Electric Potential and Electric Potential Energy

Potential is a scalar, so you ADD the contributions instead of resolving them — which is what makes it easier than the field. Two charges, one point, and this returns the net potential there, the energy of a test charge brought in from infinity, the work each party does, and the same energy in electronvolts.

Last updated 11 Nov 2026

Net potential at a point, and the energy of a charge in it

Scalar addition → V, U and the work

Charge q₁, in microcoulombs (µC)With its sign. A positive charge raises the potential everywhere around it and a negative charge lowers it, and that is the whole of the scalar addition: no angles, no components, just a signed sum.

Distance from q₁ to the point P, in centimetresStraight-line distance. It does not matter where q₂ is relative to q₁ — only how far each charge is from P. That is the practical advantage of working with potential, and it is why the geometry of these questions is so much simpler than the field version of the same question.

Charge q₂, in microcoulombs (µC)With its sign. Set it to zero if your question has only one charge; the page then just reports q₁’s own potential, and the second contribution row reads zero rather than disappearing.

Distance from q₂ to the point P, in centimetresStraight-line distance again. If q₂ is zero this field changes nothing, but it still has to hold a number above zero: the engine refuses to compute at all while any field is blank or zero where a division would follow.

Test charge brought to P from infinity, in nanocoulombs (nC)This one does not affect the potential at all — the potential is set by q₁ and q₂ and by where P is, and it exists whether or not anything is brought there. The test charge only converts that potential into an ENERGY. Make it negative and watch the sign of the energy flip, which is the point of having the field here.

134,813.2768VExample

A 5 µC charge 20 cm from P, a −3 µC charge 30 cm from P, and a 4 nC test charge brought to P

A signed sum, and one multiplication that turns volts into joules

V = (1/4πε₀) ∑ q_i/r_i · U = q₀V · W_external = ΔU = +U · W_field = −ΔU = −U

V: the potential at the point, in volts, which is joules per coulomb. One volt means one joule of work per coulomb brought from infinity. It is a property of the charges and the place, and it exists whether or not there is anything at that place to feel it
q_i/r_i: one term per source charge, signed, with r the straight-line distance from that charge to the point. First power of r, NOT the square — this is the commonest confusion between potential and field. And no angles: this is the sum that makes potential easier to work with than the field it comes from
U = q₀V: the potential energy of a charge q₀ placed at a point of potential V. Signed twice over: once by q₀ and once by V. A negative test charge at a positive potential has NEGATIVE energy, which means it is bound there and you must do work to remove it
W_external: the work YOU do bringing the test charge in from infinity without changing its speed. It equals ±U, same sign as U, because all the work you do ends up as potential energy if the kinetic energy has not changed
W_field: the work the electrostatic FIELD does over the same journey, which is −U — equal and opposite. Mixing up which of the two a question is asking for is worth a sign and therefore a mark, and the wording to look for is “work done by the field” against “work done in bringing” or “work done against the field”

Worked example

A 5 µC charge 20 cm from P, a −3 µC charge 30 cm from P, and a 4 nC test charge brought to P
FIRST CONTRIBUTION. V₁ = kq₁/r₁ = (8.987551786×10⁹)(5×10⁻⁶)/0.20 = +224,688.79 V. Note the single power of r: dividing by 0.20, not by 0.04
SECOND CONTRIBUTION, WITH ITS SIGN. V₂ = kq₂/r₂ = (8.987551786×10⁹)(−3×10⁻⁶)/0.30 = −89,875.52 V
ADD THEM. No geometry, no components: V = 224,688.79 − 89,875.52 = 134,813.28 V, about 135 kV. If this had been a FIELD question you would now be resolving two vectors and you would need to know where q₂ is relative to q₁. For potential you do not
TURN IT INTO AN ENERGY. U = q₀V = (4×10⁻⁹)(134,813.28) = 5.3925×10⁻⁴ J, positive, so you must DO that work to bring the positive test charge to P and it would fly away if released. The electrostatic field does −5.3925×10⁻⁴ J over the same journey
AND IN THE UNIT ATOMIC PHYSICS USES. Divide by the elementary charge: 5.3925×10⁻⁴/1.602176634×10⁻¹⁹ = 3.37×10¹⁵ eV. That is what an electronvolt is for — not a new physical quantity, just joules measured in a unit the size of a single charge rather than the size of a coulomb
A CHECK WORTH MAKING. q₁ on its own would produce 134,813.28 V at a distance of kq₁/V = 0.3333 m. So as far as the potential is concerned, this two-charge arrangement looks from P exactly like a lone 5 µC charge 33.3 cm away. The FIELD at those two places is completely different, which is the warning in this check as well as the point of it

