The Formula and Its Meaning
Take a point charge $Q$ fixed at the origin and a point $P$ at distance $r$ from it. The electrostatic potential at $P$ is defined as the work done by an external force in bringing a unit positive charge, without acceleration, from infinity up to $P$. NCERT obtains this by integrating the force per unit charge along a radial path from $r' = \infty$ to $r' = r$:
$$ V(r) = \frac{1}{4\pi\varepsilon_0}\,\frac{Q}{r} = \frac{kQ}{r}, \qquad k = \frac{1}{4\pi\varepsilon_0} = 9\times 10^{9}\ \text{N·m}^2/\text{C}^2 $$
Two structural facts are built into this result. First, the choice of zero potential at infinity makes the formula consistent: as $r \to \infty$, $V \to 0$. Second, because the Coulomb force is conservative, the work is path-independent — any path from infinity resolves into a radial part (which contributes) and perpendicular parts (which contribute nothing), so $V$ is a single-valued scalar at every point.
A unit positive test charge carried from infinity to $P$ against the field of $Q$; the work done per unit charge is the potential at $P$.
The same derivation appears in NIOS (Section 16.1.1), where the work to move a charge from a far point to $P$ is computed and divided by the charge to recover $V = kq/r$. The SI unit is the volt: one volt is the potential at a point if one joule of work is needed to bring one coulomb of charge there from infinity.
Sign of V and the 1/r Falloff
Equation $V = kQ/r$ holds for any sign of $Q$. NCERT stresses this explicitly: the derivation used $Q > 0$, but the result is general. For a positive charge the potential is positive everywhere around it; for a negative charge it is negative everywhere. The distance $r$ is always taken positive, so the sign of $V$ is carried entirely by the sign of $Q$.
| Source charge | Sign of V | Physical reading |
|---|---|---|
| $+Q$ | $V > 0$ | External force does positive work pushing a unit positive charge in against repulsion. |
| $-Q$ | $V < 0$ | The attractive force pulls the test charge in; external work per unit charge is negative. |
| At $r \to \infty$ | $V \to 0$ | Reference level; potential vanishes far from any charge. |
The contrast with the field is the single most examined idea in this subtopic. The field of a point charge is $E = kQ/r^2$ — it is the force per unit charge. The potential is the work per unit charge, obtained by integrating that force over distance; the extra integration softens one power of $r$. Hence $E \propto 1/r^2$ but $V \propto 1/r$. Double the distance and the field drops to a quarter, while the potential only halves.
$V \propto 1/r$, not $1/r^2$
Students who memorise "inverse square" for everything electrostatic write $V \propto 1/r^2$. That is the field. The potential of a point charge falls off only as $1/r$. If $r$ is tripled, the field becomes one-ninth but the potential becomes one-third.
Field: $E = kQ/r^2$ · Potential: $V = kQ/r$. Different powers of $r$.
The V-vs-r Graph
NCERT plots both $V \propto 1/r$ and $E \propto 1/r^2$ on the same axes (Figure 2.4). The two are easy to tell apart: the potential curve is the shallower one because it decays more slowly with distance. For $-Q$ the entire potential curve is reflected below the axis, since every value of $V$ simply changes sign.
Potential of $+Q$ (above the axis) and $-Q$ (below the axis). Both magnitudes follow a $1/r$ hyperbola and approach zero at large $r$.
A practical takeaway for graph-reading questions: near the charge the curve rises (or falls) steeply because $1/r$ blows up as $r \to 0$, and far away it flattens toward the zero line. A constant-potential region, by contrast, is a flat line — and a flat $V$ means $E = 0$, the basis of NEET 2020 Q.95.
Potential of a System of Charges
Because potential is related to work, it obeys the superposition principle. NCERT (Section 2.5) states it directly: for charges $q_1, q_2, \dots, q_n$ at distances $r_{1P}, r_{2P}, \dots, r_{nP}$ from the point $P$, the resultant potential is the algebraic sum of the individual potentials.
$$ V = V_1 + V_2 + \dots + V_n = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n} \frac{q_i}{r_{i}} = \sum_i \frac{k\,q_i}{r_i} $$
The word algebraic carries the whole burden of the calculation: each term enters with the sign of its own charge, and the terms add as ordinary numbers. There is no angle to resolve and no vector to decompose — a sharp contrast with the electric field of a system, which must be added head-to-tail as a vector.
Each charge contributes $kq_i/r_i$ at $P$; the potential is the running total of signed scalars — no directions involved.
Apply the scalar sum to two opposite charges and you get the dipole result. See Potential Due to a Dipole, where $V$ falls off as $1/r^2$ and depends on direction.
Potentials add as signed scalars — never as vectors
A common error is to "resolve" potentials into components, as one would for fields. Potential has no direction. For a system you simply add $kq_i/r_i$ with each charge's sign. Forgetting the minus sign on a negative charge is the most frequent slip in two-charge problems.
$V = \sum_i \dfrac{kq_i}{r_i}$ — keep the sign of each $q_i$; no angles, no components.
$V = 0$ does not mean $E = 0$
At the midpoint of a $+q$, $-q$ pair the two potentials cancel ($+kq/r - kq/r = 0$), yet both fields point the same way and add, so $E \neq 0$. Conversely, where $E = 0$ the potential can be non-zero. The two quantities are independent: one is a scalar sum, the other a vector sum.
Scalar cancellation ($V=0$) and vector cancellation ($E=0$) are different conditions.
Worked Examples
Find the potential at a point $P$ due to a charge $4\times 10^{-7}\,\text{C}$ located $9\,\text{cm}$ away. Hence find the work done in bringing a charge $2\times 10^{-9}\,\text{C}$ from infinity to $P$.
$V = \dfrac{kQ}{r} = \dfrac{(9\times 10^{9})(4\times 10^{-7})}{0.09} = 4\times 10^{4}\,\text{V}$.
Work to bring charge $q$ from infinity: $W = qV = (2\times 10^{-9})(4\times 10^{4}) = 8\times 10^{-5}\,\text{J}$. The answer does not depend on the path, since the electrostatic force is conservative.
Two charges $+5\,\text{nC}$ and $-3\,\text{nC}$ lie $0.20\,\text{m}$ and $0.30\,\text{m}$ respectively from a point $P$. Find the net potential at $P$.
$V = k\left(\dfrac{q_1}{r_1} + \dfrac{q_2}{r_2}\right) = 9\times 10^{9}\left(\dfrac{5\times 10^{-9}}{0.20} + \dfrac{-3\times 10^{-9}}{0.30}\right)$
$= 9\times 10^{9}\,(25\times 10^{-9} - 10\times 10^{-9}) = 9\times 10^{9}\times 15\times 10^{-9} = 135\,\text{V}$. The negative charge subtracts; no direction is used.
Potential due to a point charge
- $V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r} = \dfrac{kQ}{r}$, with $k = 9\times 10^{9}\,\text{N·m}^2/\text{C}^2$; zero at infinity.
- It is the work per unit positive charge to bring it from infinity to that point; path-independent.
- The sign of $V$ follows the sign of $Q$: positive around $+Q$, negative around $-Q$.
- $V \propto 1/r$, whereas the field $E \propto 1/r^2$ — the potential decays more slowly.
- For a system, $V = \sum_i \dfrac{kq_i}{r_i}$ — an algebraic (signed scalar) sum, by superposition.
- $V = 0$ does not force $E = 0$, and vice versa.