Electrostatic potential

Push a shopping trolley up a ramp and you feel it resist you the whole way — the steeper the slope, the harder the push. Physicists compress that entire feeling into one number for every point on the ramp: its height. Height alone tells you how much work gravity will do on the trolley between any two points, no matter which zig-zag path it took to get there. Electric charges live on an invisible version of the same ramp. At every point in space around a charge or a group of charges there is a number — the electrostatic potential $V$ — that plays exactly the role height plays for the trolley. It tells you, instantly and without any messy path-tracing, how much work the electric force will do (or how much work you must do against it) in moving a unit positive charge between two points. This chapter opens Unit II of your CBSE Class 12 syllabus by building that "electrical height" idea from scratch: what potential means, why it is a plain number (a scalar) rather than an arrow (a vector) like the electric field, why it is well-defined at all — which comes down to the electric force being conservative — and how the geometric surfaces of constant potential, the equipotential surfaces, connect neatly to the field lines you already met in the previous chapter. Get this foundation solid and the rest of the chapter — potential due to a point charge, a dipole, a whole system of charges, potential energy, and eventually capacitors — becomes a straightforward extension of one clean idea: potential is work per unit charge, stored as a number at every point in space.

Key takeaways

Electric potential $V$ at a point is the work done per unit positive charge in bringing a small test charge from infinity to that point, without any change in its kinetic energy.
Potential is a scalar (just a number, with sign), unlike the electric field, which is a vector — this makes potential far easier to add up for multiple charges.
SI unit: the volt (V), where $1\ \text{V} = 1\ \text{J/C}$ ; potential at infinity is conventionally taken as zero.
The electrostatic force is conservative, so the work done moving a charge between two points does not depend on the path — this is exactly why a single well-defined potential value can exist at every point.
Potential difference drives work: $W_{AB} = q_0(V_B - V_A)$ is the work an external agent must do moving charge $q_0$ from $A$ to $B$ ; positive charges naturally drift from high to low potential.
Equipotential surfaces join all points at the same potential; they are always perpendicular to the field lines, and no work is done moving a charge along one.
Key idea: electric potential is the work done per unit positive charge bringing a test charge from infinity to a point, a scalar quantity that exists precisely because the electrostatic force is conservative, so potential difference alone — not the path taken — fixes the work $W_{AB}=q_0(V_B-V_A)$ done in moving any charge between two points.

The core insight

Electric potential is the work done per unit positive charge in bringing a test charge from infinity to a point without changing its kinetic energy, a path-independent scalar (because the electrostatic force is conservative) whose differences fix the work $W_{AB}=q_0(V_B-V_A)$ done moving any charge between two points, with equipotential surfaces — always perpendicular to field lines — marking out regions of equal "electrical height."

Where it fits in the syllabus

This is the opening topic of the CBSE Class 12 Physics chapter "Electrostatic Potential and Capacitance" (Unit II, Electrostatics), building directly on the field ideas from the previous chapter, "Electric Charges and Fields." The board syllabus expects you to define electric potential and potential difference, state the volt as the SI unit, argue why potential is scalar while the field is a vector, and connect potential to equipotential surfaces — all without yet plugging in a specific charge configuration. That next step — the actual formula for the potential of a single point charge — is deliberately kept for the next node in this chapter, so here the focus stays on the general definition, the conservative-force argument behind it, and the geometry of equipotential surfaces.

The physical picture

Consider an electric field $\vec E$ set up by some arrangement of charges. To move a small positive test charge $q_0$ from one point to another, work must generally be done — either by an external agent against the field, or by the field itself on the charge. The electrostatic potential at a point $P$ is defined as the work done per unit positive charge by an external agent in bringing the test charge from infinity to $P$ , moving it slowly (quasi-statically) so its kinetic energy never changes:

V(P) = \frac{W_{\infty \to P}}{q_0}

where: $V(P)$ is the potential at point $P$ (volts), $W_{\infty\to P}$ is the work done by the external agent (joules), and $q_0$ is the small positive test charge (coulombs). Potential is a scalar — it carries a sign (positive or negative) but no direction — and its SI unit is the volt, $1\ \text{V} = 1\ \text{J/C}$ . Since only differences in potential usually matter physically, the potential at infinity is conventionally fixed at zero, giving every other point in space a definite numerical "electrical height." The potential difference between two points $A$ and $B$ follows the same logic:

V_B - V_A = \frac{W_{AB}}{q_0}

where: $W_{AB}$ is the work done in slowly moving the test charge from $A$ to $B$ against the electric force. This equation only makes sense — gives one fixed number, regardless of the route taken — because the electrostatic force is conservative: the work done moving a charge between two fixed points is the same for every possible path between them. Compare this with a frictional force, whose work very much depends on the path (a longer path means more friction work); no single "potential" could be assigned to a point if the electric force behaved that way. Because the work is path-independent, $V$ can be treated exactly like height on a landscape — a definite number carried by the point itself, not by any particular journey to reach it. That landscape picture also explains the natural direction of motion for charges. The electric force does positive work when a positive charge moves from higher to lower potential — "downhill." Left alone, positive charges drift from high-potential regions toward low-potential regions, exactly as a ball rolls downhill under gravity; negative charges do the reverse, drifting toward higher potential. Joining up all the points that share the same potential value gives an equipotential surface. Two properties are worth memorising. First, no work is done in moving a charge along an equipotential surface, since $V_B - V_A = 0$ there — this is why moving along the surface of a charged conductor, which is itself equipotential, costs no work. Second, equipotential surfaces are always perpendicular to the electric field lines at every point; if they were not, the field would have a component along the surface, which would do work while "staying at the same height," a contradiction.

Figure

Two diagrams side by side showing a ball rolling down a height contour hill on the left and equally spaced vertical equipotential planes crossed by horizontal field lines on the right, both illustrating that motion along a surface of constant potential does no work while motion across the surfaces does

Derivation APotential difference across a uniform field

The goal of this derivation is to connect the abstract definition of potential difference to the field you can actually measure, in the simplest possible setting — a region of uniform electric field, such as between two large charged parallel plates — using nothing but the definition of work and algebra, so that the relation between electric field magnitude and potential drop becomes concrete rather than just symbolic.

Step 1. Set up the geometry. Let the field $\vec E$ point uniformly from a plate at point $A$ toward a plate at point $B$ , separated by a distance $d$ measured along the field direction. To move a small positive test charge $q_0$ from $A$ to $B$ against the field, quasi-statically, an external agent must supply a force equal in magnitude to the electric force on the charge, $F = q_0E$ , at every point along the path.

Step 2. Compute the work. Since the force $q_0E$ is constant in magnitude and directed opposite to the displacement (the agent pushes the positive charge from low field-potential-drop direction toward the plate the field points at, i.e. moves it "uphill" against the field's natural push), the work done by the external agent over the straight-line distance $d$ is simply $W_{AB} = q_0 E d$ .

Step 3. Substitute into the definition of potential difference. Using $V_B - V_A = W_{AB}/q_0$ from the definition above, the charge $q_0$ cancels directly: $V_B - V_A = Ed$ . Since $B$ is the plate the field points toward, it is the lower-potential plate, so it is more natural to write the potential drop along the field direction as $V_A - V_B = Ed$ , or in magnitude terms, $|\Delta V| = Ed$ .

Step 4. Read the physical meaning. A stronger field ( $E$ larger) packs more potential drop into the same distance $d$ — equivalently, equipotential planes crowd closer together where the field is stronger, exactly as height contours crowd together on a steep slope. This single relation, $|\Delta V| = Ed$ , is the algebraic bridge between the field you calculate from charges and the potential you calculate from work, and it will resurface directly when the parallel-plate capacitor is built later in this chapter. Check the units: $E$ in V/m multiplied by $d$ in m gives volts, matching $\Delta V$ — a quick way to confirm the relation is dimensionally consistent.