Potential against field, side by side, for the same two charges

The last row is the one examiners like, and both halves of it are worth checking by hand. Between equal and opposite charges the two potentials cancel and the two field vectors reinforce; between two equal positive charges the field vectors cancel and the potentials add. So neither quantity being zero tells you anything at all about the other, and a question that says one of them is zero is not telling you about the other.
	Electric potential V	Electric field E
Kind of quantity	scalar — a number with a sign	vector — a magnitude and a direction
Combining two sources	add the signed numbers	add the vectors, with components
Distance dependence	1/r	1/r²
Unit	volt = joule per coulomb	volt per metre = newton per coulomb
Zero at	infinity, by convention	wherever the vectors cancel
Can be zero where the other is not	yes — midway between +q and −q the FIELD is large and V is zero	yes — midway between +q and +q the field is zero and V is not

Why potential is the easier half of electrostatics, and where the sign goes wrong

The reason potential is worth learning is that it collapses the geometry. To find the FIELD of several charges at a point you must find each vector, resolve it into components, add the components and recombine. To find the POTENTIAL you add signed numbers. For three or four charges that is the difference between a page of trigonometry and a line of arithmetic, and the reason is simply that energy has no direction.

What goes wrong is almost always a sign, and almost always on the work. There are two works in every one of these questions and they are equal and opposite. The work done BY THE FIELD as a charge moves is −ΔU. The work done BY YOU, moving it slowly against the field, is +ΔU. A question that says “find the work done in bringing the charge from infinity” wants the second; one that says “find the work done by the electric field” wants the first. This page prints both, labelled, because the formula cannot tell you which one was asked for.

The second thing to be careful about is the zero. Potential has no absolute value — only differences are measurable — so the zero is a convention, and the convention for point charges is V = 0 at infinite distance. That is what makes the potential of a positive charge positive everywhere and the energy of a bound pair negative. In circuits the convention is different: there, zero is the point someone has labelled earth, and the two conventions meet in the single quantity a voltmeter reads. A voltmeter never reads a potential; it reads a potential DIFFERENCE between the two places you put the probes.

And the electronvolt is not a new quantity. It is a joule in different clothing: the energy an elementary charge gains across one volt, 1.602176634×10⁻¹⁹ J exactly, since the elementary charge has been exact by definition since 2019. Every energy on this page is printed in both units for that reason. For the related arithmetic on the force rather than the energy, see Coulomb’s law and the field of a point charge; for what happens when that energy is stored in a capacitor instead of in a pair of charges, see the parallel-plate capacitor.

Frequently asked questions

Why is the potential 1/r when the field is 1/r²?

Because potential is the field integrated over distance, and integrating 1/r² gives 1/r. The unit argument gets you there without calculus: a field is a force per charge and a potential is an ENERGY per charge, energy is force times distance, so the potential must carry one extra power of distance on top — which, in a denominator, is one power fewer. If you ever forget which is which, check the units.

Can the potential at a point be zero while the field there is not?

Yes, and the standard example is the midpoint between +q and −q. The two potentials are equal and opposite and cancel to zero; the two field vectors both point the same way and add. The reverse also happens: midway between two equal POSITIVE charges the field cancels to zero and the potential is large. Neither quantity being zero tells you anything about the other.

Do I include the sign of the charge when I work out the potential?

Always. This is the opposite of the habit that works for field magnitudes. The potential of a negative charge is negative, and the signed sum is the answer — there is no separate direction step to do afterwards, so if you drop the signs here you have thrown the information away rather than deferred it.

