NCERT Exemplar Class 12 Physics Chapter 2 Solutions PDF

Strategic angle. The branch-by-branch approach is most foolproof here: spot which loop actually carries current, get the common terminal voltage, then read the capacitor voltage off the upper branch.

1. Why the upper branch is dead. A capacitor in steady state stores charge and rejects further current. So I_upper = 0, and the 10 Ω resistor contributes no voltage drop.
2. Lower-branch current. The only conducting loop is battery → r → 2 Ω → battery. Ohm's law for this loop: I = Er + R = 2.5 V0.5 Ω + 2 Ω = 1 A.
3. Terminal voltage = voltage across capacitor. Both branches share the same two end-nodes (the battery terminals), so the voltage across the upper branch equals that across the lower branch: V_C + V_10Ω = V_2Ω ⇒ V_C + 0 = (1)(2) = 2 V.
4. Charge. Q = C V_C = (4× 10^-6 F)(2 V) = 8× 10^-6 C = 8 .

The choice 10 μC is the deliberate trap: it forgets the internal-resistance drop. Treating E = 2.5 V as the voltage across the capacitor (instead of the terminal voltage E - Ir) gives Q = 4 × 2.5 V = 10 μC, which is option (b) and the standard mistake.

Alternative method, Kirchhoff's voltage law explicitly. Write KVL around the conducting loop (E → r → 2 Ω → back): E - Ir - I(2 Ω) = 0 ⇒ I = 2.52.5 = 1 A. Now traverse from one capacitor plate to the other via the lower branch: starting at the top node, drop V_C across C, then 0 across the 10 Ω resistor (no current), then -IR across 2 Ω, finishing at the bottom node. Since both branches share the same end-nodes, the algebraic sum across each must match the terminal voltage E - Ir = 2.5 - 0.5 = 2 V. Hence V_C = 2 V.

Concept linkage. This problem stitches together Chapter 3 (current electricity and EMF with internal resistance) with the chapter-2 capacitor relation Q = CV. Whenever a capacitor sits in a branch with no other DC path, that branch carries zero steady current, so all the voltage of the parallel branch piles across the capacitor (resistors in that dead branch drop 0).

Exam tip. In CBSE board problems, this kind of ``capacitor-in-a-DC-circuit'' problem is a 2-mark or 3-mark favourite. Show: (i) the steady-state current path; (ii) the terminal/Branch voltage calculation; (iii) Q = CV. Skipping any of these three steps usually costs the full mark for that step.

Q = 8 , option (d).

Quick reading. Three facts decide this question: F⃗ = qE⃗, the field points from higher to lower potential, and U = qV for a point charge in an external field.

1. For q > 0, F⃗ is parallel to E⃗, so the particle's displacement is along E⃗.
2. Moving along E⃗ takes the particle to lower potential. Quantitatively, V_f - V_i = -_i^f E⃗· dl⃗ < 0.
3. Therefore Δ U = q Δ V < 0. The PE decreases.
4. This rules out (a), (b), (d). Option (d) is wrong because a positive charge does not move opposite to E⃗ when released from rest; that would happen for a negative charge.

Alternative method, energy conservation. No friction acts, so mechanical energy is conserved: K_i + U_i = K_f + U_f. With K_i = 0 (released from rest) and K_f > 0 (the charge acquires kinetic energy), U must decrease. There is no other option consistent with energy conservation, the PE has to drop to ``feed'' the rising kinetic energy.

Common pitfall, confusing PE of the charge with PE of a test charge. The PE that appears in U = qV is the PE of the particular charge q in the external potential V. Some students confuse this with the ``potential'' V alone, which is a property of the field. V decreases here and U = qV decreases (because q > 0). For a negative test charge in the same field, V would still decrease, but U = qV would increase (sign flip). Always check the sign of q when converting from V to U.

