Potential due to a point charge

Drop a single charged pith ball on a table and it quietly rewrites the space around it. It does not touch anything, does not move anything by contact — yet every point in the room now carries an invisible number, a potential, that measures how much work it would take to bring a unit positive charge there from far away. Walk closer to a positive charge and that number climbs; walk closer to a negative one and it falls into negative territory. This is the single most important formula in electrostatics after Coulomb's law itself: the potential due to a point charge, $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ . It looks almost too simple — one charge, one distance, one constant — but this formula is the atom out of which the rest of the chapter is built. The potential of a dipole is just this formula added twice, once for each charge, at slightly different distances. The potential of any complicated charge distribution — a charged sphere, a line of charge, a whole thunder-cloud — is this same formula summed (or integrated) over every tiny piece of charge in that distribution. Get comfortable with the single point-charge case and the rest of "potential due to a system of charges" becomes bookkeeping, not new physics. This article rebuilds the formula from the electric field by a short line integral, so you see why potential falls off as $1/r$ while the field it comes from falls off as $1/r^2$ — a distinction examiners love to test and students routinely blur. Then it works two worked numbers all the way through — one at board level, one pushed to competitive-exam speed — and closes with the sign traps, the misconception that trips up almost every first attempt, and a tight formula card you can revise from ten minutes before the exam.

Key takeaways

Potential due to a point charge at distance $r$ is $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ , a scalar, defined so that $V \to 0$ as $r \to \infty$ .
It is obtained by integrating the electric field $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ along the radial direction — one extra power of $r$ in the denominator compared to the field.
$V$ carries the sign of $q$ directly: positive charge gives positive potential everywhere, negative charge gives negative potential everywhere — there is no vector direction to worry about, only a sign.
All points at the same distance $r$ from the charge form a spherical equipotential surface, and no work is done moving a test charge along it.
Near the charge ( $r \to 0$ ) the idealised formula blows up to infinity — a signal that the point-charge model breaks down at zero distance, not that potential is actually zero there.
Potential falls off as $1/r$ , slower than the field's $1/r^2$ , so potential "reaches farther" than field — a charge can raise the potential at a point noticeably even where its field is already negligible.
Key idea: the potential of a point charge, $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ , is the line integral of its $1/r^2$ field along the radial direction, is a signed scalar that inherits the sign of the source charge, and is the single building block from which every other potential formula in the chapter — dipole, system of charges, charged sphere — is assembled by superposition.

The core insight

The potential of a point charge, $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ , is a signed scalar obtained by integrating its $1/r^2$ field radially from infinity, and it is the elementary building block that superposition assembles into every other potential formula in electrostatics.

Where it fits in the syllabus

This is the second result in the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), coming right after the general definition of electrostatic potential and potential difference, and immediately before potential due to a dipole and potential due to a system of charges. The formula $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ is explicitly listed in the prescribed scope, and the board expects both its derivation from the field (a short, low-algebra line integral) and confident numerical use. It sits squarely within syllabus scope — nothing here strays into the chapter's out-of-scope items (energy-stored-in-a-capacitor derivation, Van de Graaff device detail, dielectric molecular models), which belong to later sections of the same chapter.

The physical picture

Picture a positive point charge $q$ sitting still at the origin. Coulomb's law tells us the electric field it produces at a distance $r$ is radially outward (for positive $q$ ) with magnitude

E(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}

where: $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ is Coulomb's constant, $q$ is the source charge with its sign, and $r$ is the distance from the charge to the field point. Physically, this is the force per unit positive test charge placed at that point — it is a vector, pointing away from a positive $q$ and toward a negative one. Now ask a different question: how much work would an external agent have to do, against this field, to bring a unit positive test charge from far away (where we agree $V=0$ ) up to a distance $r$ from $q$ ? That work-per-unit-charge is, by definition, the electric potential at that point:

V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}

where: the constant and $q$ are exactly as before, $r$ is again the distance from the charge to the point where we want the potential, and $V$ is measured in volts (joules per coulomb) when SI units are used throughout. Unlike $E$ , potential $V$ is a scalar — it has a magnitude and a sign, but no direction — which is precisely what makes it so much easier to add up for many charges than the field is: field addition needs vector components, potential addition is plain arithmetic (with sign). The picture below shows why the two formulas look so similar yet differ by exactly one power of $r$ , and why the equipotential surfaces of a point charge are simply concentric spheres.

Figure

A point charge at the centre with concentric dashed circles marking equal potential values that get closer together near the charge and a radial arrow labelled with the field falling faster than the potential as distance increases

Two features of this picture are exam gold. First, because $V$ depends only on $r$ (not on direction), every point on a sphere of radius $r$ centred on the charge sits at the same potential — these spheres are the equipotential surfaces of a point charge, and no work is needed to slide a test charge along one of them. Second, the two formulas share the same constant and the same numerator $q$ , differing only in the power of $r$ ; that single extra power in the denominator of $E$ compared to $V$ is not a coincidence — it is exactly what integrating $E$ with respect to $r$ produces, as the derivation below shows.

