Electrostatic Potential and Capacitance

Chapter 1 described the electric field by the force it exerts. This chapter describes the same field by the work it does. The electrostatic force is conservative: the work done in taking a charge from one point to another does not depend on the path, and the work done around any closed path is zero. That is exactly the property that lets us define a potential energy, just as we do for gravity.

Everything in the chapter grows from that one fact. Potential energy per unit charge is the electrostatic potential, a scalar, and it is far easier to add scalars from many charges than to add field vectors. Surfaces of constant potential give a picture of the field. The potential of a conductor is constant, and that leads to the capacitor, a device that stores charge and energy, and to the effect of a dielectric between its plates.

For the exam the chapter is a set of derivations to be written in full (dipole potential, parallel plate capacitance, energy density, series and parallel combinations) plus numericals on capacitor networks and on what changes when a dielectric goes in. One thing to note before you start: the current CBSE syllabus asks for the energy-stored-in-a-capacitor result, U = Q²/2C = ½CV² = ½QV, as a formula only — its derivation is not examined, even though it is shown in full in Section 2.15 for understanding. Both the derivations and the numericals are covered section by section below.

Take a fixed charge configuration and bring a test charge q from a point R to a point P slowly, so that it gains no kinetic energy. The external force just balances the electric force, and the work it does, W_RP, is stored as the increase of potential energy U_P − U_R. Because the force is conservative, this work is the same along every path from R to P. Only differences of potential energy are physically meaningful, so we choose a zero: the potential energy is taken as zero at infinity.

Since the work done is proportional to q, we divide it out. Electrostatic potential V at a point is the work done by an external force in bringing a unit positive charge from infinity to that point, without acceleration. Potential difference between two points is the work done per unit positive charge in taking it from one point to the other: V_P − V_R = W_RP/q. The SI unit is the volt, 1 V = 1 J C⁻¹. Potential is a scalar and can be positive, negative or zero. Its dimensional formula is [ML²T⁻³A⁻¹].

In the exam write the definition with the three qualifying phrases: unit positive charge, from infinity, without acceleration. A question that says the work done by the field is W means the potential energy has fallen by W, so watch which agent the question is talking about.

This is the first derivation of the chapter and it is short enough to be asked for 2 marks. Place a charge Q at the origin. The field at a distance r' from it is E = Q/(4πε₀r'²), directed radially outward for positive Q. Bring a unit positive test charge from infinity to the point P at distance r. At each step the external force is equal and opposite to the electric force, so the work done by the external force in a small displacement Δr' (towards Q) is ΔW = −E Δr'.

W = −∫_∞^r (Q/4πε₀r'²) dr' = (Q/4πε₀) [1/r']_∞^r = Q/(4πε₀r).

By definition this work is the potential: V(r) = Q/(4πε₀r). It is positive for positive Q, negative for negative Q, and it falls as 1/r while the field falls as 1/r². The graph of V against r is a rectangular hyperbola starting high near the charge; the E graph lies below it and drops faster. Both are asked as a 2-mark sketch.

Two checks the examiner likes. First, the potential of a positive charge is positive everywhere, because you have to do work against repulsion to bring a positive test charge in. Second, the potential depends only on r, so all points at the same distance from Q have the same potential; that is the equipotential sphere of section 2.6.

A dipole has charges −q and +q separated by 2a, dipole moment p = q × 2a along the axis from −q to +q. Take the centre O as origin and a point P at distance r from O, the line OP making angle θ with the axis. Let r₁ be the distance of P from +q and r₂ its distance from −q. Potential is a scalar, so

V = (q/4πε₀) (1/r₁ − 1/r₂).

From the geometry, r₁² = r² + a² − 2ar cos θ and r₂² = r² + a² + 2ar cos θ. For a point far from the dipole (r ≫ a), drop a²/r² and use the binomial approximation: 1/r₁ ≈ (1/r)(1 + a cos θ/r) and 1/r₂ ≈ (1/r)(1 − a cos θ/r). Subtracting, 1/r₁ − 1/r₂ ≈ 2a cos θ/r², so

V = (1/4πε₀) · q · 2a cos θ/r² = (1/4πε₀) · p cos θ/r² = (1/4πε₀) · p·r̂/r².

