Electrostatic Potential and Capacitance — NEET Notes

Electrostatic potential — work per unit charge

The Coulomb force is a conservative force. That single fact — proved by noting its inverse-square form, identical in structure to Newton's law of gravitation — guarantees that the work required to move a charge between two points depends only on the endpoints, never on the path taken. NCERT opens the chapter by exploiting precisely this property. If work is path-independent, then for every position in space we can assign a single number that represents the stored energy of a unit charge placed there. That number is the electrostatic potential V.

Formally, the electrostatic potential at a point P is the work done by an external force, against the electrostatic force, in bringing a unit positive test charge without acceleration from infinity to P:

V_P = W_{∞ → P} / q   —   work per unit charge to bring a test charge from infinity to P

Definition of electrostatic potential — a scalar field over space

Three points worth absorbing right away. First, V is a scalar — it has magnitude, no direction. Electric field E is the vector; potential V is the scalar. Second, only the difference V_B − V_A is physically meaningful; the absolute zero of potential is a matter of convention. The standard convention takes V(∞) = 0. Third, V has SI units of volt, where 1 V = 1 J / 1 C — one joule of work per coulomb of charge transferred.

The work-energy relation that follows is the workhorse of the rest of the chapter. To move a charge q from point A to point B in an external field, the work done by the external agent is:

W_{A → B} = q (V_B − V_A)

This is the equation NEET 2017 Q.179 tested almost verbatim — four diagrams showing different equipotential layouts and asking which requires the most work to move a charge from A to B. The answer was "the same in all" because the work depends only on V_B − V_A, never on the path between them.

Potential due to a point charge

The simplest charge configuration — a single point charge Q at the origin — gives the most useful single result of the chapter. By integrating the work done against the Coulomb force from infinity to a point at distance r, NCERT derives:

V(r)  =  ¹⁄_4πε₀  ·  ^Q⁄_r  =  ^kQ⁄_r

Potential at distance r from a point charge Q — the master equation

The constant k = 1/(4πε₀) ≈ 9 × 10⁹ N·m²/C². Notice that V varies as 1/r — slower than the field, which falls as 1/r². The sign of V tracks the sign of Q: positive charges produce positive potential, negative charges produce negative potential. At infinity, V → 0, consistent with the chosen zero.

One application worth memorising — the potential of a uniformly charged spherical shell (or, by extension, a conducting sphere) of radius R carrying total charge q. Outside the shell (r ≥ R), the shell behaves as if all its charge were concentrated at its centre, so V = kq/r. Inside the shell, the field is zero, so the potential cannot change — it stays at the value it had at the surface, V = kq/R. The potential is continuous across the surface; the field is discontinuous (jumps from zero inside to σ/ε₀ outside).

Potential due to a dipole and a system of charges

An electric dipole consists of two equal and opposite charges +q and −q separated by 2a, with dipole moment vector p of magnitude p = q × 2a, pointing from −q to +q. By the superposition principle, the potential at a point P with position vector r from the centre of the dipole is the algebraic sum of the potentials due to the two charges. Working out the geometry to first order in a/r (valid for r ≫ a), NCERT obtains:

V(r)  =  ¹⁄_4πε₀  ·  ^p·r̂⁄_r²  =  ^{p cos θ}⁄_{4πε₀ r²}

Two contrasts with the single-charge potential leap out. The dipole potential falls as 1/r², not 1/r — faster, because at large distances the dipole increasingly looks like a charge-neutral object. And unlike the spherically symmetric point-charge potential, the dipole potential depends on the angle θ between r and p. Two special cases:

• Axial position (θ = 0 or π): V = ± p / (4πε₀r²) — positive on the +q side, negative on the −q side.
• Equatorial plane (θ = π/2): cos θ = 0, so V = 0 everywhere on the perpendicular bisector. The equatorial plane of a dipole is an equipotential at zero potential.

For a system of n point charges q₁, q₂, …, qₙ at positions r₁, r₂, …, rₙ, the superposition principle for potentials is even simpler than for fields — because V is scalar, no vector addition is needed. At a point P:

V_P  =  ¹⁄_4πε₀  ·  ( q₁/r₁_P + q₂/r₂_P + ... + qₙ/rₙ_P )

where r_iP is the distance from charge q_i to the point P. The simplicity of scalar addition is what makes potential so much more tractable than field for multi-charge problems — and NEET exploits it. The dipole question in NEET 2023 Q.37 was a direct application: two charges at known distances from point P, find the net potential by scalar addition.

Equipotential surfaces — where field meets potential

An equipotential surface is a surface on which the potential takes the same value at every point. Such surfaces give us a complete alternative visualisation of any electrostatic situation. For a single point charge, every concentric sphere is equipotential (because V = kq/r depends only on r). For a uniform field, equipotentials are flat planes perpendicular to the field. For a dipole, they are surfaces of constant p·r̂/r² — more elaborate, but the perpendicularity rule always holds.

