NCERT Solutions for Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance
All 11 exercise questions of the latest NCERT textbook (2026–27 reprint) solved like a topper’s answer-sheet — electrostatic potential of charges and dipoles, equipotential surfaces, conductors, capacitors in series and parallel, dielectrics and the energy stored in a capacitor — with diagrams linked to the exact NCERT pages, comparison tables, complete revision notes and exam tips.
Chapter at a glance — complete revision notes
2.1 Introduction
- In Class XI (Chapters 5 and 7) the idea of potential energy was introduced: when an external force does work in taking a body from one point to another against a conservative force (gravity, spring force), the work is stored as potential energy and can be recovered as kinetic energy when the body is released. The electrostatic force is also a conservative force, so we can associate a potential energy with it.
- In Chapter 1 we studied the electrostatic force through Coulomb’s law and the electric field. This chapter describes the same physics in terms of energy — electric potential, a scalar that is often simpler to work with than the vector field. It also introduces capacitors, devices that store electrostatic energy.
2.2 Electrostatic potential
- Work against a field (Fig. 2.1): for a test charge q in the field of a source charge Q, the work done by an external force (equal and opposite to the electric force, so that the charge moves without acceleration) in moving q from R to P is WRP = q(VP − VR) = UP − UR — the change in potential energy. This work is independent of the path (the electrostatic force is conservative); it depends only on the end points.
- Potential energy at a point: the work done by the external force in bringing the charge q from infinity to that point. Only potential differences are physically significant; the zero of potential energy is arbitrary and is conveniently chosen at infinity.
- Electrostatic potential V at a point = work done by the external agency in bringing a unit positive test charge (without acceleration) from infinity to that point: V = W/q. Potential difference VP − VR = work per unit charge in moving it from R to P. Strictly, use an infinitesimal test charge dq and the ratio dW/dq so that the source configuration is not disturbed.
- SI unit: volt (V) = 1 J C⁻¹, named after Count Alessandro Volta (1745–1827), inventor of the voltaic pile (first electric battery).
2.3 Potential due to a point charge
- For a charge Q at the origin, work done in bringing a unit positive charge from infinity to a point at distance r (Fig. 2.2, 2.3): V(r) = Q / (4πε₀ r) (Eq. 2.7), taking V = 0 at infinity. The work is the same along any path — a general path can be resolved into a radial part (does work) and a perpendicular part (does none, since the field is radial).
- V is positive for Q > 0 and negative for Q < 0; it falls as 1/r (whereas E falls as 1/r²) (Fig. 2.4). For Q > 0, the work done in bringing a positive charge from infinity is positive; for Q < 0 the work done by the external agent on a positive test charge is negative.
- Example 2.1: a charge 4 × 10⁻⁷ C at 9 cm: V = 9 × 10⁹ × 4 × 10⁻⁷/0.09 = 4 × 10⁴ V; work to bring 2 × 10⁻⁹ C from infinity = qV = 8 × 10⁻⁵ J — independent of the path.
2.4 Potential due to an electric dipole
- For a dipole (−q at one end, +q at the other, separation 2a, moment p = 2qa) the potential at a point with position vector r making angle θ with p (Fig. 2.5), for r ≫ a: V = (1/4πε₀) p cos θ / r².
- On the axis (θ = 0° or 180°): V = ± p/(4πε₀r²) (positive on the +q side, negative on the −q side). On the equatorial plane (θ = 90°): V = 0. The dipole potential falls as 1/r² (a single charge: 1/r) and, unlike a single charge, depends on the angle θ. The result is exact for a point dipole at any r.
2.5 Potential due to a system of charges
- Potential is a scalar, so by the superposition principle the potential at P due to charges q₁, q₂, …, qₙ is simply the algebraic sum (Fig. 2.6): V = (1/4πε₀) [q₁/r₁P + q₂/r₂P + … + qₙ/rₙP]. Contrast with the vector addition required for E.
- Example 2.2: charges 3 × 10⁻⁸ C and −2 × 10⁻⁸ C, 15 cm apart: V = 0 at 9 cm from the positive charge (between the charges) and also at 45 cm from the positive charge (30 cm from the negative charge) on the extended line — outside, on the side of the smaller charge.
- Example 2.3: uses the field lines of a positive and a negative point charge (Fig. 2.8) to decide the signs of potential differences, potential-energy differences and the work done by the field or by an external agency; the key rules are: potential falls along the direction of the field (so it is higher nearer a positive source charge); a negative charge has higher potential energy where the potential is lower; and the work done by the field is positive when the charge moves so that its potential energy decreases (kinetic energy increases).
2.6 Equipotential surfaces
- An equipotential surface is a surface with a constant value of potential at all points. For a point charge, concentric spheres centred on the charge (Fig. 2.9); for a uniform field (say along x) the planes perpendicular to the field (Fig. 2.10); Fig. 2.11 shows those for a dipole, two like charges, etc.
- Properties: (i) no work is done in moving a charge between two points on an equipotential surface; (ii) E is always perpendicular to the equipotential surface at every point (otherwise a component of E along the surface would do work); (iii) E points in the direction of the steepest decrease of potential; (iv) the spacing between equipotentials in a given increment ΔV shows the field strength — closer surfaces mean stronger field; (v) two equipotential surfaces never intersect.
2.6.1 Relation between field and potential
For two close equipotentials V and V + δV separated by δl (normal to the surface): E = −dV/dl — the component of E along any direction is the negative rate of change of potential in that direction. Hence E can also be expressed in V m⁻¹ (= N C⁻¹).
2.7 Potential energy of a system of charges
- The potential energy of a system = the work done (by an external agency) in assembling the charges at their locations, bringing them from infinity one at a time. Work depends only on the final arrangement, not on the order.
