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"At what points is the electric potential zero?" is a standard question in electrostatics, and NCERT's own example stops at two points on a line. The full answer is a whole surface, and knowing its shape helps you avoid the classic mistakes. This post works a fresh example, finds the surface and shows why a point of zero potential usually has a non-zero field.
Where is this in the syllabus?
The NTA Syllabus for JEE (Main) 2026 lists "electric potential and its calculation for a point charge, electric dipole and system of charges", and the NEET (UG) 2026 syllabus has the same line. The NCERT Class 12 chapter Electrostatic Potential and Capacitance (Reprint 2026-27) works this as Example 2.2, with 3 × 10⁻⁸ C and −2 × 10⁻⁸ C placed 15 cm apart, and finds zero potential at 9 cm and 45 cm from the positive charge. Exercise 2.1 asks the same with different values. We use a new pair here.
How do you set up the condition?
Potential is a scalar, so you add the contributions with their signs, as OpenStax University Physics, section 7.3 does for a system of point charges:
V = kq₁/r₁ + kq₂/r₂ = 0, which means r₁/r₂ = |q₁/q₂|.
Three things follow at once:
Like charges: both terms have the same sign, so V is never zero at a finite distance.
Opposite charges: on the joining line there is one point between them and one outside, on the side of the smaller charge. Beyond the larger charge, the larger charge is always closer and always wins.
Equal and opposite charges: r₁ = r₂, the whole perpendicular bisector plane. Our post on dipole potential and equipotentials covers that case.
Worked example: +8 nC and −2 nC, 30 cm apart
Put +8.0 nC at x = 0 and −2.0 nC at x = 30 cm (sample values). The condition is r₁ = 4r₂.
Between the charges: 8/x = 2/(30 − x), so 240 − 8x = 2x and x = 24 cm. Check: 8/24 − 2/6 = 0.
Outside, beyond the −2 nC charge: 8/x = 2/(x − 30), so 8x − 240 = 2x and x = 40 cm. Check: 8/40 − 2/10 = 0.
Both points are on the side of the smaller charge, as expected.
What does the full zero-potential surface look like?
Every point where the distance to one charge is four times the distance to the other lies on a sphere. Here 24 cm and 40 cm are the two ends of its diameter, so its centre is at x = 32 cm and its radius is 8 cm. It encloses the smaller charge.
Check a point off the line: (32 cm, 8 cm). Then r₁ = √(32² + 8²) = 32.98 cm and r₂ = √(2² + 8²) = 8.25 cm. The ratio is 4.00, so V = 0 there too.
For charges with |q₁/q₂| = m and separation d, the centre is at d m²/(m² − 1) from the larger charge and the radius is d m/(m² − 1). With m = 4 and d = 30 cm: 32 cm and 8 cm. As m approaches 1 the sphere grows until it becomes the flat bisector plane.
Where is the field zero, then?
Field adds as vectors and uses r², so its zero is somewhere else. It must lie outside, beyond the smaller charge: 8/x² = 2/(x − 30)², which gives x = 2(x − 30) and x = 60 cm. The potential there is k(8 × 10⁻⁹/0.60 − 2 × 10⁻⁹/0.30) = 60 V with k = 9.0 × 10⁹ N m²/C², not zero. Our post on placing a third charge in equilibrium uses the same field condition.
Which mistakes do students make?
Using r² for potential. Potential goes as 1/r; field as 1/r².
Dropping signs. The condition needs the signed charges. Searching for zero with two positive charges wastes time.
Looking beyond the bigger charge. There is no zero there.
Stopping at the line. If the question says "points" in space, the answer is the sphere.
Assuming V = 0 means E = 0, or the reverse.
A practice routine
Redo the example with +9 nC and −1 nC, 20 cm apart. (Check: 18 cm and 22.5 cm from the +9 nC charge.)
Find the centre and radius of the zero-potential sphere for that pair.
Find the zero-field point for the same pair and its potential.
Do NCERT Exercise 2.1 and check that both points lie on the smaller charge's side.
If writing the condition is fine but the sign logic keeps tripping you up, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning at each step.
Start with practice step 1 now; it takes ten minutes and covers the line, the sphere and the field.