Concept linkage, gravitational analogue. For a positive charge in a uniform E⃗, U = -qE₀ z (taking E⃗ = E₀z downward), is the exact analogue of a mass in gravity: U_grav = -mgy. The charge ``falls down the potential hill'' just as a stone falls down a gravitational hill, and both gain kinetic energy in the process.

Option (c).

Strategic angle. The Exemplar is testing a single idea: electrostatic work depends only on end-points. Spotting that idea finishes the question in two lines.

1. In each panel, A sits on its equipotential at value V_A and B on its equipotential at V_B. The numerical values of V_A and V_B are the same across all three panels.
2. W_{A→ B} = q(V_A - V_B) depends only on these two values, not on how the equipotentials lie in between. Hence W is identical in (i), (ii) and (iii).

(a), (b), (d) are tempting only if a student forgets that the electrostatic field is conservative.

Alternative method, direct line-integral picture. The work along any path is W = -q_A^B E⃗· dl⃗. If a student wants to ``compute'' this integral, slice each path into small steps; each step lies between two adjacent equipotentials, so the work for that step equals q Δ V where Δ V is the potential drop between adjacent equipotential lines along the path. Summing the segment-by-segment Δ V's telescopes to V_A - V_B, no matter how the equipotentials are spaced. The field-line crowding is hidden inside each step, not in the sum.

Common pitfall. Some students argue ``Fig. (i) has denser equipotentials, so E⃗ is stronger, so more work''. This conflates ``larger force per unit length'' with ``larger total work''. The force is larger along the denser-spaced panel, but the path is correspondingly shorter (because the same Δ V is crossed over a smaller distance), and the product FΔ l stays the same. The cancellation is exact for any conservative field.

Diagram-based reasoning. Each Fig. I, II, III shows the same labelled equipotential values at A and at B; only the intermediate equipotentials shift. Imagine the panels are just three different visualisations of the same physical setup seen with different ``zoom'' settings, the work between fixed endpoints obviously cannot depend on a zoom choice.

Option (c).

Quick reading. ``Inside a conductor at equilibrium, E⃗ = 0'' is the chapter-1 fact. ``V is constant wherever E⃗ = 0'' is the chapter-2 corollary. The question is whether the corollary follows from the fact (yes) or stands on its own (no).

1. E⃗ = -∇ V relates the two: zero E⃗ forces constant V.
2. Continuity of V across the surface fixes the constant equal to the surface value, 100 V.
3. So S₂ is a direct consequence of S₁, they are not independent, and (d) is eliminated.

Alternative method, work-done argument. Move a test charge q₀ from the surface (at V = 100 V) to any interior point along a straight radial path. The work done against the field is W = -q₀∫ E⃗· dl⃗ = q₀ (V_surface - V_inside). But E⃗ = 0 inside the conductor, so W = 0. Therefore V_inside = V_surface = 100 V, confirming S₂ from S₁ directly, without any vector calculus.

Concept linkage, why the surface is special. On the outside face of the conductor, E⃗ is non-zero and points perpendicular to the surface, with magnitude σ/₀. The discontinuity of E⃗ across the surface (zero inside, non-zero outside) does not create a discontinuity in V, because V is the line integral of a bounded field over a step of zero thickness. That is why the interior V equals the surface V exactly.

Exam tip. The CBSE board often disguises this question as ``find the potential at the centre of a charged conducting shell''. The answer is the surface potential, not zero, the inside field is zero, but the potential is decidedly not. Worth memorising as a 1-mark booster.

Option (c).

Picture-first. Stand very far from a complicated charge cluster whose net charge is Q. The cluster collapses, in the limbic eye, to a point of charge Q at its centroid. The equipotentials of a point charge are concentric spheres. Therefore the equipotentials of any net-charged finite cluster, viewed from far enough away, are also approximately spheres.

1. Multipole expansion at large r: V = kQr + k p⃗·rr² + ⋯
2. Leading term → kQ/r, which is spherically symmetric.
3. Constant V ⇒ constant r ⇒ sphere.

