In short

The electric field is the negative gradient of the potential:

\vec{E} \;=\; -\nabla V \;=\; -\left(\frac{\partial V}{\partial x}\hat{x} + \frac{\partial V}{\partial y}\hat{y} + \frac{\partial V}{\partial z}\hat{z}\right).

In one dimension, E = -dV/dr — the field at a point is the negative slope of the V-versus-position graph there. The "negative" is what makes a positive test charge roll downhill — it accelerates toward lower potential. Running the relation backward gives the line integral

V(\vec{b}) - V(\vec{a}) \;=\; -\int_{\vec{a}}^{\vec{b}} \vec{E}\cdot d\vec{\ell},

which is path-independent for electrostatic fields (you will see why). Knowing V everywhere gives you \vec{E} everywhere (one derivative per coordinate); knowing \vec{E} everywhere gives you V everywhere (up to a constant). Units: [V/\text{m}] = [\text{N}/\text{C}] — so "volts per metre" is literally the same unit as "newtons per coulomb," just two different names for what the field is doing.

Look at any capacitor in an electronics lab drawer at IIT Madras and you will see stamped on its side something like "250 V / 3 mm" — the voltage it is rated for, and the separation of its plates. That one label is secretly telling you the strength of the electric field inside: divide the voltage by the gap, 250\ \text{V}/(3\times 10^{-3}\ \text{m}) \approx 83 kV per metre. The field is what ionises air if you overshoot the rating; the field is what a charge inside the gap feels; the field is what does work on an electron beam as it flies through a CRT tube. The stamped number on the nameplate is a scalar. The field is a vector. And between them there is this remarkable translation rule:

\vec{E} \;=\; -\nabla V,

the negative gradient of the potential.

This is the bridge between two ways of describing electrostatics. One language says "there is a push at every point in space" — that is the field. The other language says "every point has a scalar value assigned to it" — that is the potential. The two languages carry exactly the same physics: whichever one you know, you can compute the other with one line of calculus. And the translation is not a postulate you accept; it falls out of the definitions of work and energy that you already have. This article derives it, shows how to use it in both directions, and pauses at the few spots where students routinely trip — the minus sign, the "slope of V" reading of dV/dr, the reason equipotentials sit perpendicular to field lines everywhere.

Deriving \vec{E} = -\nabla V from work and energy

Start with what you already know. The potential difference between two nearby points A and B is defined by

V_B - V_A \;=\; \frac{-W_{AB}^\text{(by field)}}{q_0} \;=\; \frac{W_{AB}^\text{(by you, against the field)}}{q_0},

where W_{AB} is the work needed to move a positive test charge q_0 from A to B (against the field). Alternatively, the electric field does work W_{AB}^\text{field} = \int_A^B \vec{F}\cdot d\vec{\ell} = q_0\int_A^B \vec{E}\cdot d\vec{\ell} on the test charge; when you move "against" the field you do the negative of that, which is what the definition above records.

The one-dimensional case

Set up the simplest geometry: a uniform field \vec{E} along +\hat{x}, and two points A and B separated by dx in the same direction. The test charge q_0 moves from A to B — a tiny displacement d\vec{\ell} = dx\,\hat{x}.

The work done by the field on q_0 is

dW^\text{by field} \;=\; \vec{F}\cdot d\vec{\ell} \;=\; q_0\vec{E}\cdot d\vec{\ell} \;=\; q_0 E_x\,dx.

Why: force on the test charge is \vec{F} = q_0\vec{E}; the dot product \vec{E}\cdot d\vec{\ell} = E_x\,dx when both \vec{E} and d\vec{\ell} are along \hat{x}. When they align, the field does positive work; when they oppose each other, the field does negative work.

By the definition of potential difference,

V_B - V_A \;=\; -\frac{dW^\text{by field}}{q_0} \;=\; -E_x\,dx.

Why: work done by the field reduces the electric potential energy of the test charge; equivalently, it drops the potential V seen by the charge. The minus sign encodes this energetic fact — when the field pushes you downhill (positive work), you end up at lower potential.