- Two charges: W = U = q₁q₂ / (4πε₀ r₁₂) (positive for like charges, negative for unlike). Three charges (Fig. 2.12): U = (1/4πε₀)[q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃]. Generally, sum over all pairs. If U is negative the charges are bound — work must be supplied to separate them to infinity.
- Example 2.4: charges +q, −q, +q, −q at the corners A, B, C, D of a square of side d: total energy = −(q²/4πε₀d)(4 − √2). Bringing a fourth charge q₀ to the centre E needs no extra work because the potential at E is zero (equal and opposite charges at equal distances).
2.8 Potential energy in an external field
2.8.1 Single charge
In an external potential V(r), the potential energy of a charge q is qV(r); V(r) is due to external charges, not the potential due to q itself (the self-potential is undefined/infinite at its own location).2.8.2 Two charges
U = q₁V(r₁) + q₂V(r₂) + q₁q₂/(4πε₀r₁₂) — external-field terms plus mutual interaction.- Example 2.5: charges 7 μC and −2 μC at (−9 cm, 0, 0) and (9 cm, 0, 0): mutual energy = −0.7 J; work to separate them to infinity = +0.7 J; in an external field E = A/r² (A = 9 × 10⁵ N C⁻¹ m²) the interaction with the field adds +70 J − 20 J = 50 J, so the total energy is 49.3 J.
2.8.3 Potential energy of a dipole in an external field (Fig. 2.16)
Work done by the external torque in rotating the dipole from θ₀ to θ₁ in a uniform field = pE(cos θ₀ − cos θ₁). With the zero of energy at θ₀ = 90°: U(θ) = −pE cos θ = −p·E. Minimum (−pE) at θ = 0° (stable equilibrium), maximum (+pE) at θ = 180° (unstable equilibrium), zero at 90°.- Example 2.6: molecules with p = 10⁻²⁹ C m; a mole in E = 10⁶ V m⁻¹ with the field direction suddenly rotated by 60°: total p = 6 × 10⁻⁶ C m, ΔU = −3 J − (−6 J) = 3 J, so about 3 J of heat is released as the dipoles re-align (100% polarisation assumed).
2.9 Electrostatics of conductors
- A conductor has free charge carriers (electrons in metals) that move under any electric field. In electrostatic equilibrium (no net motion of charge):
- (1) The electrostatic field E is zero everywhere inside a conductor — whether it is neutral or charged, solid or hollow. Reason: free charges would otherwise keep moving; in an external field the charges redistribute to produce an induced field that exactly cancels the applied field inside.
- (2) At the surface of a charged conductor E is normal to the surface at every point (a tangential component would drive surface charges).
- (3) No net charge in the interior — any excess charge resides only on the surface (Gauss’s law: E = 0 inside ⇒ zero enclosed charge for any Gaussian surface inside, Fig. 2.17).
- (4) The potential is constant throughout the volume of the conductor and has the same value as on its surface — the conductor is an equipotential (since E = 0 inside, dV/dl = 0; and E has no tangential component on the surface).
- (5) Electric field at the surface: E = (σ/ε₀) n̂, where σ is the local surface charge density and n̂ the outward unit normal (derived with a small Gaussian “pill-box”, E = 0 inside). Note: this is twice the field σ/2ε₀ of an infinite sheet, because the field of the rest of the conductor also contributes equally; the field is discontinuous across the surface (0 inside, σ/ε₀ outside).
- (6) Electrostatic shielding — a cavity inside a conductor (Fig. 2.18) is shielded from outside electrical influences: the field inside the cavity is zero whatever the outside charges (provided the cavity has no charge). This is used in Faraday cages, and to protect sensitive instruments; it does not work the other way round — charges placed inside the cavity do not shield the exterior from their field.
- Example 2.7 (applications): (a) a dry comb attracts bits of paper by polarising them; on a rainy day or with wet hair friction is reduced and charge leaks away, so it does not; (b) aircraft tyres are made slightly conducting so that accumulated charge leaks away to the ground; (c) metallic ropes dragging on the ground carry away static charge from vehicles carrying inflammable material, avoiding sparks; (d) a bird sits on a bare high-power line with both feet at the same potential, so no current flows through its body; a man touching it while standing on the ground is at a large potential difference, causing a fatal shock.
2.10 Dielectrics and polarisation
- Dielectrics are non-conducting substances (insulators) with no free charge carriers (only bound charges). In an external field the charges are only slightly displaced.
- Non-polar molecules (centres of positive and negative charge coincide, e.g. O₂, CO₂, H₂) have zero permanent dipole moment; an external field induces a dipole moment along the field. Polar molecules (centres do not coincide, e.g. H₂O, HCl) have a permanent dipole moment, randomly oriented (net zero) because of thermal agitation; an applied field aligns them, giving a net dipole moment (Figs 2.20, 2.21).
- Polarisation (Fig. 2.22): in either case the dielectric develops a net dipole moment per unit volume (polarisation P) in the direction of the field. For a slab, this gives equal and opposite induced surface charges and no net bulk charge. These surface charges set up an induced field opposite to the applied field, so the net field in the dielectric is reduced: E = E₀/K, where K is the dielectric constant. For many dielectrics P ∝ E (linear isotropic).
2.11 Capacitors and capacitance
- A capacitor is a system of two conductors separated by an insulator (Fig. 2.24), carrying charges +Q and −Q. The field is localised between the conductors (a single charged conductor is shown in Fig. 2.23). The potential difference V between them is proportional to Q, so C = Q/V (Eq. 2.31) — capacitance, determined purely by the geometry (size, shape and separation of the conductors) and the medium, not by Q or V.
- SI unit: farad (F) = 1 C V⁻¹ (after Faraday). 1 F is huge, so we use μF (10⁻⁶ F), nF (10⁻⁹ F), pF (10⁻¹² F).
- A capacitor confines the field lines in a small region so that even a large field gives only a modest potential difference between the conductors.
