PhysicsCBSE Class 12 · Electrostatic Potential and Capacitance

Electric Potential and Potential Energy Class 12: Charges and Dipoles

Q: What is the difference between electric potential and potential energy?

Potential belongs to a point in the field: work per unit charge, in volt. Potential energy belongs to a charge (or system of charges) placed there: U = qV, in joule.

Q: Why is the potential of a dipole zero on its equatorial line?

Every point on the equatorial line is equally far from +q and −q, so their potentials kq/r and −kq/r cancel exactly.

Q: Can potential be zero where the field is not zero?

Yes. On the equatorial line of a dipole, V = 0 but E is not zero. The field depends on how V changes from point to point, not on the value of V.

Q: What does a negative potential energy mean?

The system is bound. You would have to supply energy equal to |U| to separate the charges to infinity, as for an electron and a proton.

Q: Why is the work done independent of path?

The electrostatic force is conservative. The work it does between two points depends only on the potentials there, so any path gives W = q(V_B − V_A).

Updated 15 October 2026 · Fact-checked

Electric potential at a point is the work done per unit positive charge to bring a test charge there from infinity: V = kq/r for a point charge, and the plain sum of such terms for many charges. Potential energy of two charges is kq₁q₂/r, and of a dipole in a uniform field −pE cos θ.

Bringing +q from infinity to P costs work, giving V = kQ/r: the potential falls as 1/r and tends to zero far away.

Illustration drawn for QuestionBuddy.

Understand Electric Potential and Potential Energy

Lifting a bag up a staircase stores gravitational potential energy; letting it fall gives the energy back. Pushing a positive charge towards another positive charge is similar: you do work against the repulsion, and it is stored as electrostatic potential energy. Because the Coulomb force is conservative, this work depends only on the start and end points, not on the path.

Potential and potential difference. The potential at a point is the work done by an external agent, per unit positive test charge, in bringing it slowly (without acceleration) from infinity to that point: $V = \frac{W _{\infty \to P}}{q}$ . The potential difference between two points is $V_{B} - V_{A} = \frac{W _{A \to B}}{q}$ . The SI unit is the volt: 1 V = 1 J/C. Potential is a scalar, so potentials from several charges simply add with their signs.

Point charge. Bringing a unit charge from infinity to distance r against the field of Q gives $V = \frac{1}{4 π ε _{0}} \frac{Q}{r}$ . It is positive near a positive charge, negative near a negative one, falls as $\frac{1}{r}$ (not $\frac{1}{r ^{2}}$ ), and is zero at infinity.

Dipole. For a point at distance r from the centre of a dipole, with $r$ making angle $θ$ with $p$ , $V = k q (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ , where $r_{1}$ and $r_{2}$ are the distances from $+ q$ and $- q$ . For $r ≫ a$ , $r_{1} \approx r - a cos θ$ and $r_{2} \approx r + a cos θ$ , giving $V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$ . On the axis ( $θ = 0°$ ) $V = \frac{k p}{r ^{2}}$ ; on the equatorial line ( $θ = 90°$ ) $V = 0$ , even though the field there is not zero. A dipole's potential falls as $\frac{1}{r ^{2}}$ , one power faster than a single charge.

System of charges. $V = k \sum_{i} \frac{q _{i}}{r _{i}}$ , adding with signs. No vectors are needed, which is why potential problems are often easier than field problems. A point where V = 0 need not have E = 0, and vice versa.

Potential energy of a system of charges. The work done to assemble charges from infinity is stored as potential energy. For two charges, $U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}$ : positive for like charges (work had to be done to push them together), negative for unlike charges (they attract, so the system is bound). For three charges, add the energy of every pair: $U = k (\frac{q _{1} q _{2}}{r _{12}} + \frac{q _{2} q _{3}}{r _{23}} + \frac{q _{1} q _{3}}{r _{13}})$ . To pull the charges apart to infinity, an external agent must supply $- U$ .