Now the left-hand side is V_B - V_A = dV (the change in V over the step dx). Dividing both sides by dx:

\boxed{\;E_x \;=\; -\frac{dV}{dx}\;} \tag{1}

This is the one-dimensional form of the field–potential relation. The field at a point is the negative slope of the V-versus-x graph at that point.

Two features of (1) to register:

Generalising to three dimensions

For a general displacement d\vec{\ell} = dx\,\hat{x} + dy\,\hat{y} + dz\,\hat{z}, repeat the derivation. The work by the field is

dW^\text{by field} \;=\; q_0\vec{E}\cdot d\vec{\ell} \;=\; q_0(E_x\,dx + E_y\,dy + E_z\,dz).

Why: the dot product of two vectors expands component-wise. Each term contributes the work done by one component of the field over its corresponding displacement.

And

dV \;=\; -\frac{dW^\text{by field}}{q_0} \;=\; -(E_x\,dx + E_y\,dy + E_z\,dz). \tag{2}

But V is a function of position, V(x,y,z), so multivariable calculus tells you

dV \;=\; \frac{\partial V}{\partial x}\,dx + \frac{\partial V}{\partial y}\,dy + \frac{\partial V}{\partial z}\,dz. \tag{3}

Why: the total differential of a function of three variables is the sum of each partial derivative times its corresponding infinitesimal. If V changes with x at rate \partial V/\partial x, a step dx changes V by (\partial V/\partial x)\,dx, and similarly for y and z.

Set (2) equal to (3) and compare coefficients of dx, dy, dz:

\boxed{\;E_x \;=\; -\frac{\partial V}{\partial x}, \quad E_y \;=\; -\frac{\partial V}{\partial y}, \quad E_z \;=\; -\frac{\partial V}{\partial z}\;} \tag{4}

Or, in compact vector notation,

\vec{E} \;=\; -\left(\frac{\partial V}{\partial x}\hat{x} + \frac{\partial V}{\partial y}\hat{y} + \frac{\partial V}{\partial z}\hat{z}\right) \;=\; -\nabla V. \tag{5}

Why: the symbol \nabla (pronounced "del" or "nabla") is a shorthand for the vector of partial derivatives \nabla = \hat{x}\,\partial_x + \hat{y}\,\partial_y + \hat{z}\,\partial_z. Applied to a scalar field V, it produces the vector of its partial derivatives. "\nabla V" is called the gradient of V, and -\nabla V is the field. The one-dimensional version (1) is just (5) restricted to motion along one axis.

Reading the gradient geometrically

The gradient \nabla V at a point has two geometric meanings that are worth keeping together:

So -\nabla V points in the direction V decreases fastest, and its magnitude is how fast V is dropping along that direction. The field, in other words, is the "steepest descent" direction of the potential.

If you imagine V(x,y) plotted as a landscape — valleys at low potential, hills at high potential — then the electric field \vec{E} at any point is the direction a ball placed there would roll, with magnitude proportional to the steepness of the slope. Positive test charges roll downhill in this landscape; negative charges roll uphill.

Equipotential contours with field arrows pointing down the gradient A set of nested equipotential curves labelled V = 4, V = 3, V = 2, V = 1 around a central high-potential region, with small red arrows at various points pointing perpendicular to the contours from high to low potential, and longer arrows where the contours are closely spaced. V=4 V=3 V=2 V=1 arrows: direction of E = −∇V contours: equipotentials
Nested equipotential contours around a positive charge-like region. Field arrows point perpendicular to each contour, from higher $V$ to lower $V$. The closer the contours, the larger $|\vec{E}|$ — the rate at which $V$ changes with distance.

One consequence pops out immediately: the field is always perpendicular to equipotential surfaces. Along an equipotential, dV = 0 by definition; equation (2) becomes \vec{E}\cdot d\vec{\ell} = 0 for any d\vec{\ell} lying in the equipotential — so \vec{E} and d\vec{\ell} are perpendicular, no matter which direction along the surface you move. This is the geometric picture you will meet head-on in the equipotential surfaces chapter.