2.12 The parallel plate capacitor
- Two large plane plates, each of area A, separated by a small distance d (Fig. 2.25), carrying +Q and −Q. Neglecting edge effects: plate surface charge density σ = Q/A; field between the plates E = σ/ε₀ = Q/(ε₀A) (fields of both plates add; zero outside); potential difference V = Ed = Qd/(ε₀A). Hence C = ε₀A/d (Eq. 2.34). C increases with plate area and decreases with separation.
2.13 Effect of dielectric on capacitance
- With a dielectric of constant K filling the space: field in dielectric E = E₀/K; V = V₀/K for the same charge; C = K C₀ = K ε₀ A / d. K = ε/ε₀ is a pure number (>1); ε = Kε₀ is the permittivity of the medium. A dielectric increases capacitance K-fold (for the entire space filled).
- Example 2.8: a slab of constant K and thickness (3/4)d inserted: V = E₀(d/4) + (E₀/K)(3d/4) = V₀(K+3)/(4K), so the capacitance rises to C = 4K C₀/(K + 3).
2.14 Combination of capacitors
2.14.1 Series (Figs 2.26, 2.27)
The same charge Q on each capacitor; voltages add: V = V₁ + V₂ + … ⇒ 1/C = 1/C₁ + 1/C₂ + 1/C₃ + …; C is smaller than the smallest individual capacitance. For n identical capacitors in series, C = C₁/n.2.14.2 Parallel (Fig. 2.28)
The same potential difference V across each; charges add: Q = Q₁ + Q₂ + … ⇒ C = C₁ + C₂ + C₃ + … (equivalent to a larger plate area).- Example 2.9: four 10 μF capacitors on 500 V (C₁, C₂, C₃ in series, in parallel with C₄): C′ = 10/3 μF; total C = 13.3 μF; charge on each of C₁–C₃ = (10/3 μF)(500 V) = 1.67 × 10⁻³ C; on C₄ = 5 × 10⁻³ C.
2.15 Energy stored in a capacitor
- Charging a capacitor means transferring charge from one plate to the other against the field. When charge q is already on it, the work to move dq is dW = (q/C) dq (Fig. 2.30), so U = ∫ q dq/C = Q²/2C = ½ C V² = ½ QV — the energy stored in the electric field.
- Energy density (energy per unit volume) in a region with field E: u = ½ ε₀ E² (for a parallel plate capacitor U/(Ad) with E = V/d); true in general for any electric field in vacuum.
- Example 2.10: 900 pF charged by 100 V: Q = 9 × 10⁻⁸ C, U = 4.5 × 10⁻⁶ J. Disconnected and connected to an uncharged 900 pF capacitor: V′ = 50 V and U = 2.25 × 10⁻⁶ J — half the original energy is lost (as heat and electromagnetic radiation in connecting wires/sparks).
Summary & “Points to ponder” (NCERT)
- The electrostatic force is conservative; potential = work per unit charge by an external agent in bringing a charge from infinity; only potential differences matter. Point charge: V = Q/4πε₀r; dipole: V = p·r̂/4πε₀r²; system: scalar sum; potential energy of a charge in an external field = qV; dipole in field U = −p·E.
- E is perpendicular to equipotentials and points in the direction of steepest decrease of potential. In a conductor E = 0 inside, E is normal and equals σ/ε₀ just outside, charge only on the surface, potential constant; a cavity is shielded.
- Capacitor: C = Q/V (geometry only); parallel plate C = ε₀A/d; with a dielectric C = Kε₀A/d; series 1/C = Σ1/Cᵢ, parallel C = ΣCᵢ; U = ½CV² = Q²/2C; u = ½ε₀E².
- Points to ponder: (a) electrostatic force acts between charges at rest only because each is held by some unspecified force; (b) a capacitor confines field lines in a small region so a large field exists with small potential difference; (c) E is discontinuous across a charged spherical shell (0 inside, σ/ε₀ n̂ outside) but the potential is continuous, equal to q/4πε₀R at the surface; (d) the torque p × E makes a dipole oscillate about E — they settle only with a dissipative mechanism; (e) the potential due to a charge at its own location is undefined, and in qV(r) the V must be due to external charges.
Two charges 5 × 10⁻⁸ C and −3 × 10⁻⁸ C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
Concept and context. The electrostatic potential V at a point is the work done per unit positive charge in bringing a test charge (without acceleration) from infinity to that point. For a point charge q at distance r, V = q/(4πε₀r) — positive for a positive source and negative for a negative source. Because potential is a scalar, the potential due to several charges is the algebraic sum of the individual potentials (superposition principle). Hence the net potential can vanish at points where the positive potential of one charge is exactly cancelled by the negative potential of the other. Because |q₁| > |q₂| here, such points exist on both the segment between the charges and on the extension of the line beyond the smaller charge.
Given: q₁ = +5 × 10⁻⁸ C at A, q₂ = −3 × 10⁻⁸ C at B, AB = 16 cm = 0.16 m. Take the origin at A and the x-axis along AB, so B is at x = 0.16 m.
Case 1 — Point P between the charges (0 < x < 0.16 m)
- Distance of P from A = x; from B = (0.16 − x). Net potential V = (1/4πε₀)[q₁/x + q₂/(0.16 − x)] = 0.
- ⇒ 5 × 10⁻⁸ / x = 3 × 10⁻⁸ / (0.16 − x) ⇒ 5(0.16 − x) = 3x.
- ⇒ 0.80 − 5x = 3x ⇒ 8x = 0.80 ⇒ x = 0.10 m = 10 cm from A (i.e. 6 cm from B).
Case 2 — Point P′ outside the segment, on the side of the smaller (negative) charge (x > 0.16 m)
- On the line beyond B, let P′ be at x from A, so its distance from B is (x − 0.16). Setting the potential to zero: 5/x = 3/(x − 0.16).
