Formula
PE of two-charge system
U = (1/4πε₀)(q₁q₂/r). Positive when same-sign charges (repulsive interaction needs work to assemble); negative when opposite signs.
-- NCERT Class 12 Physics, Ch. 2, p. 56Electrical Potential Energy Two Charges
8 MCQs9-step worked example
Source: NCERT ElectrostaticsOfficial key: NTA-verifiedLast updated: 21 Sep 2026
Electrical Potential Energy Two Charges, explained for NEET
The trap: dropping the signs of the charges and reporting the potential energy as a positive number, or dividing by r² as if it were the Coulomb force. Potential energy of two charges goes as 1/r, and its sign carries physics.
The result. NCERT Class 12 Physics Part I, Chapter 2, Section 2.7 builds the system from scratch: charges q₁ and q₂ start at infinity. Bringing q₁ in needs no work (there is no field yet). Bringing q₂ in needs q₂ times the potential of q₁ at that point. Since the electrostatic force is conservative, this work is stored as potential energy:
U = (1/4πε₀) q₁q₂ / r₁₂ (NCERT Chapter 2, Eq. 2.22, page 56)
with r₁₂ the distance between the charges and U = 0 when they are infinitely far apart. NCERT also notes (page 56) that U is the same whichever charge is brought first: work done by the electrostatic force does not depend on the path.
Reading the sign (NCERT page 56). Substitute the charges with their signs.
- Like charges (q₁q₂ > 0): the force is repulsive, positive work must be done to push them together, so U is positive.
- Unlike charges (q₁q₂ < 0): the force is attractive, a negative amount of work is needed to bring them in from infinity, so U is negative.
Bridge to NEET. Two question types recur: find U for given charges and separation, or find the work needed to change the configuration. Work done by an external agency = U(final) − U(initial). Separating the charges to infinity makes U(final) = 0, so the work required is −U. For a bound pair of unlike charges, U is negative and the work required is positive (NCERT Example 2.5, page 60).
Watch out: convert μC and cm to C and m before substituting. And keep this separate from qV(r), the energy of a charge in an external field (NCERT page 59): there V(r) is due to external charges, not due to the charge itself (Points to Ponder, page 78).
Can you answer these Electrical Potential Energy Two Charges MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Taking the potential energy to be zero when the charges are infinitely far apart, the electrostatic potential energy of a system of two point charges q₁ and q₂ separated by a distance r in vacuum is:
Show answer and why every option is right or wrong
Answer: A. NCERT Class 12 Physics Part I, Chapter 2, Eq. 2.22 (page 56): U = q₁q₂ / (4πε₀ r₁₂), directly proportional to the product of the charges and inversely proportional to the distance between them.
Why B is wrong: B is wrong because it divides by r², which is the distance dependence of the force between the charges, not of their potential energy.
Why C is wrong: C is wrong because it adds the charges; the energy depends on the product q₁q₂, which is why its sign follows the signs of the two charges.
Why D is wrong: D is wrong because it multiplies by r, which would make the energy grow as the charges move apart; U varies as 1/r.
MCQ 2Easy RecallPractice
Two point charges of opposite sign are held at a finite distance apart, with potential energy taken as zero at infinite separation. According to NCERT, the electrostatic potential energy of this system is:
Show answer and why every option is right or wrong
Answer: D. NCERT Class 12 Physics Part I, Chapter 2, page 56: for unlike charges (q₁q₂ < 0) the force is attractive, a negative amount of work is needed to bring them from infinity to their locations, so the potential energy is negative.
Why A is wrong: A is wrong because positive work is needed only for like charges, which repel; for unlike charges the work needed to bring them in from infinity is negative.
Why B is wrong: B is wrong because NCERT states the potential energy is the same whichever charge is brought first; electrostatic work is path independent.
Why C is wrong: C is wrong because U depends on the product q₁q₂, not on the sum of the charges; the product of a positive and a negative charge is negative, not zero.
MCQ 3Easy RecallPractice
A system of two point charges is assembled by bringing q₁ from infinity first and then q₂. If instead q₂ were brought first and q₁ second, to the same final positions, the potential energy of the system would be:
Show answer and why every option is right or wrong
Answer: B. NCERT Class 12 Physics Part I, Chapter 2, page 56: if q₂ is brought first and q₁ later, U is the same; Eq. 2.22 is unaltered whatever way the charges are brought, because work done by the electrostatic force is path independent.