Charge in an external field. A charge q at a point where the external potential is V has potential energy $q V$ . Moving it from A to B, an external agent does work $W = q (V_{B} - V_{A})$ , which becomes kinetic energy if the field does the work instead: $\frac{1}{2} m v^{2} = q Δ V$ . This defines the electron volt: $1 eV = 1.6 \times 1 0^{- 19}$ J, the energy an electron gains through 1 V.

Dipole in a uniform field. Rotating a dipole from $θ_{1}$ to $θ_{2}$ needs $W = pE (cos θ_{1} - cos θ_{2})$ . Taking $U = 0$ at $θ = 90°$ , the potential energy is $U = - pE cos θ = - p \cdot E$ : least ( $- pE$ ) when lined up with the field and greatest ( $+ pE$ ) when opposite.

Electric Potential and Potential Energy: formulas and key facts

Potential (definition): $V = \frac{W}{q}, V_{B} - V_{A} = \frac{W _{A \to B}}{q}$; Work by an external agent, moving the charge without acceleration. 1 V = 1 J/C.
Potential of a point charge: $V = \frac{1}{4 π ε _{0}} \frac{q}{r}$; Scalar; keep the sign of q.
Potential of a dipole: $V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}} = \frac{1}{4 π ε _{0}} \frac{p \cdot r ^}{r ^{2}}$; For r ≫ a. Axial: kp/r². Equatorial: 0.
System of charges: $V = \frac{1}{4 π ε _{0}} \sum_{i} \frac{q _{i}}{r _{i}}$; Add algebraically, no vectors.
Potential energy of two charges: $U = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r _{12}}$; For three or more charges, sum over every pair once.
Work and kinetic energy: $W = q (V_{B} - V_{A}), \frac{1}{2} m v^{2} = q Δ V$; $1 eV = 1.6 \times 1 0^{- 19}$ J.
Dipole in a uniform field: $U = - p \cdot E = - pE cos θ$; Work to rotate: $pE (cos θ_{1} - cos θ_{2})$ .

How to solve Electric Potential and Potential Energy questions

Use this method for potential, work and energy questions.

1List the charges with signs and the distance of each from the point (or from each other). Convert to C and m.
2For potential at a point, add $\frac{k q _{i}}{r _{i}}$ with signs; there is no direction to worry about.
3For potential energy of a system, list every pair once (3 charges: 3 pairs; 4 charges: 6 pairs) and add $\frac{k q _{i} q _{j}}{r _{ij}}$ .
4For work by an external agent moving q from A to B, use $W = q (V_{B} - V_{A})$ ; the path does not matter.
5For speed gained or lost, use energy conservation: loss in potential energy = gain in kinetic energy.
6For zero-potential points on a line, set $\frac{q _{1}}{x} = \frac{∣ q _{2} ∣}{d}$ in each region and keep only answers that lie in that region.
7State the answer with its unit (V, J or eV) and sign.

Electric Potential and Potential Energy in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Potential difference and the electron volt

Suppose an electron gun in an old cathode-ray display accelerates electrons from rest through a potential difference of 2000 V. Take $m_{e} = 9.1 \times 1 0^{- 31}$ kg.

(a) Find the kinetic energy gained by each electron in eV and in joule. (b) Find its final speed. (c) If the voltage were doubled, by what factor would the speed change?

Try it first, then show the solutionSolution

Energy gained = charge × p.d. = e × 2000 V = 2000 eV.
In joule: $2000 \times 1.6 \times 1 0^{- 19} = 3.2 \times 1 0^{- 16}$ J.
$v = \frac{2 K}{m} = \frac{2 \times 3.2 \times 1 0 ^{- 16}}{9.1 \times 1 0 ^{- 31}} \approx 2.65 \times 1 0^{7}$ m/s.
$K \propto V$ and $v \propto K$ , so doubling V multiplies the speed by $2 \approx 1.41$ .

Answer: (a) 2000 eV = $3.2 \times 1 0^{- 16}$ J; (b) about $2.65 \times 1 0^{7}$ m/s; (c) $2$ times.