- ⇒ 5(x − 0.16) = 3x ⇒ 5x − 0.80 = 3x ⇒ 2x = 0.80 ⇒ x = 0.40 m = 40 cm from A (i.e. 24 cm beyond B).
- Why not beyond A? To the left of A the point is closer to the larger charge (+5 × 10⁻⁸ C), so its positive potential always exceeds the magnitude of the negative potential due to the farther smaller charge; the sum can never vanish there.
| Position | Distance from A | Distance from B | Check: q₁/rA + q₂/rB (in 10⁻⁸ C/m) |
|---|---|---|---|
| P (between A and B) | 10 cm | 6 cm | 5/0.10 − 3/0.06 = 50 − 50 = 0 ✓ |
| P′ (beyond B) | 40 cm | 24 cm | 5/0.40 − 3/0.24 = 12.5 − 12.5 = 0 ✓ |
Note: although the potential is zero at these points, the electric field is not zero there (both fields point the same way between the charges). Zero potential at a point does not imply zero field, nor vice versa.
Conclusion: the potential is zero at two points on the line: 10 cm from the +5 × 10⁻⁸ C charge (between the charges) and 40 cm from it (24 cm beyond the −3 × 10⁻⁸ C charge on the extended line).
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the centre of the hexagon.
Concept and context. The electrostatic potential at a point due to a system of charges is the algebraic (scalar) sum of the potentials due to the individual charges: V = (1/4πε₀) Σ qᵢ/rᵢ. There is no direction to worry about — unlike the electric field, which for a symmetric arrangement like a regular hexagon would cancel to zero by vector addition. Here the centre of a regular hexagon of side a is at a distance equal to the side from every vertex, since a regular hexagon is made of six equilateral triangles meeting at the centre.
Given: side a = 10 cm = 0.10 m; charge on each vertex q = 5 μC = 5 × 10⁻⁶ C; number of charges n = 6; distance of the centre from each vertex r = a = 0.10 m (each of the six triangles formed with the centre is equilateral).
- Potential due to one charge at the centre: V₁ = q/(4πε₀ r) = (9 × 10⁹ × 5 × 10⁻⁶)/0.10 = 4.5 × 10⁵ V.
- Superposition: all six charges are positive and at the same distance, so their potentials simply add: V = 6V₁ = 6 × 4.5 × 10⁵ V = 2.7 × 10⁶ V.
- Compact formula: V = 6q/(4πε₀a) = 6 × (9 × 10⁹) × (5 × 10⁻⁶)/(0.10) = 2.7 × 10⁶ V.
Vcentre = 2.7 × 10⁶ V
| Quantity | Value |
|---|---|
| Distance of each vertex from the centre | 0.10 m (equals side) |
| Potential due to one charge | 4.5 × 10⁵ V |
| Number of charges | 6 |
| Total potential at the centre | 2.7 × 10⁶ V |
| Electric field at the centre | zero (vector sum of six equal fields 60° apart cancels) |
Interesting contrast. The field at the centre is zero by symmetry (the six equal forces/fields arranged at 60° intervals add to zero), but the potential is not zero, because potentials of like charges add as scalars and cannot cancel. This shows clearly that V = 0 does not mean E = 0 and E = 0 does not mean V = 0.
Conclusion: the electrostatic potential at the centre of the hexagon is 2.7 × 10⁶ V, obtained by adding the six equal potentials of the like charges algebraically, while the electric field at the centre is zero.
Two charges 2 μC and −2 μC are placed at points A and B 6 cm apart. (a) Identify an equipotential surface of the system. (b) What is the direction of the electric field at every point on this surface?
Concept and context. An equipotential surface is a surface on which the electrostatic potential has the same value at every point; no work is done in moving a charge along it, and the electric field is always perpendicular to it at every point, pointing from higher towards lower potential. Two equal and opposite charges form an electric dipole, whose equipotential surfaces are shown in the NCERT text (Fig. 2.11a).
(a) Identify an equipotential surface
- Potential at a point P: V = (1/4πε₀)[q/r_A + (−q)/r_B] = (q/4πε₀)(1/r_A − 1/r_B), with q = 2 μC.
- When is V = 0? The potential vanishes when r_A = r_B, i.e. when the point is equidistant from A and B. The locus of all points equidistant from two fixed points is the plane that is the perpendicular bisector of the line AB (passing through its midpoint O, 3 cm from each charge).
- At every point of this plane the positive potential of the +2 μC charge is exactly cancelled by the negative potential of the −2 μC charge, so V = 0 on the whole plane, a constant. Hence the plane normal to AB at its mid-point is an equipotential surface (with potential zero, same as at infinity).
(b) Direction of the electric field on this surface
- The electric field at any point is perpendicular to the equipotential surface through that point (if E had a component along the surface, work would be done in moving a charge along it and the potential would not be constant).
- The surface here is perpendicular to AB, so E is parallel to AB at every point of the plane. By symmetry, the components of the two fields perpendicular to AB cancel and the components along AB add, and the net field points from the positive charge (A) towards the negative charge (B) — i.e. it is normal to the plane, directed along AB from A to B.
- (The same conclusion follows from Chapter 1: on the equatorial plane of a dipole E is antiparallel to the dipole moment p, which points from −q to +q, so E points from +q to −q.)
| Feature | Answer |
|---|---|
| Equipotential surface | plane perpendicular to AB through its midpoint |
| Potential on the surface | zero (V = 0) |
| Direction of E | normal to the surface, along AB from the +2 μC charge (A) towards the −2 μC charge (B) |
| Reason | E ⟂ equipotential surface; field directed from higher to lower potential |
Conclusion: the plane that perpendicularly bisects the line joining the charges (potential = 0) is an equipotential surface, and the electric field at every point on it is normal to the surface, directed along AB from the positive charge towards the negative charge.
