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A question shows three or four charges at the corners of a shape and asks for the work needed to put them there. Students often add potentials, or count a pair twice, or drop a minus sign. There is one reliable method: list every pair, write kq₁q₂/r for each, and add. This post explains why it works, then uses it on a triangle, a square and a rearrangement.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus both list "electrical potential energy of a system of two point charges". The CBSE Physics curriculum for 2026-27 has the same wording. Section 2.7 of the NCERT Class 12 Physics textbook (2026-27 reprint) goes further: it derives the result for three charges and works a four-charge square in Example 2.4. So the syllabus names two charges, but the textbook, and therefore exam questions, extend it to more. For potential and dipoles, read our guide to electric potential and dipoles.
Why is it a sum over pairs?
Bring the charges in one at a time from infinity. The first one needs no work, because nothing is there to push against. The second needs kq₁q₂/r₁₂. The third has to be pushed against both: kq₁q₃/r₁₃ + kq₂q₃/r₂₃. NCERT adds these to get
U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃)
and notes that, because the electrostatic force is conservative, the total "is independent of the manner in which the configuration is assembled". Each new charge interacts with every charge already present, so every pair appears exactly once.
For n charges there are n(n − 1)/2 pairs: 3 for three charges, 6 for four, 10 for five.
Use k = 1/(4πε₀) ≈ 9 × 10⁹ N m² C⁻². With the CODATA 2022 value of ε₀ from NIST, 8.8541878188 × 10⁻¹² F m⁻¹, k works out to 8.99 × 10⁹, so 9 × 10⁹ is accurate to about 0.1%.
Worked example 1: a triangle with unlike charges
Charges +q, +q and −2q sit at the corners of an equilateral triangle of side a = 0.10 m, with q = 2 μC. Find the work needed to assemble them.
Three pairs, all at distance a:
(+q, +q): +q²
(+q, −2q): −2q²
(+q, −2q): −2q²
Sum of products: q² − 4q² = −3q². So U = −3kq²/a = −3 × 9 × 10⁹ × (2 × 10⁻⁶)² / 0.10 = −1.08 J.
The negative sign means the system is bound: you would have to supply 1.08 J to pull the charges back apart to infinity. A useful check: the total charge is zero, and the attraction between unlike pairs outweighs the one repulsion, so negative makes sense.
Worked example 2: four equal charges on a square
Four charges of +2 μC sit at the corners of a square of side 0.10 m. NCERT's Example 2.4 uses alternating signs; here they are all equal, which changes the answer.
Six pairs: four along the sides at distance a, and two across the diagonals at distance a√2.
U = (kq²/a)(4 + 2/√2) = (kq²/a)(4 + √2).
kq²/a = 9 × 10⁹ × 4 × 10⁻¹² / 0.10 = 0.36 J, so U = 0.36 × 5.414 = 1.95 J.
Forgetting the two diagonal pairs is the most common error. It would give 1.44 J.
Worked example 3: adding a charge at the centre
How much more work is needed to bring a +1 μC charge from infinity to the centre of that square, with the four corner charges held fixed?
The new charge pairs with each corner charge at distance a/√2 = 0.0707 m. Work = Q × (potential at centre) = Q × 4kq/(a/√2) = 4√2 kqQ/a = 4 × 1.414 × 9 × 10⁹ × 2 × 10⁻⁶ × 1 × 10⁻⁶ / 0.10 = 1.02 J.
Only the four new pairs count. The six existing pairs were already paid for.
Worked example 4: rearranging the triangle
Starting from example 1, the triangle is stretched so its side becomes 0.20 m. How much work must an external agent do?
U scales as 1/a, so the new energy is −1.08/2 = −0.54 J. Work done by the agent = U(final) − U(initial) = −0.54 − (−1.08) = +0.54 J. Positive, because you are pulling a bound system apart.
Which mistakes cost marks?
Counting pairs twice. If you sum over every charge and every other charge, divide by two. Listing pairs avoids this.
Dropping signs. Put each charge in with its sign; the sign of U is physics, not decoration.
Adding potentials and calling it energy. Potential belongs to a point; potential energy belongs to a pair or a system.
Missing diagonal pairs in squares and the long pairs in lines of charges.
A practice routine
Redo examples 1 and 2 from a blank page, starting with a list of pairs.
Solve NCERT Example 2.4 yourself, then compare with the textbook solution.
Try three equal charges on a straight line, spaced 0.10 m apart. Predict first: the outer pair is twice as far apart, so it contributes half as much as each neighbour pair.
Mix in one question per session from our list of common electrostatics mistakes.
If the idea of potential energy belonging to a system rather than to one charge still feels slippery, the Fermi Concept Library covers every concept in physics with its own lesson.
Today, find the energy of five equal charges at the corners of a regular pentagon, starting with a list of all ten pairs.