Memory hook: An electron through V volts gains V electron volts, every time.

Work stored as potential energy

Imagine two small spheres, each carrying +3 nC, which you slowly push from 30 cm apart to 10 cm apart on a smooth insulating table.

(a) How much work do you do? (b) Where does this energy go? (c) If you release the spheres, what happens to this energy?

Try it first, then show the solutionSolution

$U = \frac{k q ^{2}}{r}$ . Work = $U_{10} - U_{30} = 9 \times 1 0^{9} \times (3 \times 1 0^{- 9})^{2} (\frac{1}{0.1} - \frac{1}{0.3})$ .
$= 8.1 \times 1 0^{- 8} \times 6.67 \approx 5.4 \times 1 0^{- 7}$ J.
It is stored as electrostatic potential energy of the pair, because the electric force is conservative.
On release they fly apart, and the stored energy turns into their kinetic energy.

Answer: (a) about $5.4 \times 1 0^{- 7}$ J; (b) stored as electrostatic potential energy; (c) it changes into kinetic energy as they move apart.

Memory hook: Pushing like charges together is like compressing a spring: the work is stored and comes back.

Tricks to remember Electric Potential and Potential Energy

Short memory aids. Say each one aloud once, then use it on a question the same day.

Potential: 1/r, field: 1/r²
Point charge: V ∝ 1/r, E ∝ 1/r². Dipole: V ∝ 1/r², E ∝ 1/r³. Each step is one power higher.
Use it when: Ratio and graph questions.
Scalars just add
Potential has no arrows: add kq/r with signs and you are done.
Use it when: Speed in multi-charge potential questions.
Count the handshakes
Potential energy of n charges has one term per pair, like handshakes: 3 charges → 3 terms, 4 charges → 6 terms.
Use it when: Not missing or double-counting pairs.
Zero V is not zero E
On a dipole's equatorial line V = 0 but E ≠ 0; between two equal like charges at the midpoint E = 0 but V ≠ 0.
Use it when: Assertion-reason traps.

Electric Potential and Potential Energy: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Potential at the corner of a square · 2 marks · Very short answerBoard level

Charges of +2 μC, −3 μC and +4 μC are placed at corners A, B and C of a square ABCD of side 20 cm. Find the potential at corner D.

Try it first, then show the solutionSolution

A and C are adjacent to D (0.2 m away); B is diagonally opposite ( $0.2 2 \approx 0.283$ m away). (½)
$V_{D} = 9 \times 1 0^{9} (\frac{2 \times 1 0 ^{- 6}}{0.2} + \frac{4 \times 1 0 ^{- 6}}{0.2} - \frac{3 \times 1 0 ^{- 6}}{0.283})$ . (1)
$= 9 \times 1 0^{9} \times (1.94 \times 1 0^{- 5}) \approx 1.75 \times 1 0^{5}$ V. (½)

Answer: About $1.75 \times 1 0^{5}$ V.

Example 2: Derivation: potential due to a dipole · 3 marks · Short answerBoard level

Derive an expression for the electric potential at a point at distance r from the centre of a short electric dipole, the line to the point making angle θ with the dipole moment. Hence write the potential on the axial and equatorial lines.

Try it first, then show the solutionSolution

Charges −q and +q are 2a apart. Let $r_{1}$ and $r_{2}$ be the distances of P from +q and −q. By superposition $V = \frac{k q}{r _{1}} - \frac{k q}{r _{2}}$ . (1)
For $r ≫ a$ , by geometry $r_{1} \approx r - a cos θ$ and $r_{2} \approx r + a cos θ$ , so $\frac{1}{r _{1}} - \frac{1}{r _{2}} = \frac{2 a c o s θ}{r ^{2} - a ^{2} c o s ^{2} θ} \approx \frac{2 a c o s θ}{r ^{2}}$ . (1)
So $V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$ . Axial (θ = 0°): $V = \frac{k p}{r ^{2}}$ ; equatorial (θ = 90°): V = 0. (1)

Answer: $V = \frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$ ; kp/r² on the axis, 0 on the equatorial line.

