite separation.
The pull is quasi-static (no kinetic-energy bookkeeping — this is the potential energy alone); $k \approx 8.99\times10^{9}$ N·m²/C².

The F–r graph, fully annotated

A static rendering (Matplotlib): the shaded area under F is the accumulated integral ΔU, and the slope of ΔU is F. 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):