Two properties of equipotential surfaces are pure NEET fodder. Both follow from a single argument: if you move a charge along an equipotential surface, the potential does not change, so no work is done; but W = qE·dl, so if W = 0 for any dl along the surface, then E must have no component along the surface. The two consequences are NEET 2022 Q.8 and the cousin asked in 2017 Q.179.

The quantitative connection between field and potential follows from considering two close equipotential surfaces of values V and V + dV separated by perpendicular distance dl. Moving a unit charge against the field from the lower-V surface to the higher-V surface does work |E|·dl, and this equals the potential difference dV. So:

E = − dV / dr

Two important consequences: (i) electric field points in the direction of steepest decrease of potential (the minus sign), and (ii) where potential is constant (dV/dr = 0), the field is zero. This second point is exactly what NEET 2020 Q.95 asked — a region with constant potential throughout a volume has E = 0 in that region.

Potential energy of a system of charges

The potential energy U of a system of charges is the work that must be done by an external agent to bring all the charges from infinity to their final configuration, against the electrostatic forces among them. For two charges q₁ and q₂ separated by distance r₁₂:

U  =  ¹⁄_4πε₀  ·  ^{q₁ q₂}⁄_r₁₂

The sign matters. For like charges (q₁q₂ > 0), U is positive — you had to do positive work against repulsion to assemble them. For unlike charges (q₁q₂ < 0), U is negative — the charges attract, and the system is at a lower energy than when separated to infinity. For a three-charge system, you add the pairwise interaction energies for all three pairs (q₁q₂, q₂q₃, q₁q₃); for n charges, you sum over all n(n−1)/2 distinct pairs.

Two related but distinct quantities come up frequently. The potential energy of a charge q in an external field at the point with potential V(r) is simply qV(r). The potential energy of an electric dipole of moment p in a uniform external field E, taking U(θ = π/2) = 0, is U(θ) = −p·E = −pE cos θ. The dipole has minimum energy when aligned with the field (θ = 0, U = −pE) and maximum when antiparallel (θ = π, U = +pE).

The unit electron-volt (eV) is born here: it is the energy gained by an electron of charge e when accelerated through a potential difference of 1 V. Numerically, 1 eV = 1.6 × 10⁻¹⁹ J. This is the working unit in atomic, nuclear, and particle physics — keV, MeV, GeV are all multiples.

Electrostatics of conductors — five locked-in properties

Conductors contain free charge carriers (electrons in metals). When you place a conductor in an electric field — or charge it — the free electrons redistribute themselves until the situation is static, meaning no current flows. In this static state, five properties are guaranteed. NEET draws on each of them in different question patterns, and all five must be at instant recall.

A consequence worth carrying into problem-solving: when two charged conductors are connected by a wire, charge redistributes until both are at the same potential. NEET 2021 Q.4 used this exact idea — two spheres of radii R₁ and R₂ connected by a wire — and asked for the ratio of surface charge densities. Same potential V means kQ₁/R₁ = kQ₂/R₂, which after expressing Q = σ × 4πR² gives σ₁/σ₂ = R₂/R₁. Smaller sphere holds higher surface density. NEET 2022 Q.27 ran the same logic the other way — two hollow spheres with equal charge but different radii — same Q means V ∝ 1/R, so the smaller sphere is at higher potential.

Dielectrics and polarisation

Dielectrics are non-conductors — substances with no (or negligibly few) free charge carriers. Unlike a conductor, a dielectric placed in an external field cannot zero out the field through electron drift. Instead, the molecules of the dielectric respond by developing dipole moments, and the collective effect is a partial cancellation of the external field — never complete, only reduced.

The molecules of a dielectric come in two flavours. Non-polar molecules (O₂, H₂, CO₂) have coincident centres of positive and negative charge in the absence of any field — zero permanent dipole moment. An external field stretches them apart, inducing a dipole moment in the direction of the field. Polar molecules (H₂O, HCl, NH₃) have a permanent dipole moment due to inherent asymmetry in their charge distribution; without an external field, thermal motion randomises their orientations, giving zero net dipole moment. An external field partially aligns them along its direction. NEET 2021 Q.2 tested this distinction directly: polar molecules have a permanent dipole moment.

In either case — non-polar via induction or polar via alignment — the dielectric acquires a net dipole moment per unit volume in the field. This is called the polarisation P:

P = χ_e ε₀ E

where χ_e is the electric susceptibility of the dielectric — a dimensionless constant characteristic of the material. The polarisation produces bound surface charges of density ±σ_p on the dielectric faces normal to the field. These bound charges produce a field opposite to the external field, reducing — but not cancelling — the net field inside the dielectric.

Capacitors and capacitance

A capacitor is a system of two conductors separated by an insulator. In practice, the two conductors carry equal and opposite charges +Q and −Q (a configuration achieved by connecting them across the terminals of a battery — electrons flow from one plate to the other until the potential difference matches the battery EMF). The total charge of the capacitor is zero; the "charge of the capacitor" by convention means the magnitude Q on one plate. The potential difference V = V₁ − V₂ between the plates is proportional to Q, and the ratio defines the capacitance:

C  =  Q / V

Capacitance — charge stored per unit potential difference (SI: farad)

Three crucial properties of capacitance follow from this definition. First, C is independent of Q and V — increase the charge, the voltage rises proportionally and the ratio stays fixed. Second, C depends only on geometry (shape, size, separation of the conductors) and on the dielectric medium between them. Third, the SI unit is the farad: 1 F = 1 C / 1 V. A farad is enormous; practical capacitors are usually in μF (10⁻⁶ F), nF (10⁻⁹ F), or pF (10⁻¹² F).