What is the difference between potential and potential energy?

A factor of charge. The potential V belongs to the source charges and the place; the potential energy U = q₀V belongs to a particular charge sitting at that place. Volts against joules. The giveaway in a question is whether a second charge is mentioned: if there is a charge AT the point, you are being asked about energy, and if there is only a point, you are being asked about potential.

Is it the same V as the one in V = IR?

Same quantity, different convention for the zero. In electrostatics the zero is at infinity; in a circuit it is whichever node has been called earth, and what V = IR gives you is the DIFFERENCE between the two ends of the resistor. Both are joules per coulomb, both read in volts, and a voltmeter measures a difference in either case. Potential differences in a circuit are what emf, internal resistance and terminal voltage is about.

Related calculators

References

2022 CODATA Recommended Values of the Fundamental Physical Constants (wallet card), NIST Committee on Data for Science and Technology, physics.nist.gov/cuu/pdf/wallet_2022.pdf (read 11 November 2026). Every constant on this page is taken from here, and the reason to read the card rather than a textbook is the bracket. The elementary charge is printed as 1.602 176 634 × 10⁻¹⁹ C with no bracket at all, because since the 2019 revision of the SI it is exact by definition — the coulomb is now defined from it, not the other way round. The same is true of c, h, k and N_A. But the vacuum electric permittivity is printed 8.854 187 8188(14) × 10⁻¹² F/m and the vacuum magnetic permeability 1.256 637 061 27(20) × 10⁻⁶ N/A²: both now have an uncertainty, about 1.6 parts in 10¹⁰, where before 2019 μ₀ was exactly 4π × 10⁻⁷ and ε₀ exactly 1/μ₀c². Almost every school book still prints the old exact forms. At school precision nothing changes; the logic of which constants are definitions changed completely.
Annual Syllabus Session 2023-24, Class XII Physics (Code 042), Central Board of Secondary Education, edustud.nic.in/edu/SYLLABUS_2023_24/12/12_physics_eng_2023_24.pdf (read 11 November 2026). The board’s own document, cited and not endorsed by it. Unit I Chapter 1 asks for “Coulomb’s law-force between two point charges; forces between multiple charges, superposition principle”; Chapter 2 for “electrical potential energy of a system of two point charges” and “capacitance of a parallel plate capacitor with and without dielectric medium between the plates, energy stored in a capacitor.(No derivation, formulae only)”. Unit II asks for “drift velocity, mobility and their relation with electric current”, “Internal resistance of a cell, potential difference and emf of a cell, combination of cells in series and in parallel, kirchhoff’s rules, Wheatstone bridge”. Unit III asks for “Biot – Savart law and its application to the current carrying circular loop”, “Ampere’s law and its applications to infinitely long straight wire. Straight solenoid (only qualitative treatment)” and “force between two parallel current-carrying conductors-definition of ampere”.
Syllabus: Cambridge International AS & A Level Physics 9702, for examination in 2025, 2026 and 2027 (Version 1), Cambridge Assessment International Education, cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf (read 11 November 2026). Cited for the outcomes a Cambridge candidate is examined on, which are worded differently from the Indian board’s: 18.3.2 “recall and use Coulomb’s law F = Q1Q2 /(4πε0r2) for the force between two point charges in free space”, 18.5.3 “use V = Q/(4πε0r) for the electric potential in the field due to a point charge”, 9.1.4 “use, for a current-carrying conductor, the expression I = Anvq, where n is the number density of charge carriers”, 10.1.5 “understand the effects of the internal resistance of a source of e.m.f. on the terminal potential difference”, 19.1.3 “derive, using C = Q/V, formulae for the combined capacitance of capacitors in series and in parallel”, 20.2.2 “recall and use the equation F = BIL sin θ, with directions as interpreted by Fleming’s left-hand rule” and 20.3.2 “recall and use F = BQv sin θ”. The copy read here was truncated before the data-and-formulae sheet, so what is on THAT sheet could not be established from it.