Relation between field and potential

Stand a metre from a charged sphere and hold a tiny positive test charge. Two questions can be asked about that spot. "What force per unit charge acts here?" is answered by the electric field $\vec E$ . "How much work per unit charge would it take to bring a charge here from far away?" is answered by the electric potential $V$ . They sound like separate quantities — one a vector telling you push-and-direction, the other a single number telling you stored capacity for work — yet they are two views of the very same electrostatic landscape, and one determines the other completely. The bridge between them is astonishingly simple in one dimension: the field is the negative rate of change of potential with distance, $E = -dV/dr$ . Read that sentence physically and a whole cluster of facts falls into place at once — why field lines always run from high potential to low potential, why potential is flat (and field zero) on the surface of a charged conductor, why the unit "volt per metre" is exactly the same thing as "newton per coulomb," and why the space between two capacitor plates, wherever the field is uniform, obeys the beautifully plain relation $E = V/d$ . This article builds that bridge from the definition of potential difference itself, using nothing beyond a small-displacement argument and one line of calculus-free reasoning suited to Class 12. It then puts the relation to work on exactly the kind of problems that appear in CBSE board papers and in the parallel-plate capacitor questions that dominate this chapter — reading off a field from a potential graph, and finding the field strength inside a real capacitor from its rated voltage and plate gap.

Key takeaways

The electric field measures how fast the potential changes with position: $E = -\dfrac{dV}{dr}$ , the rate of change taken along the field direction.
The minus sign is not decorative — it says the field points from high potential toward low potential, i.e., in the direction of decreasing $V$ .
Field and potential share a single combined SI unit: $1\ \text{N/C} = 1\ \text{V/m}$ , and this relation is exactly why.
On an equipotential surface $dV = 0$ in every direction along the surface, so the field must be perpendicular to it everywhere — the two ideas (equipotentials and field-from-potential) are really one idea seen twice.
For a uniform field, such as between the plates of a parallel-plate capacitor separated by distance $d$ carrying potential difference $V$ , the relation collapses to the arithmetic-friendly $E = V/d$ .
A steep potential graph (large slope) means a strong field there; a flat stretch of potential (zero slope) means zero field there, however large the potential value itself may be.
Key idea: the electric field at a point is the negative slope of the potential with distance, $E = -dV/dr$ , which reduces to the constant-field formula $E = V/d$ for a uniform field such as inside a parallel-plate capacitor.

The core insight

Electric field is the negative rate at which potential falls with distance along the field direction, $E = -dV/dr$ , so a uniform field between two surfaces separated by $d$ and held at potential difference $V$ has magnitude $E = V/d$ , pointing from the higher-potential surface to the lower.

Where it fits in the syllabus

This relation sits inside the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), right after equipotential surfaces and just before the discussion of potential energy of a system of charges. The prescribed treatment is the one-dimensional relation $E = -dV/dr$ along the field direction — its sign, its unit consequence (V/m), and its use in simple settings such as the uniform field of a parallel-plate capacitor. The general three-dimensional gradient form of this relation belongs to a later, university-level vector-calculus treatment and is explicitly outside the board syllabus; nothing in this article goes beyond the one-dimensional version that CBSE expects.

The physical picture

Electric field and potential describe the same charge configuration from two different, complementary angles. The field $\vec E$ at a point is a force-per-unit-charge — how hard, and in which direction, a positive test charge there would be pushed. The potential $V$ at a point is a work-per-unit-charge — how much energy per coulomb it took (or would take) to assemble a charge at that location, starting from a reference point (usually infinity) where $V=0$ . Because force and work are linked by the calculus of "work equals force times distance," field and potential must be linked the same way. Consider two points a very small distance $dr$ apart, measured along the direction in which the field $\vec E$ actually points. Moving a unit positive test charge through this tiny displacement, the field does work $E\,dr$ on it (force per unit charge times distance, since the charge is unity). Work done by the field always comes at the expense of potential energy, so the potential itself must fall by exactly that amount over the same small step:

dV = -E\,dr \qquad\Longrightarrow\qquad E = -\frac{dV}{dr}

where: $E$ is the magnitude of the electric field along the chosen direction, $dV$ is the (small) change in potential over the small displacement $dr$ , and the displacement is taken along the field's own direction. The physical reading: field is literally how steeply potential is sloping downward at that point, in the direction the field points. Where the slope of $V$ against distance is steep, $E$ is large; where the graph of $V$ is flat, $E$ is zero, no matter how large or small $V$ itself happens to be there.

Figure

A graph of potential falling with distance next to a matching diagram of field lines, showing that the slope of the potential curve equals the magnitude of the electric field and the field always points toward lower potential

For a uniform electric field — the field between the two plates of a parallel-plate capacitor is the standard example — the slope $dV/dr$ is the same everywhere in the gap, so the small-displacement relation collapses to plain division: if the plates are a distance $d$ apart and carry a potential difference $V$ between them,

E = \frac{V}{d}

where: $E$ is the (constant) field magnitude in the gap, $V$ is the potential difference between the plates, and $d$ is the plate separation. This single formula is one of the most heavily used relations in this chapter, because most capacitor numericals hand you exactly $V$ and $d$ and ask for $E$ , or the reverse. This same relation also explains, from a fresh angle, why field lines always meet equipotential surfaces at right angles. Move along an equipotential surface and, by definition, $V$ does not change — $dV = 0$ in that direction. But $E = -dV/dr$ is (in general) not zero along the field's own direction. A direction of zero change and a direction of maximum change at the same point can only be perpendicular to each other, which is exactly the geometric fact used when equipotential surfaces are drawn as closed shells around a charge, always crossing the radial field lines at ninety degrees.

Derivation AGetting E = -dV/dr from the definition of potential difference

The goal here is to build the relation between electric field and potential purely from the definition of potential difference as work done per unit charge, using only a small displacement and no vector calculus beyond what "rate of change" already means — this is exactly the level expected in the CBSE treatment of this topic, and it is worth doing carefully once so that the sign and the direction convention never get confused again in a numerical.

Step 1. Recall the meaning of potential. The potential $V$ at a point is defined as the work done, per unit positive charge, in bringing that charge from infinity (taken as the zero of potential) to that point, against the electric field, without any change in kinetic energy. Potential difference between two points $A$ and $B$ is $V_A - V_B = W_{B\to A}/q$ , the work done per unit charge in moving the charge from $B$ to $A$ .

Step 2. Set up a small displacement along the field. Take two points separated by a very small distance $dr$ , chosen along the direction in which the electric field $E$ actually points at that location. Moving a unit positive test charge through this small step, the electrostatic force on it is $E$ (force per unit charge, since the charge is one coulomb), so the work done by the field on the charge over this small displacement is $dW = E\,dr$ .

Step 3. Connect that work to the change in potential. Work done by the field on a positive charge always happens at the cost of the charge's potential energy — the charge moves toward lower potential energy, hence toward lower potential (since potential is potential energy per unit charge). So the potential must decrease by exactly the amount of work done per unit charge over that step: $dV = -dW = -E\,dr$ .

Step 4. Solve for the field. Rearranging Step 3 gives $E = -\dfrac{dV}{dr}$ . The negative sign is a direct consequence of Step 3, not an arbitrary convention — it is forced by the fact that the field does positive work as the charge moves toward lower potential, so $V$ falls in the very direction $E$ points.

Step 5. Read off the two immediate consequences used everywhere in this chapter. First, since $E$ has units of $\text{V/m}$ from this equation just as surely as it has units of $\text{N/C}$ from $F = qE$ , the two units are identical: $1\ \text{N/C} = 1\ \text{V/m}$ . Second, for a uniform field (potential falling at a constant rate over a finite gap $d$ with total drop $V$ ), the small-displacement relation integrates trivially to the constant-field form $E = V/d$ , which is the version used in almost every capacitor numerical in this chapter.

