NCERT Solutions Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance

What the question is asking. We have two point charges of opposite sign on a straight line, 16 cm apart. We need every point on that line where the electric potentials from the two charges exactly cancel — i.e., where the total potential is zero (with infinity as the reference).

Given.

• q₁ = +510^-8 C at point A.
• q₂ = -310^-8 C at point B.
• Separation AB = 16 cm = 0.16 m.

Concept used — potential due to a point charge. The electrostatic potential at a distance r from a point charge q (with V∞=0) is V = 14π₀ qr. Potential is a scalar — it adds with signs, not as vectors. The sign of q is carried into the answer.

Step 1 — set up a coordinate. Let A be at x=0 and B be at x=16 cm. Let P at coordinate x on the line AB be the point where V=0. We must consider three regions: (i) between A and B (0 < x < 16), (ii) right of B (x > 16), and (iii) left of A (x < 0).

Step 2 — write the total potential. Let distances be r₁ = |x - 0| and r₂ = |x - 16| (in cm). Then V_P = kq₁r₁ + kq₂r₂ = k(q₁r₁ + q₂r₂). Setting V_P = 0: q₁r₁ + q₂r₂ = 0 q₁r₁ = -q₂r₂.

Step 3 — case A: point P lies between A and B. Then r₁ = x and r₂ = 16 - x, both positive. The equation becomes 510^-8x = 310^-816 - x. Cross-multiply: 5(16 - x) = 3x 80 - 5x = 3x 8x = 80 x = 10 cm. So one zero-potential point lies 10 cm from A equivalently 6 cm from B, between the two charges.

Step 4 — case B: point P lies outside, on the side of -310^-8 C (i.e., x > 16). Now r₁ = x and r₂ = x - 16. The equation 510^-8x = 310^-8x - 16 gives 5(x - 16) = 3x 5x - 80 = 3x 2x = 80 x = 40 cm. So a second zero-potential point lies 40 cm from A — i.e., 24 cm beyond B on the far side.

Step 5 — case C: P to the left of A (x < 0). Then r₁ = -x, r₂ = 16 - x. The equation q₁/r₁ = -q₂/r₂ gives 510^-8-x = 310^-816 - x. But the left side is positive only when x < 0, and we get 5(16-x) = -3x, i.e., 80 - 5x = -3x, so 2x = 80, x = 40 cm — which contradicts x < 0. No solution in this region. (Physically: on the far side of the larger positive charge, the positive potential from it always exceeds the negative potential from the smaller, more distant negative charge.)

Final answer. Zero potential at 10 cm from A (between the charges) and at 40 cm from A (beyond B).

What the question is asking. Six identical positive charges sit at the six corners of a regular hexagon of side 10 cm. Find the electric potential at the centre of the hexagon.

Given.

• Side of hexagon, a = 10 cm = 0.10 m.
• Charge at each vertex, q = 5 = 510^-6 C.
• Number of vertices, N = 6.

Concept used.

1. For a regular hexagon, the centre is equidistant from all six vertices. In fact, the distance from centre to vertex equals the side length: r = a = 0.10 m. (A regular hexagon is made of 6 equilateral triangles meeting at the centre — each triangle has all sides equal to a.)
2. Potential is a scalar: it adds algebraically. Six identical charges at equal distance contribute equally.

Step 1 — distance from centre to each vertex. For a regular hexagon of side a: r = a = 0.10 m.

Step 2 — potential from one charge at the centre. V₁ = 14π₀ qr = k qr.

Step 3 — total potential at the centre. Because all six contributions are equal and have the same sign, V = 6 V₁ = 6 k qr.

Step 4 — substitute numbers. V = 6 (910⁹) (510^-6)0.10. Compute the numerator step by step: 6× 910⁹ = 5410⁹, 5410⁹× 510^-6 = 27010³ = 2.710⁵.

Step 5 — divide by r. V = 2.710⁵0.10 = 2.710⁶ V.

Final answer. V = 2.710⁶ V = 2.7 MV. This is a large potential — about a million times the voltage of a household battery. It would not occur in practice unless the charges were held in place by an insulating frame, because mutual repulsion would otherwise blow the configuration apart.

What the question is asking. Two equal-and-opposite charges form an electric dipole. We need to find a surface on which every point has the same potential, and figure out which way the field points on that surface.

Given.

• q_A = +2 at A.
• q_B = -2 at B.
• Separation AB = 6 cm.