Why A is wrong: A is wrong because reversing the order does not count the interaction twice; the first charge still needs no work and the second still needs (charge) × (potential of the first).
Why C is wrong: C is wrong because the sign of U is fixed by the sign of q₁q₂, and the product is the same in either order.
Why D is wrong: D is wrong because the electrostatic force is conservative, so the stored energy depends only on the final configuration, not on the paths or order.
MCQ 4Direct ApplicationPractice
Two point charges, +4.0 μC and +5.0 μC, are 0.60 m apart in vacuum. Taking 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻², their electrostatic potential energy is:
Show answer and why every option is right or wrong
Answer: A. U = 9.0 × 10⁹ × (4.0 × 10⁻⁶)(5.0 × 10⁻⁶) / 0.60 = 0.18 / 0.60 = 0.30 J. Both charges are positive, so U is positive (NCERT Chapter 2, Eq. 2.22 and page 56).
Why B is wrong: B is wrong because 0.18 J is k q₁q₂ before dividing by the separation; the division by r = 0.60 m was left out.
Why C is wrong: C is wrong because 0.50 J comes from dividing 0.18 by r² = 0.36 instead of by r = 0.60.
Why D is wrong: D is wrong because it attaches a negative sign to a pair of like charges; q₁q₂ > 0 here, so U is positive.
MCQ 5Direct ApplicationPractice
A +4.0 nC charge and a −5.0 nC charge are 2.0 cm apart in vacuum. Taking 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻², the electrostatic potential energy of the pair is:
Show answer and why every option is right or wrong
Answer: B. U = 9.0 × 10⁹ × (4.0 × 10⁻⁹)(−5.0 × 10⁻⁹) / 0.020 = −1.8 × 10⁻⁷ / 0.020 = −9.0 × 10⁻⁶ J. Unlike charges give negative U (NCERT Chapter 2, Eq. 2.22 and page 56).
Why A is wrong: A is wrong because −4.5 × 10⁻⁴ J comes from dividing −1.8 × 10⁻⁷ by r² = 4.0 × 10⁻⁴ m² instead of by r = 0.020 m.
Why C is wrong: C is wrong because −9.0 × 10⁻⁸ J comes from dividing by 2.0 (the distance left in centimetres) instead of 0.020 m.
Why D is wrong: D is wrong because it drops the sign of the −5.0 nC charge; the product q₁q₂ is negative, so U must be negative.
MCQ 6Direct ApplicationPractice
Two point charges held a distance r apart have electrostatic potential energy 0.24 J. If their separation is increased to 3r, their potential energy becomes:
Show answer and why every option is right or wrong
Answer: D. U is proportional to 1/r (NCERT Chapter 2, Eq. 2.22, page 56). Tripling r divides U by 3: 0.24 J / 3 = 0.080 J.
Why A is wrong: A is wrong because 0.72 J = 0.24 × 3 treats U as proportional to r; U varies inversely with the separation.
Why B is wrong: B is wrong because it assumes U does not depend on the separation; for the same charges U changes as 1/r.
Why C is wrong: C is wrong because 0.027 J ≈ 0.24 / 9 uses a 1/r² dependence, which belongs to the force, not to the potential energy.
MCQ 7CalculationPractice
A +3.0 μC charge and a −6.0 μC charge are held 0.45 m apart in vacuum with no external field. Taking 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻², how much work must an external agency do to separate them to infinite distance (charges at rest at the start and the end)?
Show answer and why every option is right or wrong
Answer: C. Step 1: U = 9.0 × 10⁹ × (3.0 × 10⁻⁶)(−6.0 × 10⁻⁶) / 0.45 = −0.162 / 0.45 = −0.36 J. Step 2: W = U(final) − U(initial) = 0 − (−0.36 J) = 0.36 J (same method as NCERT Chapter 2, Example 2.5(b), page 60).
Why A is wrong: A is wrong because −0.36 J is the potential energy U itself; the work required to separate the charges is 0 − U, which reverses the sign.
Why B is wrong: B is wrong because 0.80 J comes from dividing −0.162 by r² = 0.2025 m² (giving U = −0.80 J) and then taking 0 − U.
Why D is wrong: D is wrong because 0.16 J is the size of k q₁q₂ = 0.162 before dividing by the separation of 0.45 m.