Special cases the marking scheme expects: on the axis (θ = 0° or 180°) V = ±p/(4πε₀r²), positive on the side of +q; on the equatorial plane (θ = 90°) V = 0 everywhere. Without the approximation, the axial potential is V = p/(4πε₀(r² − a²)). Two contrasts with a single charge close the answer: the dipole potential falls as 1/r², not 1/r, and it depends on the angle θ, not only on r. Because the equatorial plane is at zero potential, no work is done in moving a charge anywhere on it, a favourite 1-mark question.

For charges q₁, q₂, ..., q_n at distances r_1P, r_2P, ..., r_nP from a point P, the superposition principle gives the potential as the algebraic (scalar) sum:

V = (1/4πε₀) (q₁/r_1P + q₂/r_2P + ... + q_n/r_nP).

Signs of the charges go in; no components, no angles. This is why potential is so convenient: the potential at the centre of a hexagon with six equal charges at its vertices is simply six times the potential of one (NCERT 2.2), whereas the field there is zero by symmetry. For a continuous distribution the sum becomes an integral over charge elements.

A uniformly charged thin spherical shell of radius R and charge q is the standard example. Outside the shell (r ≥ R) the field is that of a point charge at the centre, so V = q/(4πε₀r). Inside (r < R) the field is zero, so no work is done in moving a charge about and the potential is constant, equal to its value on the surface: V = q/(4πε₀R). This is why a hollow metal sphere charged to 10 V on its surface is at 10 V at its centre too. The V–r graph is flat inside the sphere and falls as 1/r outside; the E–r graph is zero inside, jumps to σ/ε₀ at the surface and falls as 1/r².

The point where the net potential of two charges is zero is a regular 2- or 3-mark numerical. Set q₁/x = |q₂|/(d − x) for the point between them and q₁/x = |q₂|/(x − d) for the point beyond the smaller charge; there are two such points when the charges are of opposite sign.

An equipotential surface is a surface on which the potential has the same value at every point. For a single point charge the surfaces are concentric spheres centred on the charge; for a uniform field they are planes perpendicular to the field; for a dipole they are the closed curves around each charge, with the equatorial plane as the zero-potential surface; for two equal positive charges they are peanut-shaped surfaces that surround both charges at large distances.

Properties, each worth a mark when asked:

• No work is done in moving a charge on an equipotential surface, because the potential difference between any two points on it is zero.
• The electric field is always normal to the equipotential surface. If it had a tangential component, work would be done in moving a charge along the surface, contradicting the first point.
• Two equipotential surfaces can never intersect. At the crossing the potential would have two values, and the field would have two directions.
• The surfaces are crowded where the field is strong and spread out where it is weak. Around a point charge, equal steps of potential need larger and larger steps in r, so the spheres are not equally spaced.

Relation between E and V. Take two close equipotential surfaces at V and V + δV, separated by a perpendicular distance δl. The work done by the field in moving a unit positive charge from the first to the second is |E| δl. It also equals the drop in potential, −δV. So |E| δl = −δV, that is

E = −δV/δl.

Two conclusions the examiner wants stated: the field points in the direction in which the potential decreases most steeply, and its magnitude is the rate of change of potential with distance along that direction, the potential gradient, in V m⁻¹ (identical to N C⁻¹).

The potential energy of a system of charges is the work done in assembling it, bringing the charges from infinity to their positions. Start with two charges. Bring q₁ first: no field exists yet, so no work is done. Now bring q₂ to a point at distance r₁₂ from q₁. The work done is q₂ times the potential of q₁ at that point:

U = W = q₂V₁ = (1/4πε₀) · q₁q₂/r₁₂.

The result is symmetric in the two charges, so the order of assembly does not matter, and because the force is conservative the path does not matter either. Like charges give positive U (work must be done to push them together), unlike charges negative U (the field does the work).

For three charges, bring q₃ last. It moves in the field of both earlier charges, so the extra work is q₃(V₁ + V₂) and

U = (1/4πε₀) (q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃).

In general the potential energy is the sum over every pair of charges, each pair counted once. The work required to dissociate a system, that is to separate the charges to infinity, is −U. For three charges 1 μC, 1 μC and −4 μC at the corners of an equilateral triangle of side 10 cm, U = 9 × 10⁹ × 10⁻¹² × (1 − 4 − 4)/0.1 = −0.63 J, so 0.63 J of work is needed to pull the system apart; that exact numerical has been set in the board paper.