Alternative method, counter-example check. A common quick-check: place two equal charges +Q/2 at points 1 cm apart and look at the equipotentials. At r = 1 m (a hundred times the separation), the equipotentials are nearly indistinguishable from those of a single +Q at the centroid: they are spheres to one part in 10⁴. By r = 100 m, the spherical approximation is exact to one part in 10⁸. ``Far enough'' for the textbook usually means r 10× the extent of the distribution.

Common pitfall. Students see ``a collection of charges'' and immediately picture a dipole, which has plane-and-lobed equipotentials. The Exemplar deliberately specifies ``Q_tot≠ 0'' to rule out the dipolar case: with a non-zero monopole moment, the dipole contribution becomes a negligible correction at large r, not the leading term.

Concept linkage. The same logic explains why textbooks treat charged spheres ``as if'' all the charge sat at the centre (shell theorem for V in the exterior region). The shell-theorem result is the leading multipole expansion truncated after the monopole, equipotentials are exact spheres, not just approximate.

Spheres, option (a).

Structural observation. Two capacitors in series obey 1/C = 1/C₁ + 1/C₂. The composite slab is the single capacitor of the same total thickness that would have the same capacitance. So one writes 1/C_eq in two forms and solves for k.

1. C_i = ₀ k_i A / d_i for each block.
2. 1/C_eq = d₁/(₀ k₁ A) + d₂/(₀ k₂ A) = (k₂ d₁ + k₁ d₂)/(₀ k₁ k₂ A).
3. C_eq = ₀ k₁ k₂ A / (k₁ d₂ + k₂ d₁).
4. Equate to ₀ k A / (d₁ + d₂): k = k₁ k₂ (d₁ + d₂)k₁ d₂ + k₂ d₁.

Sanity check, limiting cases.

• If k₁ = k₂ = κ: k = κκ (d₁+d₂)/(κ d₂+κ d₁) = κ.
• If d₂ = 0 (only block 1): k = k₁ k₂ d₁/(k₂ d₁) = k₁.
• If k₂ → ∞ (block 2 a perfect conductor): k → k₁ (d₁+d₂)/d₂. The block-2 layer becomes irrelevant as a capacitive element, only d₂ of dielectric thickness contributes; the effective single slab is thicker than just block-1, scaled appropriately.

Alternative method, voltage-add picture. Field inside block i is E_i = σ/(₀ k_i) (the surface charge σ on the plate is shared by both blocks, since the same charge sits on the conductor). Voltage drop across block i: V_i = E_i d_i = σ d_i/(₀ k_i). Sum: V_total = σ(d₁/k₁ + d₂/k₂)/₀. Effective k for the same total thickness d₁+d₂ obeys V_total = σ(d₁+d₂)/(₀ k). Comparing gives the same formula. The ``voltage-add'' picture is often the cleaner board-exam derivation since it avoids reciprocals.

Exam tip. CBSE board frequently asks the series-dielectric derivation as a 3-mark question. Show clearly (i) same Q in both blocks, (ii) V₁ + V₂ = V, (iii) 1/C = 1/C₁ + 1/C₂, then (iv) solve. Skipping any of these steps loses a mark each.

Option (c).

Picture-first. Equipotentials are surfaces perpendicular to the field. Uniform E⃗ in z gives equipotentials as the planes z= const, i.e. the x-y planes.

1. On any plane z = z₀, every point has the same V.
2. Therefore V is constant along x (at fixed z), constant along y (at fixed z), and constant on the whole plane z = z₀.
3. V is not constant in all space because it varies with z, so (a) is wrong.

Alternative method, work-done check. Move a unit test charge from (x₁, y₁, z₀) to (x₂, y₂, z₀), both at the same z = z₀. The displacement is purely in the xy-plane, and E⃗ = E₀z is perpendicular to that displacement. So ∫E⃗· dl⃗ = 0, hence V(x₂,y₂,z₀) = V(x₁,y₁,z₀). A purely physical statement, no calculus needed, confirms (b), (c), (d).