The interactive V\to E connection

Drag the test point along the x-axis in the figure below. The top graph is a scalar potential V(x), chosen for clarity. The bottom graph is E_x(x) = -dV/dx, the negative slope of the top graph. The live readouts show the value of V, the slope dV/dx, and the corresponding field E_x at the chosen point.

Interactive V(x) and derived field E(x) Two graphs stacked vertically: the upper shows V(x) as a smooth curve; the lower shows E(x) = -dV/dx. A draggable test point on x reads off the values and places markers on both graphs. x V(x) x Eₓ(x) drag the red point along x
The top curve is a chosen potential $V(x) = 3e^{-0.4x^2}$. The bottom curve is the resulting field $E_x(x) = -dV/dx = 2.4\,x\,e^{-0.4x^2}$. At $x = 0$, $V$ is at its peak and $E_x = 0$ (flat top); for $x > 0$, $V$ decreases, so $E_x > 0$ (field points to the right); for $x < 0$, $V$ also decreases as you move away from the peak, so $E_x < 0$.

Inverting the relation — recovering V from \vec{E}

Equation (2), dV = -\vec{E}\cdot d\vec{\ell}, can be integrated along a path from a reference point \vec{a} to a field point \vec{b}:

V(\vec{b}) - V(\vec{a}) \;=\; -\int_{\vec{a}}^{\vec{b}}\vec{E}\cdot d\vec{\ell}. \tag{6}

Why: summing up the infinitesimal changes dV = -\vec{E}\cdot d\vec{\ell} along a path from \vec{a} to \vec{b} gives the total change in V over the path. This is the fundamental theorem of calculus applied to the gradient — you will meet it more abstractly in multivariable math.

With the reference taken at infinity, V(\infty) = 0, and (6) becomes

V(\vec{r}) \;=\; -\int_\infty^{\vec{r}}\vec{E}\cdot d\vec{\ell}. \tag{7}

For the electrostatic field, the integral is path-independent: the work done on a test charge in going from \vec{a} to \vec{b} does not depend on which path you take. You can trust this because Coulomb's law makes the electrostatic force conservative — the same property that lets you define a potential energy at all. (The formal proof uses Stokes's theorem and \nabla\times\vec{E} = 0 for any electrostatic field; you will meet this in the going-deeper section.)

The practical consequence: for any \vec{E} you have determined (from Gauss's law, superposition, or symmetry), integrate along the most convenient path back to infinity. The answer is V at your field point. You already used this in Potential Due to Systems of Charges when you recovered V for a solid sphere from the Gauss's-law field.

Why the field points from high to low potential

Equation (5), \vec{E} = -\nabla V, encodes the direction rule in its minus sign. Here is the physical interpretation laid out plainly.

Take a tiny positive test charge q_0 in a region where V is larger at point A and smaller at point B. The field at each point pushes the test charge in some direction; where does it push?

The potential energy of the test charge at a point is U = q_0 V. With q_0 > 0, U is larger at A (higher V) than at B. Every mechanical system tends to move toward lower potential energy (ask the apple why it fell). So the positive test charge accelerates from A toward B, from higher V to lower V.

Since the only force on the test charge is q_0\vec{E} and q_0 > 0, \vec{E} points along the direction the charge accelerates — from high V to low V. A negative charge would do the opposite: its potential energy U = qV is lower where V is higher (because q<0), so it rolls "uphill" in V toward high potential — but the field, which belongs to space not to the charge, still points from high V to low V. The charge is simply pushed in the opposite direction because it has negative charge.

This is why a capacitor "pulls" electrons from the negative plate to the positive plate internally once it starts to discharge: from the electron's point of view, the negative plate is at low V and the positive plate is at high V; the field points from positive to negative (high to low), and the electron, being negative, moves opposite to the field — toward the positive plate.

Cylindrical and spherical forms

For problems with circular or spherical symmetry, Cartesian partials are clumsy. Use the appropriate coordinates.

In cylindrical coordinates (r,\phi,z), the gradient is

\nabla V \;=\; \hat{r}\,\frac{\partial V}{\partial r} + \hat{\phi}\,\frac{1}{r}\frac{\partial V}{\partial \phi} + \hat{z}\,\frac{\partial V}{\partial z}.

