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Electrostatics questions often hand you the potential, as a graph or a formula, and ask for the field. It looks like a calculus question, but most lost marks come from the minus sign, from slopes measured in centimetres, or from assuming that zero field means zero potential. This post works through graphs, formulas and a three-dimensional case, then gives a short practice routine.
Where does E = −dV/dx come from?
The NCERT Class 12 chapter Electrostatic Potential and Capacitance (2026-27 reprint) derives it in section 2.6.1, "Relation between field and potential". Move a unit positive charge a small distance δl against the field from one equipotential surface to the next; the work done is |E|δl, and that equals the potential difference. So |E| = −δV/δl (NCERT Eq. 2.20). NCERT draws two conclusions: the field points in the direction in which the potential decreases steepest, and its size is the change in potential per unit displacement normal to the equipotential surface.
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list electric potential, potential difference and equipotential surfaces, and the CBSE 2026-27 curriculum lists the same. Neither syllabus names this relation as a separate item, but it sits in the NCERT section on equipotentials. Check your year's syllabus.
Why the minus sign?
A positive charge released in a field accelerates from high potential towards low potential, the way a ball rolls downhill. So the field points the way V falls. If V increases along +x, the slope dV/dx is positive and E points along −x. Drop the minus sign and every field you calculate points backwards.
Worked example 1: reading a V–x graph
A graph (sample numbers) shows V rising in a straight line from 0 V at x = 0 to 10 V at x = 2 cm, staying at 10 V until x = 5 cm, then falling in a straight line to −10 V at x = 6 cm.
0 to 2 cm: slope = 10 V ÷ 0.02 m = 500 V/m. E = −500 V/m, so 500 V/m pointing along −x.
2 to 5 cm: flat, so E = 0. The potential is 10 V, not zero.
5 to 6 cm: slope = −20 V ÷ 0.01 m = −2000 V/m. E = +2000 V/m, pointing along +x.
A proton placed at x = 5.5 cm feels F = eE = 1.6 × 10⁻¹⁹ × 2000 ≈ 3.2 × 10⁻¹⁶ N along +x. The steepest part of the graph is where the field is strongest.
Worked example 2: a formula for V(x)
V = 4x³ − 6x² volts, with x in metres (sample function). Find E at x = 0.5 m and x = 2 m, and where E = 0.
dV/dx = 12x² − 12x, so E = −(12x² − 12x) = 12x − 12x².
At x = 0.5 m: E = 6 − 3 = 3 V/m, along +x.
At x = 2 m: E = 24 − 48 = −24 V/m, so 24 V/m along −x. A 2 µC charge there feels 2 × 10⁻⁶ × 24 = 4.8 × 10⁻⁵ N along −x.
E = 0 where 12x(1 − x) = 0: at x = 0 and x = 1 m. These are points of equilibrium for a charge. Whether the equilibrium is stable is a potential-energy question, which our post on potential energy curves and equilibrium covers.
Worked example 3: V in three dimensions
If V depends on x, y and z, take each component separately, holding the other two coordinates fixed:
Ex = −∂V/∂x, Ey = −∂V/∂y, Ez = −∂V/∂z
V = 2x²y − 3z volts (sample function). At the point (1, 2, 1) m:
Ex = −4xy = −8 V/m
Ey = −2x² = −2 V/m
Ez = +3 V/m
|E| = √(64 + 4 + 9) = √77 ≈ 8.77 V/m. The NCERT chapter works only in one dimension, so treat this as stretch practice for harder JEE questions.
Worked example 4: equipotentials
Flat equipotential surfaces are drawn every 10 V, 5 cm apart. The field is uniform, perpendicular to the surfaces, pointing from higher to lower V, and its size is 10 ÷ 0.05 = 200 V/m. Closer spacing means a stronger field.
Going the other way, potential difference is minus the area under an E–x graph: a uniform 200 V/m along +x over 0.1 m means V drops by 20 V.
Which mistakes cost marks?
Losing the minus sign, so the field points up the potential hill.
Slopes in V/cm. Convert to metres; 10 V over 2 cm is 500 V/m, not 5.
Assuming V = 0 where E = 0, or the reverse. At the midpoint of a dipole, V = 0 but E is not; see our post on electric potential of dipoles.
Differentiating with respect to the wrong variable in V(x, y, z).
Reading direction from the sign of V instead of the sign of the slope.
Slopes and fields are easier to connect when you can see both at once. The Fermi Simulation Lab has interactive simulations for ideas that are easier to see than read.
A practice routine
Day 1: Sketch three V–x graphs of your own and draw the matching E–x graph under each.
Day 2: Four V(x) formulas: find E, where E = 0, and the force on a given charge.
Day 3: Two V(x, y, z) questions and two equipotential-spacing questions.
Day 4: Reverse: given E–x graphs, find potential differences from the areas.
Tonight, take V = 5x² − 3x volts and find where the field is zero and which way it points at x = 1 m.