Derivation AGetting V(r) from E(r) by a line integral

The goal of this derivation is to obtain the formula for the electric potential at a point at distance $r$ from an isolated point charge $q$ , starting only from Coulomb's law for the electric field of that same charge and from the definition of potential as the work done per unit charge in bringing a positive test charge from infinity to that point against the electric field, so that the resulting $1/r$ dependence of potential can be understood as a direct consequence of integrating the $1/r^2$ dependence of the field rather than as a separate rule to memorise.

Step 1. Start from the definition. The potential at a point P, at distance $r$ from charge $q$ , is the work done by an external agent, per unit charge, in moving a small positive test charge $q_0$ from infinity to P, moving against the electric force at every step (so that the test charge gains no kinetic energy — it is moved quasi-statically). In terms of the field, this work-per-unit-charge is $V(r) = -\displaystyle\int_{\infty}^{r} E\,dr'$ , where the electric field along the radial direction outward from $q$ is $E(r') = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r'^2}$ .

Step 2. Substitute the field expression and integrate. $V(r) = -\displaystyle\int_{\infty}^{r} \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r'^2}\,dr' = -\dfrac{q}{4\pi\varepsilon_0}\left[-\dfrac{1}{r'}\right]_{\infty}^{r} = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{r'}\right]_{\infty}^{r}$ . This uses the standard result $\displaystyle\int \dfrac{dr'}{r'^2} = -\dfrac{1}{r'}$ , which is the one calculus fact this derivation needs.

Step 3. Evaluate the limits. At the upper limit, $\dfrac{1}{r'} = \dfrac{1}{r}$ . At the lower limit, as $r' \to \infty$ , $\dfrac{1}{r'} \to 0$ . So the bracket evaluates to $\left(\dfrac{1}{r} - 0\right) = \dfrac{1}{r}$ , giving $V(r) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ .

Step 4. Read off the physics. The choice $V(\infty) = 0$ was built into the lower limit of the integral, which is why the formula gives exactly zero far from the charge and is not an arbitrary convention tacked on afterwards. Because the integral only ever involved the magnitude of $E$ along a straight radial path, and $q$ was carried through the whole calculation with its own sign, the result correctly gives positive $V$ for positive $q$ and negative $V$ for negative $q$ , with no separate sign rule needed. The one extra power of $r$ that appears in $V$ compared with $E$ is exactly the one power that integration of $1/r'^2$ always produces — this is the algebraic reason potential falls off more slowly with distance than the field that generates it.

Worked examples

Board level - potential at a given distance

Question: Find the electric potential at a point 30 cm from a point charge of $+4\ \mu\text{C}$ .

Step 1. Write down the formula and list the knowns. $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ , with $q = 4\times10^{-6}$ C and $r = 0.30$ m, and $\dfrac{1}{4\pi\varepsilon_0} = 9\times10^{9}\ \text{N}\,\cdot\,\text{m}^2/\text{C}^2$ .

Step 2. Substitute directly. $V = \dfrac{(9\times10^{9})(4\times10^{-6})}{0.30}$ .

Step 3. Simplify the numerator first, then divide. Numerator $= 9\times10^{9}\times4\times10^{-6} = 3.6\times10^{4}$ . Then $V = \dfrac{3.6\times10^{4}}{0.30} = 1.2\times10^{5}$ V.

Answer: the potential at 30 cm from the charge is $1.2\times10^{5}$ V, that is, 120 kV.

Competitive level - locating a target potential and comparing with the field

Question: A point charge is $-2\ \mu\text{C}$ . (a) At what distance from it is the potential exactly $-9\times10^{4}$ V? (b) At that same distance, what is the magnitude of the electric field, and how does the ratio $E/V$ compare with $1/r$ ?

Step 1. Rearrange the potential formula to solve for $r$ . From $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ , we get $r = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{V}$ . Substitute $q = -2\times10^{-6}$ C and $V = -9\times10^{4}$ V.

Step 2. Compute $r$ , tracking signs carefully. $r = \dfrac{(9\times10^{9})(-2\times10^{-6})}{-9\times10^{4}} = \dfrac{-1.8\times10^{4}}{-9\times10^{4}} = 0.2$ m. Both numerator and denominator are negative, so their ratio is positive — exactly as a physical distance must be, and a useful self-check that the arithmetic is on track.

Step 3. Now find the field at that same $r = 0.2$ m using $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q|}{r^2} = \dfrac{(9\times10^{9})(2\times10^{-6})}{(0.2)^2} = \dfrac{1.8\times10^{4}}{0.04} = 4.5\times10^{5}$ N/C, directed radially toward the charge (since it is negative).