MCQ 8CalculationPractice
Two +2.0 μC point charges are pushed closer together, from a separation of 0.60 m to 0.20 m, slowly and with no external field. Taking 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻², the work done by the external agency is:
Show answer and why every option is right or wrong
Answer: C. k q₁q₂ = 9.0 × 10⁹ × (2.0 × 10⁻⁶)² = 0.036 J m. U(initial) = 0.036/0.60 = 0.060 J; U(final) = 0.036/0.20 = 0.18 J. W = U(final) − U(initial) = 0.18 − 0.060 = 0.12 J; positive, as expected for pushing like charges together (NCERT Chapter 2, Eq. 2.22 and page 56).
Why A is wrong: A is wrong because −0.12 J is U(initial) − U(final) = 0.060 − 0.18; the external work is final minus initial, and pushing repelling charges together needs positive work.
Why B is wrong: B is wrong because 0.24 J adds the two energies (0.060 + 0.18) instead of taking their difference.
Why D is wrong: D is wrong because 0.80 J uses 1/r²: 0.036/0.040 − 0.036/0.36 = 0.90 − 0.10; potential energy varies as 1/r, not 1/r².
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How do you solve a Electrical Potential Energy Two Charges question? A worked example
- 1
Given
Two charges, 7 μC and −2 μC, are placed at (−9 cm, 0, 0) and (9 cm, 0, 0) respectively, with no external field. Take 1/(4πε₀) = 9 × 10⁹ N m² C⁻², as NCERT does.
- 2
Required
(a) The electrostatic potential energy of the system. (b) The work required to separate the two charges infinitely away from each other.
- 3
Concept
The potential energy of two point charges is the work done by an external agency to bring them from infinity to their positions. Separating them back to infinity takes U to zero, so the work required is U(final) − U(initial) = 0 − U.
- 4
Formula
U = (1/4πε₀) q₁q₂ / r₁₂ (NCERT Chapter 2, Eq. 2.22, page 56); W = U₂ − U₁.
- 5
Substitution
Both charges lie on the x-axis, so r₁₂ = 9 cm − (−9 cm) = 18 cm = 0.18 m.
(a) U = 9 × 10⁹ × (7 × 10⁻⁶ C)(−2 × 10⁻⁶ C) / 0.18 m
(b) W = 0 − U - 6
Calculation
(a) q₁q₂ = −14 × 10⁻¹² C²; 9 × 10⁹ × (−14 × 10⁻¹²) = −0.126 J m; U = −0.126 / 0.18 = −0.7 J
(b) W = 0 − (−0.7 J) = 0.7 J
The 0 for U at infinite separation is an exact reference value, and 1/(4πε₀) is taken as 9 × 10⁹ N m² C⁻² as NCERT does; neither limits the significant figures. The given data (7 μC, −2 μC) carry one significant figure, and NCERT reports the answers to one significant figure. - 7
Final answer
(a) U = −0.7 J; (b) work required = 0.7 J (NCERT Chapter 2, Example 2.5, page 60)
- 8
Common trap
Writing U = +0.7 J by ignoring the minus sign of the second charge, then saying 0.7 J is "released" on separation. The charges are unlike, so they attract: U is negative and an external agency must supply +0.7 J to pull them apart. A second slip is using 9 cm (the distance of each charge from the origin) as r instead of the 18 cm between the charges, which doubles the magnitude to 1.4 J.
- 9
Similar NEET-style question
"Charges of +5.0 μC and −3.0 μC are placed at (0, −10 cm, 0) and (0, 5.0 cm, 0) with no external field. Find the potential energy of the system and the work needed to separate the charges to infinity. Take 1/(4πε₀) = 9.0 × 10⁹ N m² C⁻²."
Strategy: r = 5.0 − (−10) = 15 cm = 0.15 m. U = 9.0 × 10⁹ × (5.0 × 10⁻⁶)(−3.0 × 10⁻⁶) / 0.15 = −0.135 / 0.15 = −0.90 J. Work to separate = 0 − (−0.90 J) = 0.90 J. Keep the sign, and do not use 10 cm or 5.0 cm alone as r.
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What to remember before solving Electrical Potential Energy Two Charges questions
More in Electrostatics: 3 exam traps and mistakes · 11 formulas · 6 question patterns from its other lessons.
Electrical Potential Energy Two Charges questions from past NEET papers
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Sources
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