In spherical coordinates (r,\theta,\phi), it is

\nabla V \;=\; \hat{r}\,\frac{\partial V}{\partial r} + \hat{\theta}\,\frac{1}{r}\frac{\partial V}{\partial \theta} + \hat{\phi}\,\frac{1}{r\sin\theta}\frac{\partial V}{\partial \phi}.

For a spherically symmetric V (depending only on r), all angular derivatives vanish and

\vec{E} \;=\; -\frac{dV}{dr}\,\hat{r}. \tag{8}

Apply (8) to the point-charge potential V = kQ/r:

E_r \;=\; -\frac{d}{dr}\left(\frac{kQ}{r}\right) \;=\; -kQ\cdot\left(-\frac{1}{r^2}\right) \;=\; \frac{kQ}{r^2}. \tag{9}

Why: the derivative of 1/r is -1/r^2; the two minus signs combine to give a plus. You recover Coulomb's law for the field, with the correct sign: positive Q gives outward-pointing field, negative Q gives inward.

Apply (8) to the solid sphere's inside potential V = kQ(3R^2 - r^2)/(2R^3):

E_r \;=\; -\frac{d}{dr}\left(\frac{kQ(3R^2 - r^2)}{2R^3}\right) \;=\; -\frac{kQ}{2R^3}\cdot(-2r) \;=\; \frac{kQr}{R^3}. \tag{10}

This is the linear-in-r field inside a uniformly charged ball, matching what Gauss's law gave you. One line of calculus recovers a formula that took a Gaussian-surface argument in the previous chapter.

Worked examples

Example 1: Axial field of a charged ring from the potential

A uniformly charged ring of radius a carries total charge Q. From the previous chapter, the potential on its axis is

V(x) \;=\; \frac{kQ}{\sqrt{x^2+a^2}},

where x is the distance along the axis from the centre. Use \vec{E} = -\nabla V to find the axial component of the electric field on the axis.

Ring of charge with axial field point Ring of radius a viewed edge-on; an axial field point at distance x from the centre. V(x) given, Ex(x) to be found. ring, radius a, charge Q field point (axis) x V(x) = kQ / √(x² + a²) Eₓ(x) = ?
$V$ on the axis is known from the previous chapter; take $-dV/dx$ to get $E_x$ without any vector integration.

Step 1. Write the potential as a function of the axial coordinate.

V(x) \;=\; kQ\,(x^2+a^2)^{-1/2}.

Why: rewriting the square root as a power makes the chain rule easier to apply.

Step 2. Differentiate with respect to x.

\frac{dV}{dx} \;=\; kQ\cdot\left(-\tfrac{1}{2}\right)(x^2+a^2)^{-3/2}\cdot 2x \;=\; -\frac{kQx}{(x^2+a^2)^{3/2}}.

Why: the chain rule with f(u) = u^{-1/2} and u = x^2 + a^2 gives df/du = -\tfrac{1}{2}u^{-3/2}, multiplied by du/dx = 2x. The factors of 2 cancel, and a minus sign survives.

Step 3. Apply \vec{E} = -\nabla V; on the axis the only non-zero component is E_x = -dV/dx.

E_x(x) \;=\; -\frac{dV}{dx} \;=\; \frac{kQx}{(x^2+a^2)^{3/2}}. \tag{11}

Why: the overall minus in (5) combines with the minus from the derivative to give a positive result when x>0 (field points away from the ring in the +\hat{x} direction, as you expect for a positively charged ring).

Step 4. Sanity checks.

  • At x=0 (centre of the ring): E_x = 0. Good — by symmetry, every element of the ring contributes an axial component equal in magnitude but opposite in direction to the element diametrically opposite; they cancel at the centre.
  • At x\to\infty: E_x \to kQx/x^3 = kQ/x^2, the field of a point charge Q. Good.
  • Maximum of |E_x|: set dE_x/dx = 0, which gives x = a/\sqrt{2}. At this distance the axial field is strongest. You can check this yourself by differentiating (11).

Result: E_x(x) = \dfrac{kQx}{(x^2+a^2)^{3/2}}, in agreement with the direct integration of Coulomb's law performed in the electric field chapter.