Step 4. Check the ratio. $\dfrac{|E|}{|V|} = \dfrac{4.5\times10^{5}}{9\times10^{4}} = 5 = \dfrac{1}{r}$ , since $r = 0.2$ m and $1/0.2 = 5$ . This is exactly the relation the derivation predicts: because $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ and $E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}$ , dividing the second by the first always gives $E/V = 1/r$ , a fast sanity check worth keeping in your back pocket. Answer: the point is 0.2 m (20 cm) from the charge, the field there has magnitude $4.5\times10^{5}$ N/C, and $E/V = 1/r$ checks out exactly. Timing note: the whole four-step chain — rearrange, substitute with signs, compute the field at the same point, and verify with the $E/V = 1/r$ shortcut — should take under two minutes in a competitive exam once the formula and its rearrangement are automatic; the ratio check is a 5-second insurance policy against a sign or arithmetic slip, not something to derive from scratch under time pressure.

Common traps

Dropping the sign of $q$ . Potential is a signed scalar, and that sign must be carried through every substitution — a negative source charge always gives a negative potential at every finite point around it, never zero and never positive.
Confusing the $1/r$ law for $V$ with the $1/r^2$ law for $E$ . This is the single most common numerical slip in this topic; if an answer for potential comes out looking like it used $r^2$ in the denominator (or vice versa for the field), stop and recheck which formula was intended.
Treating equal fields as equal potentials, or zero field as zero potential. Far from an isolated point charge both tend to zero together, but for two or more charges a point can have zero field with nonzero potential, or vice versa — that reasoning does not carry over automatically once more than one charge is involved.
Believing $V=0$ at $r=0$ . The idealised point-charge formula actually diverges to infinity as $r\to0$ ; zero potential is fixed by convention only at $r\to\infty$ , not at the location of the charge.
Forgetting the reference point. $V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ is only the potential relative to infinity; when a question later asks for potential relative to some other reference point, this default zero must be adjusted, or subtracted between two points to get a potential difference.

You should be able to…

State the formula $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r}$ for the potential due to an isolated point charge, and derive it from the field by the short radial line integral.
Correctly carry the sign of $q$ through a calculation of $V$ , for both positive and negative source charges.
Explain, using the $1/r$ versus $1/r^2$ comparison, why potential falls off more slowly with distance than the field that produces it.
Identify the equipotential surfaces of an isolated point charge as concentric spheres centred on the charge.
Distinguish the potential $V$ (a property of a point in space) from the potential energy $U$ of a pair of charges (a property of their interaction).
Solve numerically for any one of $q$ , $r$ , or $V$ given the other two, including problems that combine potential and field at the same point.

Ask the tutor: want to see how this single point-charge formula gets superposed twice, with a sign flip, to build the potential due to an electric dipole — ask the AI tutor to walk through it step by step.

Why does potential fall off slower than the field, and does that actually matter for solving problems?

It matters more than it looks. Because $V$ falls as $1/r$ while $E$ falls as $1/r^2$ , potential "reaches" much farther than the field does — at ten times the distance, the field has dropped to a hundredth of its close-up value, but the potential has only dropped to a tenth. Practically, this means that when you add up the influence of many distant charges, you often cannot ignore far-away charges' contribution to potential even after their contribution to the field has become negligible. It also explains why potential is such a convenient scalar tool for energy calculations across a whole charge distribution, while field calculations for the same distribution can get messy fast.

Is the potential due to a point charge the same thing as its potential energy?

No, and mixing them up costs marks. The potential $V(r) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ belongs to the source charge $q$ alone — it is a property of the point in space around $q$ , defined per unit test charge, and has units of volts. The potential energy of a pair of charges, $U = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r}$ , only exists once a second charge is actually placed at that point, and equals the product of that second charge with the potential the first one creates there: $U = q_2 V$ . Potential is a property of a location; potential energy is a property of an interaction between two (or more) actual charges.

Why do we choose the reference potential to be zero at infinity instead of at the charge itself?

Because $r=\infty$ is the one place we can define consistently and unambiguously, no matter what the charge distribution looks like: an electric field always decays with distance, so "infinitely far away" is a natural, distribution-independent zero. Choosing the charge's own location as the zero point is impossible for a point charge, since the field (and hence the work integral) diverges there — the formula for V itself would not even be defined. The infinity convention is simply the most universal bookkeeping choice, and it is the one the whole chapter (dipole potential, potential of a system of charges) is built on, so switching conventions midway would break every later formula's simple form.

My textbook derivation integrates E dr from r to infinity instead of infinity to r — did I learn it wrong?

Neither version is wrong; they describe the same physical work done from two equivalent bookkeeping directions and must give the same answer. One common convention (used above) is $V(r) = -\int_{\infty}^{r} E\,dr'$ , the work per unit charge done by an external agent bringing the test charge in from infinity against the field. An equally valid equivalent is to write $V(r) = \int_{r}^{\infty} E\,dr'$ , since reversing the limits of an integral flips its sign, and $-\int_{\infty}^{r} = \int_{r}^{\infty}$ . Both give exactly $\frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ for a point charge — if your version differs, check the sign convention on E and on the direction of integration rather than assuming the physics is different.