A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?
Concept and context. A charged conductor in electrostatic equilibrium has its excess charge entirely on the outer surface. For a uniformly charged spherical conductor (radius R, total charge q) the symmetry allows the use of Gauss’s law with concentric spherical Gaussian surfaces: inside (r < R) the enclosed charge is zero and so E = 0; outside (r ≥ R) the sphere behaves as a point charge q at its centre, E = q/(4πε₀r²), directed radially outward for q > 0. Just outside the surface E = σ/ε₀ = q/(4πε₀R²), consistent with the general result for the field at the surface of a conductor.
Given: R = 12 cm = 0.12 m; q = 1.6 × 10⁻⁷ C; k = 1/4πε₀ = 9 × 10⁹ N m² C⁻².
(a) Inside the sphere (r < R)
- Electrostatic field inside a conductor is zero — a Gaussian surface inside encloses no charge (charge sits on the surface), so E × 4πr² = 0. Therefore E = 0. (This is also why a cavity in a conductor is shielded from external fields.)
(b) Just outside the sphere (r = R = 0.12 m)
- E = kq/R² = (9 × 10⁹ × 1.6 × 10⁻⁷)/(0.12)² = 1440/0.0144.
- E = 1.0 × 10⁵ N C⁻¹, directed radially outward (normal to the surface, because the charge is positive).
- Check using the surface charge density: σ = q/4πR² = 1.6 × 10⁻⁷/(4π × 0.0144) = 8.84 × 10⁻⁷ C m⁻²; σ/ε₀ = 8.84 × 10⁻⁷/8.854 × 10⁻¹² = 1.0 × 10⁵ N C⁻¹ ✓.
(c) At a point 18 cm from the centre (r = 0.18 m > R)
- E = kq/r² = (9 × 10⁹ × 1.6 × 10⁻⁷)/(0.18)² = 1440/0.0324.
- E ≈ 4.4 × 10⁴ N C⁻¹, directed radially outward.
- Cross-check with the inverse-square law: E(18 cm) = E(12 cm) × (12/18)² = 1.0 × 10⁵ × 4/9 = 4.4 × 10⁴ ✓.
| Position | Field E | Direction |
|---|---|---|
| (a) inside, r < 12 cm | 0 | — |
| (b) just outside, r = 12 cm | 1.0 × 10⁵ N C⁻¹ | radially outward |
| (c) r = 18 cm | 4.4 × 10⁴ N C⁻¹ | radially outward |
Observation: the electric field jumps from 0 (inside) to 1.0 × 10⁵ N C⁻¹ (just outside) across the surface — the field is discontinuous across a charged surface (by σ/ε₀) while the potential remains continuous (V = q/4πε₀R at the surface and inside).
Conclusion: the electric field is zero inside the conducting sphere, 1.0 × 10⁵ N C⁻¹ (radially outward) just outside, and 4.4 × 10⁴ N C⁻¹ (radially outward) at 18 cm from the centre.
A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1 pF = 10⁻¹² F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Concept and context. A capacitor is a pair of conductors separated by an insulator; its capacitance C = Q/V depends only on geometry and the medium. For a parallel plate capacitor of plate area A and separation d with vacuum (or air) between the plates, C₀ = ε₀A/d. When the whole space between the plates is filled with a dielectric of dielectric constant K, polarisation of the dielectric reduces the net field to E₀/K, so for the same charge the potential difference falls by a factor K and the capacitance rises K-fold: C = Kε₀A/d. The constant K is the ratio of the permittivity of the medium to that of free space (K = ε/ε₀) and is a pure number greater than 1 (K = 1 for vacuum, ≈ 1.0006 for air).
Given: C₀ = ε₀A/d = 8 pF; new separation d′ = d/2; dielectric constant K = 6.
- Original capacitance: C₀ = ε₀A/d = 8 pF.
- Effect of halving the separation: C ∝ 1/d, so with d′ = d/2 the capacitance doubles: ε₀A/(d/2) = 2C₀ = 16 pF (air still between the plates).
- Effect of the dielectric: multiply by K = 6: C′ = K × 2C₀ = 6 × 16 pF = 96 pF.
- One-line formula: C′ = Kε₀A/d′ = 6 × ε₀A/(d/2) = 12 × (ε₀A/d) = 12 × 8 pF = 96 pF.
C′ = 96 pF (= 9.6 × 10⁻¹¹ F), i.e. 12 times the original
| Change made | Factor on C | Capacitance |
|---|---|---|
| Original (air) | 1 | 8 pF |
| Halve the plate distance | ×2 | 16 pF |
| Fill with dielectric K = 6 | ×6 | 96 pF |
Conclusion: the new capacitance is 96 pF — twelve times the original 8 pF — because halving the separation doubles C and the dielectric of constant 6 multiplies it by 6.
Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?
Concept and context. When capacitors are joined end to end in series, the same charge Q flows on to every capacitor (the inner plates are just induced charges, the pair of plates between neighbouring capacitors remaining neutral overall), while the total potential difference of the supply is shared among them: V = V₁ + V₂ + V₃. Since V = Q/C for each, the reciprocal of the equivalent capacitance is the sum of the reciprocals: 1/C = 1/C₁ + 1/C₂ + 1/C₃ + …. Consequently the equivalent capacitance of a series combination is smaller than the smallest individual capacitance.
Given: C₁ = C₂ = C₃ = 9 pF; supply voltage V = 120 V.
(a) Total (equivalent) capacitance
- 1/C = 1/9 + 1/9 + 1/9 = 3/9 = 1/3 pF⁻¹.
- Hence C = 3 pF (= 3 × 10⁻¹² F). For n identical capacitors in series, C = C₁/n = 9/3 = 3 pF ✓.