Worked examples

Board level - field from a linear potential expression

The electric potential in a region varies with position along the $x$ -axis as $V = 200 - 50x$ , with $V$ in volts and $x$ in metres. Find the electric field in this region, including its direction.

Step 1. Write down the relevant relation. Since the potential is given as a function of one coordinate only, use $E = -\dfrac{dV}{dx}$ .

Step 2. Differentiate the given expression. $V = 200 - 50x$ , so $\dfrac{dV}{dx} = -50\ \text{V/m}$ — a constant, meaning this is actually a uniform field throughout the region.

Step 3. Apply the sign. $E = -\left(\dfrac{dV}{dx}\right) = -(-50) = 50\ \text{V/m}$ . Since $E$ came out positive with $x$ increasing chosen as the positive direction, the field points along the positive $x$ -direction — consistent with $V$ falling (from $200$ toward smaller and eventually negative values) as $x$ increases, exactly as the rule "field points toward decreasing potential" requires.

Answer: the electric field is 50 V/m, directed along the positive x-axis.

Competitive level - capacitor field, then adding a dielectric slab check

A parallel-plate capacitor has its plates separated by 5 mm and connected to a 1000 V source. (a) Find the uniform electric field between the plates. (b) A student then halves the plate separation to 2.5 mm while keeping the same 1000 V across the plates. Find the new field, and state whether the field doubled, halved, or stayed the same.

Step 1. Use the uniform-field relation $E = V/d$ for the original geometry. Convert $d = 5\ \text{mm} = 5\times10^{-3}\ \text{m}$ , so $E_1 = \dfrac{1000}{5\times10^{-3}} = 2\times10^{5}\ \text{V/m}$ .

Step 2. Repeat with the halved separation, same voltage: $d_2 = 2.5\ \text{mm} = 2.5\times10^{-3}\ \text{m}$ , so $E_2 = \dfrac{1000}{2.5\times10^{-3}} = 4\times10^{5}\ \text{V/m}$ .

Step 3. Compare. $E_2 / E_1 = 4\times10^{5} / 2\times10^{5} = 2$ — the field exactly doubled when the separation was halved at fixed voltage, which is exactly what $E = V/d$ predicts (with $V$ held fixed, $E \propto 1/d$ ), and is the standard "what happens if the gap is halved" twist examiners like to add onto the plain formula question.

Answer: (a) E = 2×10^5 V/m; (b) E = 4×10^5 V/m, so the field doubled. Timing note: about a minute and a half. Both parts are a single division each — the only place marks are lost is forgetting to convert millimetres to metres before dividing, or trying to reason about the ratio without simply recomputing both fields directly.

Common traps

Dropping the negative sign in $E = -dV/dr$ : the sign is what correctly fixes the direction of the field (toward decreasing potential); in a purely magnitude-based numerical the final number may still come out right, but stating direction without the sign is a common way to lose a mark.
Believing that a large potential automatically means a large field: it is the rate of change of potential with distance that sets the field's strength, not the absolute size of $V$ itself — a region can have a huge potential and still zero field if $V$ is constant there.
Applying $E = V/d$ outside a uniform field: this shortcut is valid specifically for a uniform field, such as between capacitor plates; for a point charge or a dipole, where the field varies with position, only the differential relation $E = -dV/dr$ (or the exact point-charge/dipole potential formulas) applies.
Mixing up units: remembering that $1\ \text{N/C} = 1\ \text{V/m}$ is what lets a student move fluidly between a field given in one unit and a potential-gradient problem stated in the other, without a unit-conversion error creeping in.
Forgetting that the small displacement in the derivation must be taken along the field direction: in a general (non-uniform, non-one-dimensional) situation, the rate of change of potential in some other, arbitrary direction is not simply $-E$ ; the CBSE syllabus keeps this restricted to motion along the field, which is why the derivation and both worked examples are careful to specify the direction of $dr$ .