Now the field is produced by sources outside the system and is not changed by the charges we bring in. Single charge. The potential energy of q at a point where the external potential is V(r) is U = qV(r), the work done in bringing q from infinity to that point. One electron-volt is the energy gained by an electron accelerated through 1 V: 1 eV = 1.6 × 10⁻¹⁹ J.

Two charges. Bring q₁ to r₁: work q₁V(r₁). Bring q₂ to r₂: it works against the external field and against the field of q₁, so the work is q₂V(r₂) + q₁q₂/(4πε₀r₁₂). The total is

U = q₁V(r₁) + q₂V(r₂) + q₁q₂/(4πε₀r₁₂).

Dipole in a uniform field (the derivation most often asked from this section). A dipole of moment p at angle θ to a uniform field E experiences a torque τ = pE sin θ and no net force. To rotate it slowly from θ₀ to θ₁ an external torque does work

W = ∫_θ0^θ₁ pE sin θ dθ = pE(cos θ₀ − cos θ₁),

which is the change in potential energy U(θ₁) − U(θ₀). Choose the zero at θ₀ = 90°, the position where the dipole is perpendicular to the field. Then

U(θ) = −pE cos θ = −p·E.

U is least (−pE) at θ = 0°, the stable equilibrium, and greatest (+pE) at θ = 180°, the unstable equilibrium. The work needed to turn the dipole from the aligned position through 180° is 2pE. A typical 3-mark question gives the torque and the angle, asks you to find pE from τ = pE sin θ, and then U = −pE cos θ.

A conductor has free electrons that move until, in the static situation, there is no force on them. Every result below follows from that, and NCERT lists six. Learn them as numbered statements with the one-line reason each.

1. The electrostatic field is zero inside a conductor. If it were not, free charges would move, and the situation would not be static. This holds whether the conductor is neutral or charged, and in an external field.
2. At the surface, the field is normal to the surface at every point. A tangential component would drive surface charges along the surface. For a charged conductor the field just outside is E = (σ/ε₀) n̂, obtained by Gauss's law on a small pill-box straddling the surface with one face inside where E = 0.
3. In the static situation there is no excess charge inside the conductor. Gauss's law on any surface inside gives zero flux, so zero enclosed charge; any excess charge sits on the surface.
4. The potential is constant throughout the volume and has the same value on the surface. With E = 0 inside and no tangential field on the surface, no work is done in moving a test charge between any two points, so the potential difference is zero. This is the 2-mark question that appears again and again.
5. Surface charge density σ may vary from point to point, larger where the surface is sharply curved, but E = σ/ε₀ locally.
6. Electrostatic shielding. The field inside a cavity of a conductor is zero whatever the charges or fields outside. A sensitive instrument is protected by enclosing it in a metal box; sitting inside a car during lightning is safer for the same reason.

A conductor is therefore an equipotential volume. Two conductors at different potentials joined by a wire exchange charge until they reach a common potential; for two spheres that means Q₁/R₁ = Q₂/R₂, so the smaller sphere carries the higher surface charge density.

A dielectric is an insulator with no free charges. Placed in an external field, it does not cancel the field the way a conductor does, but it does reduce it. The mechanism depends on the kind of molecule.

Non-polar molecules (H₂, O₂, N₂, CO₂) have the centres of positive and negative charge coinciding and no permanent dipole moment. An external field pulls the two centres apart, giving each molecule an induced dipole moment in the direction of the field. Polar molecules (HCl, H₂O) have a permanent dipole moment, but thermal motion keeps the moments randomly oriented so the net moment is zero. An external field partly aligns them, more strongly at low temperature.

In both cases the dielectric acquires a net dipole moment along the field: it is polarised. The polarisation P is the dipole moment per unit volume, and for a linear isotropic dielectric P = χ_eE, where χ_e is the electric susceptibility. Inside the material the aligned dipoles cancel in the bulk, but on the two faces perpendicular to the field appear layers of uncoupled charge, the induced surface charge density ±σ_p. These layers produce a field opposite to the external one, so the net field inside the dielectric is smaller than the applied field: E = E₀/K, where K is the dielectric constant.

This is the answer to the 1-mark classic: a plastic sheet between two charged balls reduces the force between them because the sheet becomes polarised and its induced charges weaken the field. Every dielectric has a dielectric strength, the maximum field it can stand before it breaks down and conducts; for air it is about 3 × 10⁶ V m⁻¹, which sets the maximum charge a capacitor can hold.

