Mistake Master

Electric Potential

Electric potential is potential energy per unit charge, $V = U/q_0$, measured in volts, which are joules per coulomb. For a point charge $V = kq/r$, with the sign of $q$ kept, and several charges superpose by plain signed addition of $kq_i/r_i$. The field is recovered as the negative gradient, $E_x = -dV/dx$: the field is the SLOPE of the potential. Surfaces of constant $V$ are equipotentials, motion along them costs no work, and field lines pierce them at right angles. Because the electrostatic field is conservative, $\Delta V$ between two points depends only on the endpoints and never on the route.

Five errors dominate. Treating $V = 0$ and $E = 0$ as the same claim, when midway between $+q$ and $-q$ the potential is zero and the field is strong, and midway between two $+q$ the field is zero and the potential is large. Resolving potentials into components or letting two contributions partially cancel by geometry, when the only cancellation available to a scalar is between opposite signs. Exporting $E = V/d$ out of the uniform field it was derived in, which the point charge disguises because $E = V/r$ happens to hold there. Charging work for motion along an equipotential, or drawing field lines that cross contours obliquely. And letting the potential difference depend on which route was taken.

a zero of V is a CROSSING. a zero of E is a FLAT SPOT. 0 steep slope E = 8kq/d² V +q … −q : V = 0 at the midpoint and the field there is at its strongest 0 zero slope: E = 0 V +q … +q : E = 0 at the midpoint and V there is 4kq/d, far from zero
Left, the potential crosses zero on its steepest run. Right, the potential is flat at a value that is not zero. Reading the value cannot tell you the slope, and reading the slope cannot tell you the value.
every crossing is 90 degrees, or the map is wrong + correct: radial lines, 90° at every contour + 60° wrong: an oblique crossing gives E a component along the contour, so V would not be constant on it
An oblique crossing is self-contradictory: a field component along the contour would change V while moving along it, and then the contour is not a contour.

The work

Lesson live · diagnostic and drills coming soon
Lesson
Electric Potential

Adds potentials as signed scalars with no components, recovers the field as the negative slope of V rather than as V over a length, and settles equipotential work and path independence.

Skill check · 10 scenarios
Diagnostic
10-item topic check

Ten items spanning the failure modes of this topic: tying the zero of V to the zero of E, giving a scalar potential components and angles, exporting E = V/d out of the uniform field, charging work for motion along an equipotential, and making the voltage depend on the route. Take it cold to find which one is yours, or after the lesson to confirm it is not.

Not yet available · 10 items
Targeted Practice
Drill a single misconception

Pick one of the failure modes you missed and drill it on its own. The round is adaptive: two correct in a row clears it for now and moves you to the next. Two in a row is a checkpoint, not proof: if the error resurfaces later, the misconception comes back.

Unlocks from the diagnostic, which is not published yet