Common trapE = V/d only holds for a uniform field

The single most common misapplication of this topic is carrying the parallel-plate formula $E = V/d$ over to situations where the field is not uniform — for instance treating the potential due to a point charge, $V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r}$ , and then dividing that $V$ by some distance $d$ to "get" the field, which is simply wrong because the potential (and hence its slope) changes continuously with $r$ near a point charge. $E = V/d$ is a special-case shortcut that falls straight out of $E = -dV/dr$ only when the field is constant across the gap, exactly as it is between two large, closely spaced, oppositely charged parallel plates. Whenever the field is not uniform, go back to the general differential relation, or to the specific formula already known for that charge configuration — never divide a non-uniform potential by a distance.

You should be able to…

State the relation $E = -dV/dr$ and explain, physically, why the negative sign must be there.
Derive $E = -dV/dr$ from the definition of potential difference as work done per unit charge over a small displacement.
Explain why volt per metre and newton per coulomb are the same unit for electric field.
Apply the uniform-field formula $E = V/d$ correctly to parallel-plate capacitor situations, and recognise when it does not apply.
Compute the electric field, including its direction, from a given algebraic expression for potential as a function of position.
Connect a steep or flat potential-versus-distance graph to a strong or weak field respectively, and relate this to why field lines cross equipotential surfaces at right angles.

Ask the tutor: Want to see how this relation ties back to why equipotential surfaces around a point charge are always spheres perpendicular to the radial field lines, or need a hand differentiating a trickier potential expression to find the field? Ask the AI tutor for a worked walkthrough.

Why is there a negative sign in E = -dV/dr? Can I just ignore it in numericals?

The negative sign comes directly from the physics, not from convention: the electric field always does positive work on a positive charge as that charge moves toward lower potential, so potential necessarily falls in the direction the field points. If you only need the magnitude of the field (as in most capacitor numericals), you can safely drop the sign and just take the magnitude of the slope. But whenever a question asks for direction — as in the board-level worked example above — keep the sign, work it out properly, and state the direction in words (such as "along the positive x-axis") rather than guessing.

Is E = V/d the same formula as E = -dV/dr, or a different one?

They are the same relation, just applied in two different situations. $E = -dV/dr$ is the general one-dimensional statement, valid at any point where the field has some direction and potential is changing with distance along it. $E = V/d$ is what that same relation becomes specifically when the field is uniform across a finite gap $d$ carrying a fixed potential difference $V$ — the "rate of change" becomes a plain constant slope, $V$ dropped over $d$ . Use $E = V/d$ only for genuinely uniform-field setups like parallel-plate capacitors; use the differential form whenever the field varies with position.

Why is volt per metre the same unit as newton per coulomb for electric field?

Both units measure the same physical quantity from two different defining equations. From $F = qE$ , field has units of force per charge, newton per coulomb. From $E = -dV/dr$ , field has units of potential per distance, volt per metre. Since these describe the identical electric field, the units must be numerically and dimensionally equal: $1\ \text{N/C} = 1\ \text{V/m}$ . In practice, field strength in this chapter's capacitor problems is almost always quoted in V/m, precisely because it comes straight out of a potential-difference-over-distance calculation.

If I double the plate separation of a capacitor but keep the same battery voltage, does the field change?

Yes — and this is a favourite twist in exam questions. With $E = V/d$ , if $V$ (fixed by the battery) stays the same while $d$ doubles, the field halves: $E \propto 1/d$ at constant $V$ . This is the reverse of the competitive-level worked example above, where halving $d$ at fixed $V$ doubled the field instead. Always check which quantity — $V$ or $d$ — is being held fixed before deciding how $E$ responds.