Why does capacitance matter? Because high C means a capacitor can store a lot of charge at a modest voltage. Modest voltage matters: above a certain field strength called the dielectric strength (≈ 3 × 10⁶ V/m for air), the medium between the plates ionises and breaks down. So large C, achieved by clever geometry, is what allows energy storage in real devices — flashes, defibrillators, computer memory.

The parallel plate capacitor

The cleanest geometry is two large parallel plates, each of area A, separated by distance d, where d ≪ √A (so edge effects can be ignored). With vacuum between the plates and charges ±Q (surface density σ = Q/A), the field between the plates — adding the contributions of both sheets — is uniform:

E = σ/ε₀ = Q / (ε₀ A)

Outside the plates, the two contributions cancel and E = 0. The potential difference is simply V = E × d, and the capacitance follows by definition C = Q/V. Let's walk through it as a process — every step is a NEET-frequent micro-question.

So C = ε₀A/d. Three consequences for problem-solving: doubling the plate area doubles the capacitance; halving the separation doubles the capacitance; and inserting a dielectric multiplies the capacitance by K (next section). NEET 2018 Q.22 used the result E = σ/ε₀ to find the electrostatic force on a plate of a parallel-plate capacitor — it turned out to be Q²/(2Aε₀), independent of the distance between the plates (because once the charge is fixed, the field is fixed).

Effect of dielectric on capacitance

Now insert a dielectric slab of dielectric constant K into the gap, filling the volume completely. The polarisation of the dielectric produces bound surface charges ±σ_p on its faces, which oppose the field of the free charges ±σ on the plates. The net field inside the dielectric reduces from σ/ε₀ to (σ − σ_p)/ε₀, which for a linear dielectric equals σ/(Kε₀):

E_dielectric = E_vacuum / K

The potential difference V = Ed reduces by the same factor K. Since the charge Q on the plates does not change (we kept it on the plates, isolated), and C = Q/V, the capacitance increases by K:

C′  =  K · C  =  Kε₀A / d

Capacitance with a dielectric fully filling the gap

This is the universal rule: inserting a dielectric of dielectric constant K multiplies the capacitance by K. The product Kε₀ is called the permittivity ε of the medium; the dimensionless ratio K = ε/ε₀ is the dielectric constant (or relative permittivity). K > 1 for any dielectric, K = 1 for vacuum, K ≈ 1.0006 for air (so air is treated as vacuum for most purposes). Water has K ≈ 80; mica K ≈ 6; glass K ≈ 4–7. NEET 2020 Q.113 asked the converse — capacitance going from 6 μF to 30 μF means K = 5, then ε = Kε₀ = 0.44 × 10⁻¹⁰ C²N⁻¹m⁻².

Series and parallel combinations

Real circuits use multiple capacitors, and only two configurations appear in NEET — series and parallel. The reasoning is identical for both: identify what is the same on each capacitor (charge or voltage), then apply Q = CV.

NEET 2023 Q.33 was a direct two-step application: three capacitors of 3F each — two in parallel give 6F, then in series with the third 3F to give (3×6)/(3+6) = 2F equivalent. NEET 2021 Q.20 was a circuit-redraw trap: three points joined by wires are at the same potential, so the capacitor between them is short-circuited and stores no charge — the remaining two are in parallel, giving 2C.

Energy stored in a capacitor

Charging a capacitor is the process of moving charge from one plate to the other against the rising potential difference between them. Each small increment of charge δQ′ takes work δW = V′ δQ′ = (Q′/C) δQ′. Integrating from 0 to Q gives the total work — which is stored as electrostatic potential energy:

Three equivalent expressions — pick whichever has the right two known variables for your problem:

U = ½ Q V = ½ C V² = Q² / (2C)

NEET 2021 Q.5 took the third form one step further: write U = ½CV² with C = ε₀A/d and V = E·d, expand, and you get U = ½ε₀E² × (A·d). Now (A·d) is just the volume of the region between the plates. So the energy per unit volume — the energy density of the electric field — is u = ½ε₀E², a result that holds for any electric field, not just a capacitor's.

One subtle problem type — energy after redistribution. When a charged capacitor is disconnected from the battery and then connected to an uncharged capacitor (NEET 2017 Q.150 and NEET 2022 Q.46), charge redistributes until both reach a common potential. The new energy is always less than the original — the lost energy dissipates as heat and electromagnetic radiation in the connecting wires during the transient. For two equal capacitors (one charged to V, one uncharged), the final common potential is V/2, total energy halves — 50% energy lost. The 2016 Q.136 variation (2μF charged then connected to 8μF) gives 80% energy loss, because the redistribution ratio is now lopsided.