A capacitor is a pair of conductors separated by an insulator, carrying charges +Q and −Q. Q is called the charge of the capacitor, and V is the potential difference between the conductors. The field in the gap is proportional to Q, so V is proportional to Q, and the ratio

C = Q/V

is the capacitance. It measures how much charge the capacitor holds per volt of potential difference. C depends only on the geometry of the conductors (size, shape, separation) and on the medium between them; it does not depend on Q or V. The SI unit is the farad, 1 F = 1 C V⁻¹. A farad is enormous, so practical values are in microfarads (1 μF = 10⁻⁶ F), nanofarads and picofarads (1 pF = 10⁻¹² F). The dimensional formula is [M⁻¹L⁻²T⁴A²].

A single isolated conductor is also a capacitor, with the second conductor at infinity. An isolated sphere of radius R has V = Q/(4πε₀R), so C = 4πε₀R, proportional to the radius.

The charge a capacitor can hold is limited by the dielectric strength of the medium. For a given C, a large charge means a large V and hence a large field, and once the field exceeds the dielectric strength the insulation breaks down and charge leaks across. That is why high-voltage capacitors need thick or strong insulation. On a circuit diagram a capacitor is two equal parallel lines; a variable capacitor has an arrow across them.

The 3-mark numerical from this section: a capacitor charged to V holds 360 μC; reducing V by 120 V leaves 120 μC. Then C(V − 120) = 120 and CV = 360 give C = 2 μF, V = 180 V, and at 300 V the charge is 600 μC.

Two large plane parallel conducting plates, each of area A, are separated by a small distance d (d much smaller than the plate dimensions, so edge effects can be ignored). Plate 1 carries +Q and plate 2 carries −Q, so the surface charge densities are σ = Q/A and −σ.

Each plate produces a field σ/2ε₀ on both sides, directed away from the positive plate and towards the negative one (Gauss's law for an infinite plane sheet, Chapter 1). Outside the capacitor the two fields are opposite and cancel: E = σ/2ε₀ − σ/2ε₀ = 0. Between the plates they add:

E = σ/2ε₀ + σ/2ε₀ = σ/ε₀ = Q/(ε₀A),

uniform and directed from the positive plate to the negative plate. For a uniform field the potential difference is field times separation:

V = Ed = Qd/(ε₀A).

Therefore C = Q/V = ε₀A/d.

The result carries the two expected features: C grows with plate area and falls as the plates are separated, and it depends only on geometry. Numerically, ε₀ = 8.85 × 10⁻¹² F m⁻¹, so plates of 1 m² separated by 1 mm give only 8.85 nF; that is why a farad is so large. For NCERT 2.8, A = 6 × 10⁻³ m² and d = 3 mm give C = 17.7 pF.

In the exam draw the two plates with +σ and −σ, mark the field lines only between the plates, and state E = 0 outside before writing V = Ed. Questions that halve d, or double A, or do both, are direct substitutions into this formula, but read carefully whether the battery stays connected (V fixed) or is removed (Q fixed) before you say what happens to the charge or the potential.

Fill the gap of a parallel plate capacitor with a dielectric slab. The plates carry ±σ as before. The dielectric is polarised and its faces next to the plates acquire induced charge densities −σ_p (next to the positive plate) and +σ_p. The net field inside the dielectric is due to σ − σ_p:

E = (σ − σ_p)/ε₀.

For a linear dielectric σ_p is proportional to σ, so we write σ − σ_p = σ/K, where K > 1 is the dielectric constant. Then E = σ/(ε₀K) = E₀/K, the potential difference is V = Ed = σd/(ε₀K) = Qd/(Kε₀A), and

C = Q/V = Kε₀A/d = K C₀.

The capacitance increases K times. The product ε = Kε₀ is the permittivity of the medium and K = ε/ε₀ = ε_r is its relative permittivity; K is also C/C₀, the ratio of the capacitance with the dielectric to that without.

Slab of thickness t < d (the 5-mark version). The field is E₀ = σ/ε₀ in the air gap of total thickness d − t and E₀/K inside the slab of thickness t. So V = E₀(d − t) + (E₀/K)t = (σ/ε₀)(d − t + t/K), and

C = ε₀A/(d − t + t/K).