(b) Potential difference across each capacitor
- Charge on the combination: Q = C V = (3 × 10⁻¹² F)(120 V) = 3.6 × 10⁻¹⁰ C. In a series combination this same charge is on every capacitor.
- Voltage across each: V₁ = Q/C₁ = (3.6 × 10⁻¹⁰ C)/(9 × 10⁻¹² F) = 40 V. Identical capacitors share the supply equally: V₂ = V₃ = 40 V.
- Check: V₁ + V₂ + V₃ = 40 + 40 + 40 = 120 V ✓.
| Quantity | Value |
|---|---|
| Equivalent capacitance | 3 pF (< 9 pF) |
| Charge on each capacitor | 3.6 × 10⁻¹⁰ C (same for all) |
| p.d. across each capacitor | 40 V |
| Sum of p.d. | 120 V = supply |
Conclusion: the equivalent capacitance of the series combination is 3 pF, and when connected to a 120 V supply the potential difference across each capacitor is 40 V.
Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.
Concept and context. When capacitors are connected in parallel, the same potential difference V (that of the supply) exists across each, while the total charge supplied by the source divides among them: Q = Q₁ + Q₂ + Q₃. Since Q = CV for each, C = C₁ + C₂ + C₃ + … — the equivalent capacitance of a parallel combination is the sum and exceeds the largest individual capacitance (effectively the plate area is added).
Given: C₁ = 2 pF, C₂ = 3 pF, C₃ = 4 pF; V = 100 V.
(a) Total capacitance
C = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF (= 9 × 10⁻¹² F)
(b) Charge on each capacitor — each has the full 100 V across it, so Q = CV:
| Capacitor | C | Q = CV (V = 100 V) |
|---|---|---|
| C₁ | 2 pF = 2 × 10⁻¹² F | 2 × 10⁻¹² × 100 = 2 × 10⁻¹⁰ C (200 pC) |
| C₂ | 3 pF = 3 × 10⁻¹² F | 3 × 10⁻¹⁰ C (300 pC) |
| C₃ | 4 pF = 4 × 10⁻¹² F | 4 × 10⁻¹⁰ C (400 pC) |
| Total | 9 pF | 9 × 10⁻¹⁰ C = CV with C = 9 pF ✓ |
- Check: total charge Q = CV = 9 × 10⁻¹² × 100 = 9 × 10⁻¹⁰ C = Q₁ + Q₂ + Q₃ = 2 + 3 + 4 (× 10⁻¹⁰ C) ✓.
- The largest capacitor (4 pF) stores the largest charge: charges in parallel are in proportion to the capacitances (2 : 3 : 4).
Conclusion: the total capacitance is 9 pF and, at 100 V, the charges on the 2 pF, 3 pF and 4 pF capacitors are 2 × 10⁻¹⁰ C, 3 × 10⁻¹⁰ C and 4 × 10⁻¹⁰ C respectively.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10⁻³ m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Concept and context. Neglecting edge effects (plates large compared with their separation), the field between the plates of a parallel plate capacitor is uniform, E = σ/ε₀ = Q/(ε₀A), so the potential difference is V = Ed = Qd/(ε₀A). Therefore C = Q/V = ε₀A/d, which shows that capacitance depends only on the geometry (A and d) and on the medium — not on Q or V. Plates carry equal and opposite charges +Q and −Q.
Given: A = 6 × 10⁻³ m²; d = 3 mm = 3 × 10⁻³ m; V = 100 V; ε₀ = 8.854 × 10⁻¹² C² N⁻¹ m⁻².
Capacitance
- C = ε₀A/d = (8.854 × 10⁻¹²)(6 × 10⁻³)/(3 × 10⁻³).
- A/d = (6 × 10⁻³)/(3 × 10⁻³) = 2 m. So C = 8.854 × 10⁻¹² × 2 = 17.7 × 10⁻¹² F.
- C ≈ 17.7 pF (1.77 × 10⁻¹¹ F).
Charge on each plate
- Q = CV = (17.7 × 10⁻¹² F)(100 V) = 1.77 × 10⁻⁹ C.
- So each plate carries a charge of magnitude 1.77 × 10⁻⁹ C (≈ 1.8 nC): +1.77 × 10⁻⁹ C on the plate connected to the positive terminal and −1.77 × 10⁻⁹ C on the other.
C = 17.7 pF; Q = ±1.77 × 10⁻⁹ C
- Extra quantities (useful for the next question): field between the plates E = V/d = 100/(3 × 10⁻³) = 3.33 × 10⁴ V m⁻¹; energy stored U = ½CV² = ½ × 17.7 × 10⁻¹² × 100² = 8.85 × 10⁻⁸ J.
Why capacitance does not depend on Q or V. For the plates of area A carrying ±Q, the surface charge density is σ = Q/A and the field between the plates is E = σ/ε₀ = Q/ε₀A. The potential difference V = Ed = Qd/ε₀A is therefore directly proportional to Q, so the ratio C = Q/V = ε₀A/d is a constant fixed by the geometry (and by the medium between the plates). Doubling the voltage simply doubles the charge. This is exactly why capacitance is “a geometric property” as NCERT stresses.
- Effect of changing the geometry: doubling the plate area doubles C; doubling the separation halves C. Here a large A/d = 2 m is needed for even 17.7 pF — which shows how small a “farad” is for ordinary geometries (a 1 F air capacitor would need plate area ≈ 10⁸ m² at 1 mm separation, i.e. a 10 km × 10 km square).
- Check units: ε₀A/d has units (C² N⁻¹ m⁻²)(m²)/(m) = C² N⁻¹ m⁻¹ = C² J⁻¹ = F ✓ (since 1 F = 1 C V⁻¹ = 1 C²/J).