What this shows: Computing \vec{E} by taking -\nabla V of a known potential is often several times quicker than setting up a vector integral directly. The potential does the geometric bookkeeping for you — once the scalar integral is done, a single differentiation recovers the field.

Example 2: Potential from a piecewise-linear field

An electric field points along the +\hat{x} direction with magnitude that depends on position as

E_x(x) \;=\; \begin{cases}+500\ \text{V/m} & 0 \le x < 0.02\ \text{m} \\ -200\ \text{V/m} & 0.02\ \text{m} \le x < 0.05\ \text{m}\end{cases}.

This is the kind of profile you would see between the plates of a non-uniform-density capacitor or in the drift region of a discharge tube. Take V(0) = 0. Find V(x) everywhere in [0, 0.05\ \text{m}] and plot it.

Piecewise-constant field and the potential found by integration Top graph shows E(x) as a step function positive 500 V/m from x=0 to x=0.02 m, then negative 200 V/m from 0.02 to 0.05 m. Bottom graph shows V(x) as a piecewise linear function decreasing then increasing. x (m) Eₓ (V/m) +500 −200 V (V) V(0) = 0 V(0.02) = −10 V V(0.05) = −4 V x = 0.02 m
A step field gives a piecewise-linear potential. Where $E_x > 0$, $V$ decreases with $x$; where $E_x < 0$, $V$ increases with $x$.

Step 1. Use (7) for the first region.

V(x) - V(0) \;=\; -\int_0^x E_x(x')\,dx' \;=\; -(500)\int_0^x dx' \;=\; -500x.

Why: in the first region E_x is constant at +500 V/m, so the integral is simply the constant times the interval length. The minus sign in (6) turns "field pointing in +\hat{x}" into "potential decreasing with x."

With V(0) = 0: V(x) = -500x for 0\le x< 0.02 m. At the boundary:

V(0.02) \;=\; -500\times 0.02 \;=\; -10\ \text{V}.

Step 2. Start the second region from x = 0.02 m and V = -10 V.

V(x) - V(0.02) \;=\; -\int_{0.02}^x E_x(x')\,dx' \;=\; -(-200)(x - 0.02) \;=\; +200(x - 0.02).

Why: here E_x = -200 V/m is constant; the minus in (6) combines with the minus of E_x to give a plus, so V increases linearly with x in this region. The starting potential is whatever we computed at the boundary.

So V(x) = -10 + 200(x - 0.02) for 0.02 \le x < 0.05 m. At x = 0.05 m:

V(0.05) \;=\; -10 + 200\times 0.03 \;=\; -10 + 6 \;=\; -4\ \text{V}.

Step 3. Check continuity at the boundary.

From the left: V(0.02^-) = -500\times 0.02 = -10 V.

From the right: V(0.02^+) = -10 + 200\times 0 = -10 V.

Why: V must be continuous everywhere that \vec{E} is finite, because jumps in V would require an infinite field over zero distance. Both pieces give the same value at the boundary, as they must.

Step 4. Summarise.

V(x) \;=\; \begin{cases}-500x\ \text{V} & 0\le x < 0.02\ \text{m} \\ -10 + 200(x - 0.02)\ \text{V} & 0.02\le x < 0.05\ \text{m}\end{cases}.

Result: V drops linearly from 0 V to -10 V over the first 2 cm, then rises linearly from -10 V to -4 V over the next 3 cm.

What this shows: The sign of dV/dx always matches -E_x. Where the field points strongly one way, V slopes steeply the other way. A change in E_x shows up as a kink in V (piecewise-linear because the field is piecewise-constant). This is the bread-and-butter of circuit-potential problems: the moment you know \vec{E} between every pair of components, integration gives you the full V distribution.

Common confusions

If your immediate need is to compute \vec{E} from V or vice versa, you have the tools. The rest of this section clarifies why the line integral (6) is path-independent, introduces the curl condition on electrostatic fields, and sketches where the -\nabla V relation comes from in a broader energy framework.