Concept used. An equipotential surface is one on which the potential is constant. For a single point charge, equipotential surfaces are concentric spheres centred on the charge. For two charges, you must add the potentials and look for the locus of constant total V.

Step 1 — locate the obvious equipotential. Consider any point P that is equidistant from A and B: r_A = r_B = r. Then V_P = kq_Ar_A + kq_Br_B = k(+2 )r + k(-2 )r = 0. So every point equidistant from A and B has V = 0 — they all share the same potential.

Step 2 — what surface is "equidistant from two points"? The set of points in 3D equidistant from two given points is the perpendicular bisector plane of the line segment joining them. For our dipole, that's the plane perpendicular to AB and passing through its midpoint.

(a) Equipotential surface. The perpendicular bisector plane of AB — the plane normal to AB, passing through its midpoint — is the equipotential surface V = 0.

(b) Direction of the field on this surface. On an equipotential surface, the electric field is always perpendicular to the surface. Reason: if there were any component of E parallel to the surface, moving a test charge along that direction would do nonzero work — but V is constant along the surface, so work = q Δ V = 0. The only way to satisfy this for any direction tangent to the surface is for E to have no tangential component, i.e., E ⊥ surface.

Since our surface is perpendicular to AB, the perpendicular to the surface is parallel to AB. So at every point on this equipotential surface, E points along the line AB, from the positive charge (A) toward the negative charge (B).

Final answer.

• (a) Equipotential surface: the perpendicular bisector plane of AB (containing the midpoint and normal to AB). On this plane, V = 0.
• (b) The electric field at every point on this surface is perpendicular to the surface, hence parallel to AB. It points from the positive charge +2 toward the negative charge -2 (i.e., from A toward B).

Given.

• Radius of conductor, R = 12 cm = 0.12 m.
• Total charge, Q = 1.610^-7 C, uniformly distributed on the surface.
• Coulomb constant, k = 910⁹ N m² C^-2.

Concept used — Gauss's law for a spherically symmetric conductor.

• Inside a charged conductor (in electrostatic equilibrium): E = 0 — because any internal field would push the free electrons until they redistribute and kill it.
• Outside, the conductor looks (from the field's point of view) exactly like a point charge Q sitting at its centre. So E_outside(r) = 14π₀ Qr² r ≥ R.

(a) Electric field inside the sphere.

The interior of a conductor in electrostatic equilibrium has zero electric field everywhere: E_inside = 0.

(b) Electric field just outside the sphere. Set r = R = 0.12 m.

Step 1 — apply the formula. E = kQR² = (910⁹)(1.610^-7)(0.12)².

Step 2 — compute the numerator. kQ = (910⁹)(1.610^-7) = 14.410² = 1.4410³.

Step 3 — compute the denominator. R² = (0.12)² = 0.0144 m².

Step 4 — divide. E = 1.4410³0.0144 = 10⁵ N/C. E_{just outside} = 1.010⁵ N/C. The field points radially outward (since Q > 0).

(c) Electric field at r = 18 cm = 0.18 m.

Since r = 0.18 m > R = 0.12 m, the point is outside the sphere; again we use E = kQ/r².

Step 1 — substitute. E = (910⁹)(1.610^-7)(0.18)².

Step 2 — denominator. (0.18)² = 0.0324 m².

Step 3 — divide. E = 1.4410³0.0324 = 4.4410⁴ N/C. E(18 cm) ≈ 4.410⁴ N/C, radially outward.

Given.

• Original capacitance (air, separation d): C₁ = 8 pF = 810^-12 F.
• New separation: d' = d/2 (half the original).
• New medium: dielectric of constant K = 6 (replaces air).

Concept used — capacitance of a parallel-plate capacitor. C = K ₀ Ad, where A is the plate area, d is the gap, and K is the dielectric constant of the material between the plates K = 1 for vacuum or air. The capacitance:

• increases if A increases (more space for charge);
• increases if d decreases (closer plates → stronger electric coupling);
• increases if K increases (dielectric "screens" the field, allowing more charge for the same voltage).

Step 1 — write the original capacitance. C₁ = ₀ Ad = 8 pF. Note K_air = 1, so it doesn't appear explicitly.

Step 2 — write the new capacitance. With d → d/2 and K = 6: C₂ = K ₀ Ad/2 = 6 ₀ A · 2d = 12 ₀ Ad.