Worked examples

Board level - work done moving a charge between two known potentials

Question: The potential at point $A$ is 120 V and at point $B$ is 80 V. How much work is done by an external agent in slowly moving a charge of $5\ \mu\text{C}$ from $A$ to $B$ ? State whether the electric field itself does positive or negative work, and describe the direction a free positive charge released at $A$ would naturally move.

Step 1. Find the potential difference: $V_B - V_A = 80 - 120 = -40$ V.

Step 2. Apply $W_{AB} = q_0(V_B - V_A) = (5\times10^{-6})(-40) = -2\times10^{-4}$ J. This is the work done by the external agent.

Step 3. Interpret the sign. A negative value for the external agent's work means the electrostatic field itself does positive work of $2\times10^{-4}$ J on the charge as it moves from $A$ to $B$ , since a positive charge moving from higher potential (120 V) to lower potential (80 V) is moving in the field's "natural," downhill direction — no external push is needed to drive that motion; if anything, the agent must hold the charge back to keep it moving slowly.

Step 4. State the natural direction. A free positive charge released at $A$ , with no agent required, would accelerate from $A$ toward $B$ — from high potential to low potential — gaining kinetic energy exactly equal to $2\times10^{-4}$ J by the time it reaches $B$ , since the field does that much positive work on it.

Answer: The external agent does $-2\times10^{-4}$ J of work (that is, the field does $+2\times10^{-4}$ J of work on the charge), and a free positive charge at $A$ naturally drifts toward $B$ .

Competitive level - combining the uniform-field relation with potential energy

Question: Two large parallel plates are separated by $d = 4\ \text{mm}$ and maintained at a uniform field of $E = 2.5\times10^{3}\ \text{V/m}$ pointing from plate $A$ to plate $B$ . (a) Find the potential difference $V_A - V_B$ between the plates. (b) A charge of $q_0 = 4\ \mu\text{C}$ , released from rest very close to plate $A$ , moves to plate $B$ under the field alone. Find the work done by the field on it, and the change in the charge's electrostatic potential energy.

Step 1. Convert $d$ to metres: $d = 4\ \text{mm} = 4\times10^{-3}\ \text{m}$ . Apply the uniform-field relation derived above, $V_A - V_B = Ed = (2.5\times10^{3})(4\times10^{-3}) = 10$ V. So plate $A$ is at $10$ V higher potential than plate $B$ .

Step 2. Find the work done by the field on the released charge. Since the field always does positive work on a positive charge moving from high to low potential, $W_{\text{field}} = q_0(V_A - V_B) = (4\times10^{-6})(10) = 4\times10^{-5}$ J.

Step 3. Relate this to potential energy. The electrostatic potential energy of the charge changes by $\Delta U = -W_{\text{field}} = -4\times10^{-5}$ J — the charge loses potential energy exactly as it gains kinetic energy, since no external agent acts and total mechanical energy (kinetic plus electrostatic potential) is conserved for a charge moving freely in a static field.

Step 4. Sanity-check the direction. The potential energy decreases (negative $\Delta U$ ) as a positive charge moves from high potential ( $A$ ) to low potential ( $B$ ), consistent with the "ball rolling downhill" picture — potential energy, like gravitational potential energy, is highest where the "electrical height" $V$ is highest for a positive charge.

Answer: $V_A - V_B = 10$ V; the field does $4\times10^{-5}$ J of work on the charge, and its electrostatic potential energy decreases by $4\times10^{-5}$ J. Timing note: This two-part question should take about 2.5–3 minutes in a competitive exam — the unit conversion for $d$ is the most common slip; write $d$ in metres before multiplying by $E$ , and don't forget the sign convention that potential energy change is the negative of the field's work.

Common traps

Confusing electric potential (a scalar, unit volt) with electric field (a vector, unit N/C or V/m) — related by $|\Delta V| = Ed$ in a uniform field, but never the same quantity or the same unit.
Assuming a region of zero field must have zero potential, or the reverse — potential can be constant and nonzero throughout a field-free region (inside a charged conductor, for instance); a zero-field point need not have zero potential.
Forgetting the quasi-static condition in the definition — potential is work per unit charge with no change in kinetic energy, not the work done in some arbitrary fast or jerky motion.
Treating equipotential surfaces as parallel to the field lines instead of perpendicular to them — a common sketching error that also breaks the "no work along the surface" argument.
Sign confusion between the work done by the external agent and the work done by the field — they are always equal in magnitude and opposite in sign for quasi-static motion, and the potential energy change equals minus the field's work.