- Edge effects are neglected because the plate dimensions (√(6 × 10⁻³) ≈ 7.7 cm) are much larger than the 3 mm gap, so the field between the plates is practically uniform and zero outside.
| Quantity | Formula | Value |
|---|---|---|
| Capacitance | C = ε₀A/d | 17.7 pF |
| Charge (V = 100 V) | Q = CV | 1.77 × 10⁻⁹ C |
| Surface charge density | σ = Q/A | 2.95 × 10⁻⁷ C m⁻² |
| Field between plates | E = σ/ε₀ = V/d | 3.33 × 10⁴ V m⁻¹ |
Conclusion: the capacitance is 17.7 pF and, on connection to a 100 V supply, each plate carries a charge of 1.77 × 10⁻⁹ C (positive on one plate, negative on the other).
Explain what would happen if in the capacitor given in Exercise 2.8, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates, (a) while the voltage supply remained connected. (b) after the supply was disconnected.
Concept and context. A dielectric is an insulator whose molecules (polar or non-polar) get polarised in an electric field: a net dipole moment appears along the field, so equal and opposite induced (bound) charges appear on the two faces of the slab. These induced surface charges produce a field opposite to the applied field, so the net field inside the dielectric is reduced to E = E₀/K (K = dielectric constant). For a fixed charge this lowers the potential difference by the factor K, so the capacitance rises to C = KC₀. Since the mica sheet is as thick as the gap (3 mm), it fills the space completely and the simple formula C = Kε₀A/d applies. What happens next depends on what is held constant: the voltage V (supply connected) or the charge Q (supply disconnected).
Data of the capacitor of Exercise 2.8: C₀ = ε₀A/d = 17.7 pF; supply V₀ = 100 V; charge Q₀ = C₀V₀ = 1.77 × 10⁻⁹ C; field E₀ = V₀/d = 100/(3 × 10⁻³) = 3.33 × 10⁴ V m⁻¹; stored energy U₀ = ½C₀V₀² = 8.85 × 10⁻⁸ J. New capacitance with mica: C = KC₀ = 6 × 17.7 = 106 pF (1.06 × 10⁻¹⁰ F) in both cases.
(a) Supply remains connected — potential difference V = 100 V stays constant
- Capacitance rises six-fold to 106 pF (≈ 1.06 × 10⁻¹⁰ F).
- Charge increases: Q = CV = (1.06 × 10⁻¹⁰)(100) = 1.06 × 10⁻⁸ C — six times the earlier 1.77 × 10⁻⁹ C. The additional charge (≈ 8.9 × 10⁻⁹ C) flows from the battery on to the plates to keep the p.d. at 100 V: the polarised dielectric lowers the potential, so more free charge is drawn.
- Electric field in the dielectric E = V/d = 100/(3 × 10⁻³) = 3.33 × 10⁴ V m⁻¹ — unchanged (V and d are the same). The free-charge field has become 6 times larger (σ is 6 times larger) but the induced charges reduce it by the same factor 6.
- Energy stored U = ½CV² = ½ × 1.06 × 10⁻¹⁰ × 10⁴ = 5.3 × 10⁻⁷ J — six times the earlier value (extra energy comes from the battery).
(b) Supply disconnected — charge Q = 1.77 × 10⁻⁹ C stays constant
- Capacitance again rises to 106 pF.
- Charge cannot change (isolated plates): Q = 1.77 × 10⁻⁹ C.
- Potential difference falls: V′ = Q/C = (1.77 × 10⁻⁹)/(1.06 × 10⁻¹⁰) = 16.7 V (= 100/6).
- Electric field E′ = V′/d = 16.7/(3 × 10⁻³) = 5.56 × 10³ V m⁻¹ = E₀/K — reduced to one-sixth.
- Energy stored U′ = Q²/2C = (1.77 × 10⁻⁹)²/(2 × 1.06 × 10⁻¹⁰) = 1.5 × 10⁻⁸ J — one-sixth of the initial value; the “missing” energy has gone into doing work in pulling the dielectric slab into the gap (the slab is attracted into the field).
| Quantity | Initially (air) | (a) V constant (supply on) | (b) Q constant (supply off) |
|---|---|---|---|
| Capacitance C | 17.7 pF | 106 pF (×6) | 106 pF (×6) |
| Potential difference V | 100 V | 100 V (same) | 16.7 V (÷6) |
| Charge Q | 1.77 × 10⁻⁹ C | 1.06 × 10⁻⁸ C (×6) | 1.77 × 10⁻⁹ C (same) |
| Electric field E | 3.33 × 10⁴ V/m | 3.33 × 10⁴ V/m (same) | 5.56 × 10³ V/m (÷6) |
| Stored energy U | 8.85 × 10⁻⁸ J | 5.3 × 10⁻⁷ J (×6) | 1.5 × 10⁻⁸ J (÷6) |
Conclusion: on inserting the mica sheet the capacitance becomes 106 pF in both cases; with the supply connected the charge rises to 1.06 × 10⁻⁸ C at the same 100 V and E = 3.33 × 10⁴ V m⁻¹ (energy ×6), whereas with the supply disconnected the charge stays 1.77 × 10⁻⁹ C, the p.d. falls to 16.7 V, the field to 5.56 × 10³ V m⁻¹ and the energy to one-sixth.
A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
Concept and context. Charging a capacitor means transferring charge from one plate to the other against the field set up by the charges already transferred. When a charge q is already on the capacitor (potential difference q/C), moving a further charge dq needs work dW = (q/C)dq. Integrating from 0 to Q gives the energy stored in the electric field between the plates: U = Q²/2C = ½CV² = ½QV. (The corresponding energy density in the field is u = ½ε₀E².) Note that exactly half of the work done by the battery (QV) is stored in the capacitor; the other half is dissipated as heat in the circuit during charging.
Given: C = 12 pF = 12 × 10⁻¹² F; V = 50 V.
- Choose the form with C and V: U = ½CV².