Common pitfall, ``constant'' vs ``uniform''. A constant function and a uniform vector field are different things. ``V is constant in x for fixed z'' means V doesn't change as x varies. The field E⃗ is uniform (same vector everywhere) but V is not constant in z, V depends linearly on z. The two ideas often get confused on multi-correct MCQs.

Concept linkage. Parallel-plate capacitor: the field between the plates is uniform (E⃗ = E₀z), and the equipotentials are the planes z = const between the plates. Across a fixed plate, V is constant; from one plate to the other, V changes linearly with the perpendicular distance. The Exemplar question is asking about the geometry of these equipotentials in the abstract.

Options (b), (c), (d).

Strategic angle. Convert ``crowding of equipotentials'' to ``magnitude of E⃗'' via E = -dV/dn and the rest is just chasing where E⃗ is large.

1. Equipotential spacing Δ n = Δ V / E. (a) is literally this statement.
2. Sharp conductor edges concentrate charge, so σ large ⇒ E_surface = σ/₀ large ⇒ crowded equipotentials. (b) and (c) true.
3. E varies in space generally; equal spacing would require E uniform. (d) false.

Numerical illustration. For a point charge Q at origin, V = kQ/r. Equipotentials at V = 100, 200, 300, 400 V correspond to radii r = kQ/100, kQ/200, kQ/300, kQ/400. Adjacent gaps: Δ r_{1→ 2} = kQ100 - kQ200 = kQ200; Δ r_{3→ 4} = kQ300 - kQ400 = kQ1200. Spacing drops by a factor of 6 as we approach the charge, the equipotentials crowd dramatically near the source, exactly where E = kQ/r² is largest. This direct check verifies (a) explicitly.

Concept linkage, chapter-1 ``lightning-rod'' effect. In Chapter 1, the surface charge density on a conductor near a sharp edge is shown to scale as σ ∝ 1/R for radius of curvature R. Combined with E = σ/₀ just outside the conductor, this gives E ∝ 1/R, so the field at a sharp edge is huge, and by (a), the equipotentials there are crowded. Items (b) and (c) of the Exemplar question restate that chapter-1 fact in equipotential language.

Exam tip. On board-style multi-correct MCQs, ``always'' or ``necessarily'' is often the discriminator. Option (d) uses ``always'', which can only hold for a uniform field, a special case, not the general one. Train your eye to flag such absolute words when scanning options.

(a), (b), (c).

Strategic angle. Two ideas: the integral definition of work is universal (so (a) is wrong), and on equipotentials the integrand itself is zero (so the value is zero).

1. W/q = -_A^B E⃗· dl⃗ defines the work per unit charge against the field for any path. Option (a) is therefore false; option (b) is the correct defining formula.
2. For an equipotential path, E⃗⊥ dl⃗ at every point, so E⃗· dl⃗=0. The integral gives zero. Option (c) is correct, (d) is not.

Alternative method, potential-difference shortcut. Work done by the field on charge q from A to B is W_field = q(V_A - V_B). On an equipotential, V_A = V_B, so W_field = 0 and the work done by the external agent to move q is also zero. No integration required. Both option (b) (definition) and option (c) (value) are correct in any path.

Common pitfall, sign of ``work done''. Some students see the negative sign in W = -q∫E⃗· dl⃗ and assume (a) is correct because ``work done by the external agent'' is the negative of that quantity. But (a) says ``cannot be defined as'', that's a stronger claim than ``defined differently''. The defining relation works on an equipotential; both definitions just happen to give 0.

Concept linkage, orbiting an isolated point charge. Any circle centred on a point charge is an equipotential. Moving a test charge around a circle of radius r does zero work, even though the field E⃗≠ 0 everywhere on the path. The integral cancels point-by-point because E⃗ is radial and dl⃗ is tangential at every point on the circle.

Options (b), (c).

Strategic angle. Constant-V is a strong condition. It forces both E = 0 and ρ = 0 throughout the region. The rest is just reading each option against those two forced facts.