Why the electrostatic line integral is path-independent

The statement in (6) assumes that \int_{\vec{a}}^{\vec{b}}\vec{E}\cdot d\vec{\ell} depends only on the endpoints, not on the path. Why?

For any closed path C, the integral of \vec{E} around C is zero in electrostatics:

\oint_C \vec{E}\cdot d\vec{\ell} \;=\; 0. \tag{12}

This is the electrostatic "conservative field" condition. Start from the fact that the force on a test charge q_0 moving around a closed path returns the charge to its starting point. If \oint \vec{F}\cdot d\vec{\ell} \neq 0, you could extract net work from a closed cycle — a perpetual motion machine, forbidden by energy conservation. So \oint\vec{F}\cdot d\vec{\ell} = 0, which (dividing by q_0) gives (12).

By Stokes's theorem, the line integral around a closed loop equals the surface integral of the curl over any surface bounded by the loop:

\oint_C\vec{E}\cdot d\vec{\ell} \;=\; \iint_S (\nabla\times\vec{E})\cdot d\vec{A}.

For (12) to hold for every closed loop, the curl of \vec{E} must be zero everywhere in electrostatics:

\nabla\times\vec{E} \;=\; \vec{0}. \tag{13}

Any vector field with zero curl can be written as the gradient of a scalar — this is the statement that curl-free fields are gradient fields. Equation (13) is what lets you write \vec{E} = -\nabla V in the first place; without it, there would be no single-valued V.

Equation (13) fails for time-varying fields (when a changing magnetic field is present, Faraday's law gives \nabla\times\vec{E} = -\partial\vec{B}/\partial t \neq 0). That is why electric potential as a single scalar does not exist for time-dependent fields — you need the full vector and scalar potentials of electromagnetism. But for everything in this chapter and the JEE electrostatics syllabus, (13) holds and V is a well-defined function of position.

The Laplacian — where V satisfies Poisson's and Laplace's equations

Combine \vec{E} = -\nabla V with Gauss's law, \nabla\cdot\vec{E} = \rho/\varepsilon_0:

\nabla\cdot(-\nabla V) \;=\; \rho/\varepsilon_0,
\nabla^2 V \;=\; -\rho/\varepsilon_0. \tag{14}

This is Poisson's equation. The symbol \nabla^2 (the Laplacian) is the divergence of the gradient — the sum of second partial derivatives: \nabla^2 V = \partial^2 V/\partial x^2 + \partial^2 V/\partial y^2 + \partial^2 V/\partial z^2.

In regions where \rho = 0 (away from any charge), Poisson's equation reduces to Laplace's equation:

\nabla^2 V \;=\; 0. \tag{15}

This is the governing equation for the electrostatic potential in empty space. Its solutions have a famous property — the value of V at any interior point equals the average of V on any sphere surrounding that point. This "mean-value property" is why V cannot have local maxima or minima in a charge-free region; extrema can only sit on the boundary or on the charges themselves. You see this in everyday problems: inside a hollow conductor, V is a constant (the interior has no charge; the boundary is an equipotential; so V equals its boundary value everywhere inside).

Solving Poisson's and Laplace's equations subject to boundary conditions is the core of advanced electrostatics (images, separation of variables, Green's functions). You will meet these ideas in detail if you take a full electromagnetism course at IIT or ISI. For now, the one-line takeaway is: everything you know about V from Coulomb's law and superposition is captured in one partial differential equation, and every electrostatics boundary-value problem is a special case of solving (14) or (15).

The uniqueness theorem — one V, one physics

Two potentials V_1 and V_2 that satisfy the same boundary conditions (same values on all the conductor surfaces and at infinity) and the same charge distribution in between must be equal. This is the uniqueness theorem: the physical configuration determines the potential uniquely, up to the additive constant you fix by a reference choice.

Why does this matter? It gives you licence to guess. If you can find any V that satisfies Laplace's equation and the boundary conditions — even by inspired guessing or symmetry — uniqueness guarantees it is the answer, with no further proof needed. This is the logic behind the method of images, where you replace a conducting plane with a mirror image charge and solve a much simpler problem; the uniqueness theorem tells you the answer is rigorous for the original geometry.

Where this leads next