Step 3 — relate C₂ to C₁. Since ₀ A/d = C₁ = 8 pF, C₂ = 12 · C₁ = 12 × 8 = 96 pF.

Final answer. C₂ = 96 pF. The capacitance has gone up by a factor of 12 — a combined gain of × 2 from halving the gap and × 6 from the dielectric.

Given.

• Three identical capacitors, each C = 9 pF = 910^-12 F, in series.
• Supply voltage V = 120 V.

Concept used — capacitors in series. 1C_eq = 1C₁ + 1C₂ + 1C₃ + … In series, the same charge Q sits on every capacitor (charge cannot accumulate on the isolated central plates). The voltage divides between them inversely as their capacitance.

(a) Equivalent capacitance.

Step 1 — three equal capacitors in series. 1C_eq = 1C + 1C + 1C = 3C.

Step 2 — invert. C_eq = C3 = 9 pF3 = 3 pF.

(b) Potential difference across each capacitor.

Step 1 — charge on the equivalent capacitor. When 120 V is applied, Q = C_eq V = (310^-12)(120) = 36010^-12 C = 3.610^-10 C. This same charge sits on every one of the three series capacitors.

Step 2 — voltage across each individual capacitor. V_each = QC = 3.610^-10910^-12 = 40 V.

Sanity check. The three voltages should add to the supply: 40 + 40 + 40 = 120 V.

Final answers.

• (a) C_eq = 3 pF.
• (b) V_each = 40 V across each capacitor.

Given.

• C₁ = 2 pF, C₂ = 3 pF, C₃ = 4 pF, in parallel.
• Supply voltage V = 100 V.

Concept used — capacitors in parallel. C_eq = C₁ + C₂ + C₃ + … In parallel, every capacitor sits across the same voltage V. Each acquires a charge Q_i = C_i V, and the total charge from the supply is the sum of the individual charges.

(a) Equivalent capacitance. C_eq = C₁ + C₂ + C₃ = 2 + 3 + 4 = 9 pF.

(b) Charge on each capacitor. Same voltage V = 100 V across each.

Step 1 — capacitor 1. Q₁ = C₁ V = (210^-12)(100) = 210^-10 C = 200 pC.

Step 2 — capacitor 2. Q₂ = C₂ V = (310^-12)(100) = 310^-10 C = 300 pC.

Step 3 — capacitor 3. Q₃ = C₃ V = (410^-12)(100) = 410^-10 C = 400 pC.

Sanity check. Total charge supplied: Q₁ + Q₂ + Q₃ = (2 + 3 + 4)10^-10 = 910^-10 C. Compare with C_eqV = 910^-12(100) = 910^-10 C.

Final answers.

• (a) C_eq = 9 pF.
• (b) Q₁ = 210^-10 C, Q₂ = 310^-10 C, Q₃ = 410^-10 C.

Given.

• Plate area, A = 610^-3 m².
• Plate separation, d = 3 mm = 310^-3 m.
• Medium between plates: air K = 1.
• Supply voltage, V = 100 V.
• Permittivity of free space, ₀ = 8.85410^-12 C² N^-1 m^-2.

Concept used — capacitance of a parallel-plate capacitor. C = ₀ Ad. And the charge stored at applied voltage V: Q = C V.

Step 1 — write the formula with numbers. C = (8.85410^-12)(610^-3)310^-3.

Step 2 — simplify the area-over-distance factor. Ad = 610^-3310^-3 = 2. Dimensionless ratio with units of m; the m²/m = m absorbs into the prefactor.

Step 3 — multiply by ₀. C = (8.85410^-12)× 2 = 17.7110^-12 F ≈ 17.7 pF.

Step 4 — compute the charge. Q = C V = (17.7110^-12)(100) = 1.77110^-9 C ≈ 1.7710^-9 C.

Final answers.

• Capacitance: C ≈ 17.7 pF.
• Charge on each plate: Q ≈ 1.7710^-9 C = 1.77 nC. One plate is +Q, the other is -Q.

Setup recap. From Q 2.8: C_air = 17.7 pF, V = 100 V, Q_air = 1.77 nC. Now a mica sheet of K = 6 and thickness equal to the plate gap 3 mm fills the gap completely.