1. ∇ V = 0 ⇒ E⃗ = 0. Hence (b).
2. ∇·E⃗ = ρ/₀ = 0 ⇒ ρ = 0. Hence (c).
3. (a): zero field is not the same as ``uniform field'' in Exemplar usage. False.
4. (d): external charges induce surface charges on the boundary and leave the interior unchanged at the same constant V with E⃗ = 0 inside. False.

Alternative method, Gauss-law sanity. Take any closed Gaussian surface inside the constant-V region. Since E⃗ = 0 on that surface, ∮E⃗· dA⃗ = 0, so the enclosed charge is zero. Because this is true for any closed surface inside the region, including arbitrarily small ones, the volume charge density must vanish everywhere: ρ = 0. This confirms (c) without invoking the differential form of Gauss's law.

Common pitfall, (a) ``uniform'' field. A field that is identically zero is sometimes loosely called ``uniform with E = 0''. The Exemplar convention is stricter: ``uniform field'' means a non-zero constant vector field. With that reading, (a) is false because E⃗ = 0 is not a uniform non-zero field. If a student picks (a) thinking ``zero is a special uniform field'', they lose the mark.

Concept linkage, Faraday cage. The interior of a hollow conducting shell is a region of constant potential (the conductor itself). Per (b) and (c), the interior has zero field and no volume charge. Per (d), external charges (lightning, radio waves) cannot affect the inside. This is exactly the principle behind a microwave-oven door grille, an aircraft fuselage during lightning strike, and an electrostatic shielded chamber.

Options (b), (c).

Quick reading. Phase 1 charges C₁ to E with charge Q = C₁ E. Phase 2 isolates C₁ plus C₂ from the battery; they share that total charge Q with equal voltage.

1. V₁ = V₂ because the two are in parallel after K₂ closes ⇒ (a).
2. Charge conservation on the isolated C₁, C₂ system: Q₁' + Q₂' = Q = C₁ E ⇒ (d).
3. Option (c), C₁ V₁ + C₂ V₂ = C₁ E, mixes the pre- and post-switching voltage labels and is treated as incorrect in the official Exemplar key.
4. Q₁' = Q₂' would require C₁ = C₂, not given, so (b) is not forced.

Energy check. Before Phase 2, total energy stored is U_i = 12C₁E². After the redistribution, V = C₁E/(C₁+C₂), so U_f = 12(C₁+C₂)V² = 12(C₁+C₂)· C₁²E²(C₁+C₂)² = C₁²E²2(C₁+C₂). Ratio U_f/U_i = C₁/(C₁+C₂) < 1: energy decreases. The ``missing'' energy 12 C₁ C₂C₁+C₂ E² is dissipated as heat in the connecting wires (and as electromagnetic radiation during the transient). This is a classic illustration that joining two unequally charged capacitors is never a lossless process.

Common pitfall, order of switching matters. If the order were reversed (close K₂ first, then open K₁), the steady-state charges would be completely different: with both keys closed, C₁ and C₂ would be in parallel during charging, each holding E C_i. The Exemplar emphasises ``in this order'' to highlight that the redistribution problem is only triggered when the battery is disconnected first.

Exam tip, charge-conservation principle. The single phrase ``no external circuit path'' tells you to write ∑ Q<sub>i</sub> = const. Combined with V<sub>1</sub> = V<sub>2</sub> from parallel connection, this gives two equations for two unknowns. Always identify ``the isolated system'' before writing charge conservation, or the equation will be wrong.

Options (a), (d).

Strategic angle. The standard pair of conductor facts (E⃗_inside = 0, _bulk = 0) decides everything.

1. Conductor at V ≠ 0 with no external charges ⇒ the conductor itself must hold some charge. That establishes (a).
2. In equilibrium, no net volume charge sits in the bulk of the conductor. That establishes (b).
3. (c) overstates: ``only on the surface'' rules out the legitimate possibility of charge on the inner cavity surface of a hollow conductor; the official key marks it incorrect.
4. (d) directly contradicts (b) and is false.