Common trapPotential is not "how strong the field is at a point" — it is accumulated work, and it can be large where the field is weak or vice versa

Students often expect potential to track field strength point by point, the way both seem to "come from the same charges." They do not track each other locally. Potential at a point depends on the entire history of work done bringing a charge in from infinity — effectively an accumulation (an integral, informally) of the field along the whole path — while the field is the local push felt right at that point. A region can have a strong field but modest potential differences over a short distance, or a weak field spread over a long distance building up a large potential difference, exactly like a landscape where a short cliff (strong local slope, small height change) is compared with a long, gentle ramp (weak local slope, large height gain by the top). Always reason about potential using $|\Delta V| = Ed$ or the work-energy relation for the specific geometry given, never by eyeballing field arrows.

You should be able to…

State the definition of electric potential at a point, in words and as an equation, and give its SI unit.
Explain, using the work-energy theorem, why the electrostatic force being conservative is essential for potential to be well-defined.
Relate potential difference to the work done in moving a charge between two points, including correct sign reasoning for the agent's work versus the field's work.
Derive and apply $|\Delta V| = Ed$ for a uniform field, and use it alongside potential energy relations in numerical problems.
Describe the two key properties of equipotential surfaces — zero work along the surface, and perpendicularity to field lines — and sketch them for a uniform field region.
Predict the natural direction of motion of a released positive or negative charge given the potential values at two points.

Ask the tutor: Want to see how this same "work per unit charge" idea turns into the exact formula for the potential of a single point charge — try walking through it with the AI tutor next.

If potential at infinity is zero, does that mean nothing is happening out there?

Not quite — the zero at infinity is a reference choice, not a physical statement about "nothing happening." Just as we could measure height from sea level or from the floor of your room without changing how a ball actually rolls, choosing $V_\infty = 0$ is simply a convenient, universally agreed baseline so that every other point gets a definite, comparable number. What is physically meaningful is always a potential difference between two points, which tells you the actual work done moving a charge between them — the choice of zero point never affects that difference, only the individual numbers you write down for each point.

Why is potential a scalar when the electric field, which seems related, is a vector?

Because potential is built from work, and work is a scalar — force dotted with displacement collapses direction information into a single number at each point. The field, by contrast, tells you the direction and magnitude of the force per unit charge right there, so it must be a vector to carry that directional information. The payoff of potential being scalar is huge for problem-solving: to find the net potential from several charges you just add ordinary numbers (with sign), whereas to find the net field you must add vectors component by component. This is exactly why later in the chapter, potential due to a system of charges is so much easier to compute than the corresponding field.

Is the work done by the external agent always exactly equal and opposite to the work done by the field?

Yes, for the quasi-static motion used in this definition. Because the charge is moved slowly with no net change in kinetic energy, the work-energy theorem forces the total work done on it — by the agent plus by the field — to be zero. So $W_{\text{agent}} = -W_{\text{field}}$ always, for this idealised, no-acceleration kind of motion. If the charge were instead released and allowed to accelerate freely under the field alone (no external agent), then all of the field's work goes into kinetic energy, and the potential energy lost equals the kinetic energy gained — that's the situation in the competitive worked example above.

Does the actual path I imagine the test charge taking from infinity matter for calculating V?

No, and that is the entire point of the conservative-force argument. You can imagine the test charge arriving along a straight line, a curvy spiral, or any wild route you like — the work done, and therefore the potential you calculate, comes out identical every time, as long as the start and end points are the same and the charge configuration producing the field doesn't change while you do it. This path-independence is what lets physicists assign potential as a property of the point itself, printed once on a map of space, rather than a property of any particular journey to reach it.