Example 3: Potential energy of three charges · 3 marks · Short answerHarder board level

Charges of +1 μC, +2 μC and −3 μC are placed at the corners of an equilateral triangle of side 10 cm. (a) Find the potential energy of the system. (b) How much work is needed to separate the charges to infinity? (c) Interpret the sign of U.

Try it first, then show the solutionSolution

(a) $U = \frac{k}{a} (q_{1} q_{2} + q_{2} q_{3} + q_{1} q_{3}) = \frac{9 \times 1 0 ^{9}}{0.1} (2 - 6 - 3) \times 1 0^{- 12}$ . (1)
$= 9 \times 1 0^{10} \times (- 7 \times 1 0^{- 12}) = - 0.63$ J. (1)
(b) Work needed = $0 - U = + 0.63$ J. (c) The negative U means the system is bound: energy must be supplied to pull it apart. (1)

Answer: (a) −0.63 J; (b) 0.63 J; (c) the system is bound.

Example 4: Zero-potential points between unlike charges · 3 marks · Short answerHarder board level

Charges of +4 nC and −6 nC are placed at x = 0 and x = 10 cm. Find all points on the x-axis where the electric potential is zero. Is the electric field zero at these points?

Try it first, then show the solutionSolution

Between the charges (0 < x < 10 cm): $\frac{4}{x} = \frac{6}{10 - x}$ gives $40 - 4 x = 6 x$ , x = 4 cm. (1)
Outside, on the side of the smaller charge (x < 0), at distance d from +4 nC: $\frac{4}{d} = \frac{6}{d + 10}$ gives d = 20 cm, so x = −20 cm. Beyond the −6 nC charge, the larger charge is always nearer, so V cannot be zero there. (1)
The field is not zero at either point: between unlike charges both fields point the same way, and at x = −20 cm the field of −6 nC ( $\frac{k \cdot 6}{0. 3 ^{2}}$ ) and of +4 nC ( $\frac{k \cdot 4}{0. 2 ^{2}}$ ) are unequal. (1)

Answer: V = 0 at x = 4 cm and x = −20 cm; E is not zero at either point.

Example 5: Energy conservation with a released proton · 5 marks · Long answerHOTS / competency

A proton is held at rest 1 cm from a fixed charge of +2 μC and released. Take $m_{p} = 1.67 \times 1 0^{- 27}$ kg.
(a) Find its speed when it is very far away. (2)
(b) At what distance from the fixed charge does it have half of that final speed? (2)
(c) Why can we not use $v^{2} = u^{2} + 2 a s$ here? (1)

Try it first, then show the solutionSolution

(a) Loss of potential energy = gain in KE: $\frac{1}{2} m v^{2} = \frac{k Q e}{r _{0}} = \frac{9 \times 1 0 ^{9} \times 2 \times 1 0 ^{- 6} \times 1.6 \times 1 0 ^{- 19}}{0.01} = 2.88 \times 1 0^{- 13}$ J. (1)
$v = \frac{2 \times 2.88 \times 1 0 ^{- 13}}{1.67 \times 1 0 ^{- 27}} \approx 1.86 \times 1 0^{7}$ m/s. (1)
(b) Half the speed means a quarter of the final KE: $k Q e (\frac{1}{r _{0}} - \frac{1}{r}) = \frac{1}{4} \frac{k Q e}{r _{0}}$ . (1)
$\frac{1}{r} = \frac{3}{4 r _{0}}$ , so $r = \frac{4}{3} r_{0} \approx 1.33$ cm. (1)
(c) The force (and acceleration) changes as $\frac{1}{r ^{2}}$ , so the equations for constant acceleration do not apply; energy conservation does. (1)

Answer: (a) about $1.86 \times 1 0^{7}$ m/s; (b) at about 1.33 cm; (c) the acceleration is not constant.

Example 6: Work done in the field of a point charge · 4 marks · Case-basedHarder board level

Read the passage and answer.