- U = ½ × (12 × 10⁻¹²) × (50)² = ½ × 12 × 10⁻¹² × 2500 = 6 × 10⁻¹² × 2500.
- U = 1.5 × 10⁻⁸ J (= 15 nJ).
U = ½CV² = 1.5 × 10⁻⁸ J
| Quantity | Value |
|---|---|
| Charge on the capacitor Q = CV | 12 × 10⁻¹² × 50 = 6 × 10⁻¹⁰ C |
| Energy U = ½CV² | 1.5 × 10⁻⁸ J |
| Check ½QV | ½ × 6 × 10⁻¹⁰ × 50 = 1.5 × 10⁻⁸ J ✓ |
| Check Q²/2C | (6 × 10⁻¹⁰)²/(2 × 12 × 10⁻¹²) = 1.5 × 10⁻⁸ J ✓ |
Step-by-step derivation of U = ½CV² (full NCERT logic).
- Charging in small steps: suppose the capacitor already carries charge q, so the potential difference across it is v = q/C. To transfer a further small charge dq from the negative plate to the positive plate, work dW = v dq = (q/C) dq must be done against the field of the charges already present.
- Integrate from 0 to Q: U = ∫₀^Q (q/C) dq = (1/C)[q²/2]₀^Q = Q²/2C.
- Use Q = CV: U = Q²/2C = ½CV² = ½QV. The graph of V against q is a straight line of slope 1/C, and the energy is the area of the triangle under it (Fig. 2.30 of NCERT shows the work done in a small step as a thin strip).
- Where is the energy? It is stored in the electric field between the plates. For a parallel plate capacitor (C = ε₀A/d, V = Ed): U = ½(ε₀A/d)(Ed)² = ½ε₀E²(Ad), so the energy density (energy per unit volume) is u = ½ε₀E² — a result valid for any electric field in vacuum.
- Energy delivered by the battery: the battery moves charge Q through a potential difference V, doing work QV = 6 × 10⁻¹⁰ × 50 = 3 × 10⁻⁸ J, but only half of it (1.5 × 10⁻⁸ J) is stored; the other half is dissipated as heat in the connecting wires and the battery’s internal resistance (independent of the resistance value).
- Practical significance: capacitors store energy that can be released very rapidly — as in the flash of a camera, defibrillators and pulsed lasers. A 12 pF capacitor stores only a minuscule amount (15 nJ), which is why practical energy-storage capacitors use microfarad and farad ranges.
| Form of the formula | When it is convenient | Value here |
|---|---|---|
| U = ½CV² | C and V given | 1.5 × 10⁻⁸ J |
| U = ½QV | Q and V given | 1.5 × 10⁻⁸ J |
| U = Q²/2C | Q and C given | 1.5 × 10⁻⁸ J |
| u = ½ε₀E² | field given (energy per unit volume) | — |
Conclusion: the electrostatic energy stored in the 12 pF capacitor charged by the 50 V battery is 1.5 × 10⁻⁸ J (15 nJ).
A 600 pF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Concept and context. When a charged capacitor is connected to an uncharged one, charge flows from the first to the second until both have the same potential difference. The total charge is conserved (isolated system), but the energy is not: part of the initial energy is lost as heat in the connecting wires and as electromagnetic radiation (a small spark). This question is the same as NCERT Example 2.10(b) with different numbers.
Given: C₁ = C₂ = 600 pF = 6 × 10⁻¹⁰ F; initial voltage V = 200 V on C₁; C₂ uncharged.
Initial energy
- Charge on C₁: Q = C₁V = 6 × 10⁻¹⁰ × 200 = 1.2 × 10⁻⁷ C.
- Energy U₁ = ½C₁V² = ½ × 6 × 10⁻¹⁰ × (200)² = ½ × 6 × 10⁻¹⁰ × 4 × 10⁴ = 1.2 × 10⁻⁵ J.
After connecting to the uncharged capacitor
- The positive plates are joined together (and so are the negative plates), so the two capacitors are effectively in parallel: equivalent capacitance C′ = 600 + 600 = 1200 pF = 1.2 × 10⁻⁹ F.
- Conservation of charge: the total charge stays Q = 1.2 × 10⁻⁷ C, shared equally (0.6 × 10⁻⁷ C on each). Common potential difference V′ = Q/C′ = (1.2 × 10⁻⁷)/(1.2 × 10⁻⁹) = 100 V.
- Final energy U₂ = ½C′V′² = ½ × 1.2 × 10⁻⁹ × (100)² = 6 × 10⁻⁶ J (check: Q²/2C′ = (1.2 × 10⁻⁷)²/(2 × 1.2 × 10⁻⁹) = 6 × 10⁻⁶ J ✓).
Energy lost
ΔU = U₁ − U₂ = 1.2 × 10⁻⁵ − 6 × 10⁻⁶ = 6 × 10⁻⁶ J (half of the original energy)
- Where does the energy go? Charge flows through the connecting wires (which have resistance) and the sudden current produces heat, plus a tiny spark/electromagnetic radiation. Energy is still conserved overall; only the electrostatic energy of the capacitors decreases.
- For identical capacitors the loss is always 50 %: ΔU = ½ U₁ regardless of the values.
| Stage | Charge | Voltage | Energy |
|---|---|---|---|
| Before (C₁ alone) | 1.2 × 10⁻⁷ C | 200 V | 1.2 × 10⁻⁵ J |
| After (C₁ ∥ C₂) | 1.2 × 10⁻⁷ C (conserved) | 100 V | 6 × 10⁻⁶ J |
| Loss | 0 | −100 V | 6 × 10⁻⁶ J |
Conclusion: the electrostatic energy lost is 6 × 10⁻⁶ J — exactly half of the initial 1.2 × 10⁻⁵ J — dissipated as heat and electromagnetic radiation when the charge redistributes between the two capacitors.