New capacitance (in both cases). C_mica = K C_air = 6 × 17.7 pF = 106.2 pF ≈ 1.0610^-10 F. The capacitance increases by the dielectric constant, regardless of whether the battery is on or off — C is a geometric/material property of the capacitor itself.

(a) Battery still connected (V held at 100 V).

When the supply stays connected, the voltage across the capacitor is fixed by the battery: V = 100 V. Anything else (Q, E, energy) must adjust.

Step 1 — new charge. Q' = C_mica V = (106.210^-12)(100) = 1.06210^-8 C.

Step 2 — change in charge. Δ Q = Q' - Q = 1.06210^-8 - 1.7710^-9 ≈ 8.8510^-9 C. The extra charge Δ Q ≈ 8.85 nC flows from the battery onto the plates. The battery had to supply this extra charge.

Step 3 — electric field. E = V/d = 100/0.003 = 3.3310⁴ V/m. Unchanged (because V and d are unchanged).

(b) Battery disconnected before mica inserted (Q held at 1.77 nC).

With no battery, the charge on the plates is trapped — Q is fixed. The capacitance still goes up by K = 6, but now it's the voltage that adjusts.

Step 1 — new voltage. V' = QC_mica = 1.7710^-9106.210^-12 ≈ 16.67 V. The voltage drops to one-sixth of its previous value i.e., V/K = 100/6 ≈ 16.67 V.

Step 2 — electric field. E' = V'd = 16.670.003 ≈ 5.5610³ V/m. The field also drops by a factor of K = 6 — because the polarised dielectric partially cancels the field of the plate charges.

Summary table.

• Both cases: C_new = 106.2 pF = 6 C_old.
• (a) Battery on: V constant at 100 V; Q increases 6×; E unchanged; battery does extra work.
• (b) Battery off: Q constant at 1.77 nC; V drops 6× to 16.67 V; E drops 6×.

Given.

• Capacitance, C = 12 pF = 1210^-12 F.
• Applied voltage, V = 50 V.

Concept used — energy stored in a charged capacitor. A capacitor charged to voltage V stores an electrostatic potential energy U = 12 C V². Equivalent forms using Q = CV: U = 12 Q V = Q²2C. Why the factor of 12: when you charge a capacitor, you push the first electron across 0 voltage, the last across the full V, and the average across V/2. So total work is Q× V/2 = 12 QV.

Step 1 — plug into U = 12 C V². U = 12 (1210^-12) (50)².

Step 2 — square the voltage. V² = (50)² = 2500.

Step 3 — combine. U = 12 × 1210^-12 × 2500. First 12 × 12 = 6. Then 6× 2500 = 15,000. So U = 15,00010^-12 = 1.510^-8 J.

Final answer. U = 1.510^-8 J = 15 nJ. This is a very small amount of energy — about a billionth of a joule. A AA battery stores roughly 10 kJ — a trillion times more.

Given.

• Capacitor 1: C₁ = 600 pF, initially charged to V₀ = 200 V.
• Capacitor 2: C₂ = 600 pF, initially uncharged V = 0.
• The two are then connected together; charge redistributes.

Concept used.

1. Charge is conserved. Total charge on the two plates (the upper plates, say) before and after is the same.
2. After connection, both capacitors share the same final voltage V_f (because they're connected in parallel).
3. Energy stored in each capacitor: U = 12 CV².

Step 1 — initial charge on C₁. Q₀ = C₁ V₀ = (60010^-12)(200) = 1.210^-7 C.

Step 2 — initial energy. U_i = 12 C₁ V₀² = 12 (60010^-12)(200)². Compute: (200)² = 410⁴; then 12 · 60010^-12· 410⁴ = 1.210^-5 J.

Step 3 — final voltage after connection. When the two capacitors share the charge Q₀, they end up at the same voltage V_f, and the total capacitance is the parallel sum C₁ + C₂ = 1200 pF. V_f = Q₀C₁ + C₂ = 1.210^-7120010^-12 = 100 V. Half the original — makes sense, since the same charge now sits on twice the capacitance.

Step 4 — final energy. U_f = 12 (C₁ + C₂) V_f² = 12 (120010^-12)(100)². Compute: (100)² = 10⁴; then 12· 120010^-12· 10⁴ = 610^-6 J.

Step 5 — energy lost. Δ U = U_i - U_f = 1.210^-5 - 610^-6 = 610^-6 J.

Final answer. Δ U = 610^-6 J = 6 . Exactly half the initial energy is lost.