A charge of +5 μC is fixed at O. Point A is 10 cm from O and point B is 30 cm from O.
(a) Find the potentials at A and B. (1)
(b) Find the work an external agent must do to move a charge of −2 nC slowly from A to B. (1)
(c) Would the work be different if the charge went from A to B along a curved path? Why? (1)
(d) How much work is done in carrying the −2 nC charge once around a circle of radius 10 cm centred on O? (1)

Try it first, then show the solutionSolution

(a) $V_{A} = \frac{9 \times 1 0 ^{9} \times 5 \times 1 0 ^{- 6}}{0.1} = 4.5 \times 1 0^{5}$ V; $V_{B} = 1.5 \times 1 0^{5}$ V. (1)
(b) $W = q (V_{B} - V_{A}) = (- 2 \times 1 0^{- 9}) (1.5 - 4.5) \times 1 0^{5} = + 6 \times 1 0^{- 4}$ J (the agent must pull the negative charge away from the positive one). (1)
(c) No: the electrostatic force is conservative, so the work depends only on the end points. (1)
(d) Zero: the circle is an equipotential (all points at 10 cm), and the start and end points are the same. (1)

Answer: (a) $4.5 \times 1 0^{5}$ V and $1.5 \times 1 0^{5}$ V; (b) $+ 6 \times 1 0^{- 4}$ J; (c) no, the force is conservative; (d) zero.

Where students lose marks in Electric Potential and Potential Energy

The slips examiners see most often, and what to write instead.

Mistake: Adding potentials as vectors.
Habit from field questions.
Wrong working
V = √(V₁² + V₂²)
Right working
V = V₁ + V₂ (with signs)
Do this: Potential is a scalar: add algebraically.
Usually costs: The full numerical.
Mistake: Dropping the sign of a negative charge in V or U.
Using magnitudes as in Coulomb force calculations.
Wrong working
U = k(1)(3)/a for +1 μC and −3 μC
Right working
U = k(1)(−3)/a = −3k/a (×10⁻¹²)
Do this: Keep signs in every potential and energy term.
Usually costs: 1 to 2 marks.
Mistake: Counting a pair twice in potential energy.
Writing q₁q₂ and q₂q₁ as separate terms.
Do this: List pairs: 12, 13, 23 for three charges; 12, 13, 14, 23, 24, 34 for four.
Usually costs: The final answer mark.
Mistake: Assuming E = 0 wherever V = 0.
Mixing up the two quantities.
Do this: Remember E is the rate of change of V, not V itself.
Usually costs: 1 mark in assertion-reason and reasoning questions.
Mistake: Approximating too early in the dipole derivation.
Writing $\frac{p c o s θ}{r ^{2}}$ without the $r_{1}$ , $r_{2}$ steps.
Do this: Show $r_{1} \approx r - a cos θ$ and $r_{2} \approx r + a cos θ$ , then simplify.
Usually costs: 1 mark of a 3-mark derivation.

Exam tips

Define potential with all three parts: work done per unit positive charge, bringing it from infinity, without acceleration.
In energy questions, write the conservation equation in words first: 'loss in PE = gain in KE'.
For a system of charges, a small table of pairs and distances avoids slips.
Board and sample papers have set the potential of a system of charges and the dipole potential (for example 2023 55/1/1 Q26 and 2024 55/1/1 Q31(b)); these are quick marks if signs are handled carefully.
State units: V for potential, J (or eV) for energy and work.