Put t = d and this returns Kε₀A/d; put K → ∞ (a conducting slab) and it gives ε₀A/(d − t). A slab of thickness 3d/4 gives C/C₀ = 4K/(K + 3).

What changes when a dielectric is inserted. With the battery still connected, V is fixed: C rises to KC₀, Q rises to KQ₀, E = V/d is unchanged, and U = ½CV² rises K times. With the battery disconnected first, Q is fixed: C rises to KC₀, V falls to V₀/K, E falls to E₀/K, and U = Q²/2C falls to U₀/K. Write these as a two-column table in the answer; it is asked in almost every alternate year in one form or another.

Series. Capacitors C₁ and C₂ are joined end to end and the free ends are connected across V. The battery puts +Q on the first plate of C₁; induction puts −Q on its other plate and hence +Q on the first plate of C₂, and so on. Every capacitor in series carries the same charge Q. The potential differences add: V = V₁ + V₂ = Q/C₁ + Q/C₂. Defining the equivalent capacitance by V = Q/C,

1/C = 1/C₁ + 1/C₂, and in general 1/C = 1/C₁ + 1/C₂ + ... + 1/C_n.

The equivalent capacitance is smaller than the smallest capacitor; n identical capacitors C in series give C/n. The larger the capacitor, the smaller its share of the voltage.

Parallel. Capacitors are connected between the same two points, so each has the same potential difference V. The charges add: Q = Q₁ + Q₂ = C₁V + C₂V, so with Q = CV,

C = C₁ + C₂ + ... + C_n.

n identical capacitors in parallel give nC, so three identical capacitors have a series-to-parallel ratio of 1 : 9, the source of the 'net series capacitance is 1 pF, find the parallel value' question (answer 9 pF, and the energies across the same source are in the same ratio 1 : 9).

Networks. Reduce step by step: series groups first, then parallel, redrawing each time. Then work back outwards: the charge on a series group is the group's equivalent capacitance times the voltage across it, and that charge sits on every member of the group; the voltage across a parallel group is shared by every member. For the standard network of C₁, C₂, C₃ in series and that chain in parallel with C₄, all 10 μF across 500 V: the chain is 10/3 μF, the total is 40/3 ≈ 13.3 μF, each of C₁, C₂, C₃ holds (10/3 μF)(500 V) = 1.67 × 10⁻³ C and C₄ holds 5 × 10⁻³ C.

Charging a capacitor means moving charge from one plate to the other against the potential difference that already exists, and the work done is stored as electrostatic energy. Suppose at some stage the plates carry ±Q' so that the potential difference is V' = Q'/C. Moving a further charge δQ' from the negative to the positive plate needs work δW = V' δQ' = (Q'/C) δQ'. Adding all such steps from 0 to the final charge Q:

W = ∫₀^Q (Q'/C) dQ' = Q²/2C.

So U = Q²/2C = ½CV² = ½QV. Use the form that keeps the fixed quantity: Q²/2C when the battery has been removed, ½CV² when it stays connected. Graphically, on a plot of V against Q the area under the straight line is the energy. The current CBSE syllabus lists this topic as 'formulae only, no derivation', so the integral above will not be asked for in the exam — but working through it once is what makes the three equivalent forms of U easy to recall under pressure, which is why it is shown in full here.

Where the energy is. It resides in the field between the plates. For a parallel plate capacitor U = ½CV² = ½(ε₀A/d)(Ed)² = ½ε₀E² × (Ad). Dividing by the volume Ad gives the energy density

u = ½ε₀E² (J m⁻³),

a result that holds for any electric field in vacuum; in a dielectric it becomes ½Kε₀E².

Sharing charge. A capacitor C₁ charged to V₁ is connected to C₂ at V₂. Charge flows until both reach the common potential V = (C₁V₁ + C₂V₂)/(C₁ + C₂). The total charge is conserved but energy is not: the loss C₁C₂(V₁ − V₂)²/2(C₁ + C₂) is dissipated as heat in the connecting wires and as electromagnetic radiation. In the special case of an uncharged identical capacitor, V becomes V₁/2, the energy of the combination is half the initial energy, so exactly half is lost (NCERT 2.11: 600 pF at 200 V loses 6 × 10⁻⁶ J). That 1 : 2 ratio is a standing 2-mark question.