Practice questions on Electric Potential and Potential Energy

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The electric potential at a point on the equatorial line of a short dipole is
(a) $\frac{k p}{r ^{2}}$ (b) $\frac{2 k p}{r ^{2}}$ (c) $- \frac{k p}{r ^{2}}$ (d) zero
Hint
cos 90° = 0.
Answer
(d) zero.
Q2 · 1 mark · Very short answerBoard level
How much work is needed to move a charge of 1 μC between two points that differ in potential by 200 V?
Hint
W = qΔV.
Answer
$W = 1 0^{- 6} \times 200 = 2 \times 1 0^{- 4}$ J.
Q3 · 2 marks · Very short answerHarder board level
In a hydrogen atom, the electron and proton are about $0.53 \times 1 0^{- 10}$ m apart. Find the electrostatic potential energy of the pair in joule and in eV.
Hint
Unlike charges: U is negative.
Answer
$U = - \frac{9 \times 1 0 ^{9} \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{0.53 \times 1 0 ^{- 10}} \approx - 4.35 \times 1 0^{- 18}$ J $\approx - 27.2$ eV.
Q4 · 2 marks · Very short answerHarder board level
Charges +q, +q, −q and −q are placed at the corners A, B, C and D (in order) of a square. Show that the potential at the centre is zero, and say whether the field there is zero.
Hint
All four charges are the same distance from the centre.
Answer
Each charge is $\frac{a}{2}$ from the centre, so $V = \frac{k}{a / 2} (q + q - q - q) = 0$ . The field is not zero: it is $\frac{4 2 k q}{a ^{2}}$ , pointing from side AB towards side CD.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): No work is done in moving a charge along the equatorial line of an electric dipole.
Reason (R): The electric field at every point of the equatorial line of a dipole is zero.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
What is V on the equatorial line? What is E there?
Answer
(c). V = 0 at every point of the equatorial line, so W = qΔV = 0. But the field there is not zero; it is perpendicular to the line.
Q6 · 3 marks · Short answerHOTS / competency
Charges of +3 nC and −1 nC are 8 cm apart. Find the points on the line joining them where the potential is zero.
Hint
Look between the charges and beyond the smaller charge.
Answer
Between them, at x from +3 nC: $\frac{3}{x} = \frac{1}{8 - x}$ gives x = 6 cm. Beyond −1 nC: $\frac{3}{x} = \frac{1}{x - 8}$ gives x = 12 cm from +3 nC (4 cm beyond −1 nC). V cannot vanish on the other side of +3 nC, as the larger charge is nearer there.
Q7 · 3 marks · Short answerHOTS / competency
Three equal charges q are brought from infinity to the corners of an equilateral triangle of side a. (a) Find the work done. (b) What charge Q placed at the centroid makes the total potential energy of the four charges zero?
Hint
The centroid is $\frac{a}{3}$ from each corner.
Answer
(a) $W = U = \frac{3 k q ^{2}}{a}$ . (b) Add three pairs with Q: $\frac{3 k q ^{2}}{a} + \frac{3 k q Q}{a / 3} = 0$ , so $Q = - \frac{q}{3}$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Define electric potential at a point.Show answer
Work done per unit positive test charge in bringing it from infinity to that point without acceleration.
2.Potential of a short dipole at angle θ?Show answer
$\frac{1}{4 π ε _{0}} \frac{p c o s θ}{r ^{2}}$ .
3.Potential energy of two point charges?Show answer
$\frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r}$ .
4.What is 1 eV in joule?Show answer
$1.6 \times 1 0^{- 19}$ J.
5.Potential energy of a dipole in a uniform field?Show answer
$U = - pE cos θ$ .
6.Work done in moving a charge around a closed path in an electrostatic field?Show answer
Zero.

Electric Potential and Potential Energy: frequently asked questions

What is the difference between electric potential and potential energy?

Potential belongs to a point in the field: work per unit charge, in volt. Potential energy belongs to a charge (or system of charges) placed there: U = qV, in joule.

Why is the potential of a dipole zero on its equatorial line?

Every point on the equatorial line is equally far from +q and −q, so their potentials kq/r and −kq/r cancel exactly.

Can potential be zero where the field is not zero?

Yes. On the equatorial line of a dipole, V = 0 but E is not zero. The field depends on how V changes from point to point, not on the value of V.

What does a negative potential energy mean?

The system is bound. You would have to supply energy equal to |U| to separate the charges to infinity, as for an electron and a proton.

Why is the work done independent of path?

The electrostatic force is conservative. The work it does between two points depends only on the potentials there, so any path gives W = q(V_B − V_A).