Alternative method, Gauss-law sanity. Draw any closed Gaussian surface entirely inside the bulk of the conductor. The field is zero on it (electrostatic equilibrium), so the enclosed charge is zero. Since this is true for every closed surface inside the conductor, even one of infinitesimal volume, the volume charge density vanishes throughout the bulk. The total non-zero charge must therefore reside on the conductor's surface.

Common pitfall, ``inside the surface'' wording. Option (d) says ``inside the surface''. In Exemplar language, this means ``in the bulk volume''. The phrase is sometimes misread as ``on the inside face of the surface'' (e.g. inner surface of a hollow conductor). The correct reading is the bulk; the bulk is charge-free in a conductor at equilibrium.

Concept linkage, uniqueness theorem. Once the surface charge distribution is specified and there are no external charges, the potential everywhere is determined. So a conductor at a given V has a unique surface-charge pattern, no freedom for charge to hide in the bulk. The surface is the only ``degree of freedom'' that fixes V.

(a), (b).

Strategic angle. Decide what is held fixed (battery in ⇒ V; battery out ⇒ Q), then use Q = CV to read off how the other quantity reacts.

1. A: V = E fixed. C = ₀ A/d decreases as d grows. Q = CV drops with C. Option (c) is correct; option (a) is wrong.
2. B: Q fixed. C decreases. V = Q/C rises with falling C. Option (d) is correct; option (b) is wrong.

Energy comparison. It is worth tracking the energy in each case.

• A (battery in). U_A = 12CV² with V fixed and C decreasing ⇒ U_A decreases. The battery actually absorbs energy: as the plates move apart, the battery pulls some charge back through the circuit, doing positive work on the battery (recharging it), while the external agent does positive work pulling the plates apart. Energy bookkeeping: W_agent = Δ U + W_battery, with Δ U < 0 and W_battery > 0 enough to overcome.
• B (battery out). U_B = Q²/(2C) with Q fixed and C decreasing ⇒ U_B increases. All the external work pulling the plates apart goes into the field energy; no battery is around to absorb anything.

Common pitfall, ``what happens to σ''. Some students assume the surface charge density σ is fixed when the plates move. It isn't: σ = Q/A stays constant only if Q is constant and the area A is constant. In situation A, Q drops, so σ drops. In situation B, Q is fixed, A is fixed, so σ stays the same, and so does the field E = σ/₀ between the plates! In situation B, E doesn't change; only V = Ed rises because d increases.

Exam tip. CBSE board problems pair this question with ``in which case does the field between the plates change?''. Quick answer: only in A (because V is pinned and d grew, so E = V/d drops). In B, σ is pinned by isolation, so E = σ/₀ stays the same.

(c), (d).

Strategic angle. Equate V₁ = V₂ and read off charge ratios.

1. V₁ = V₂ ⇒ Q₁/R₁ = Q₂/R₂ ⇒ Q ∝ R.
2. σ = Q/(4π R²), so σ ∝ R/R² = 1/R.
3. Smaller R ⇒ larger σ. So _small > _large.

Alternative method, field at the surface. At the surface of an isolated conducting sphere, E = σ/₀. But also E = V/R for an isolated sphere of radius R at potential V. Equating: σ = ₀ V/R, so for equal V, σ ∝ 1/R. The smaller sphere has the larger surface field and hence the larger surface charge density.

Numerical example. Take V = 1 kV, R₁ = 10 cm, R₂ = 1 cm. Then ₁ = (8.854× 10^-12)(1000)/(0.1) ≈ 8.85× 10^-8 C/m², and ₂ = 10 ₁ ≈ 8.85× 10^-7 C/m². The smaller sphere has ten times the density.

Concept linkage, corona discharge. Air breaks down at E_air ≈ 3× 10⁶ V/m. For an isolated sphere at V, breakdown occurs at V ≈ 3× 10⁶ R volts. A 1 cm sphere can hold 30 kV before breakdown; a 1 mm sphere only 3 kV. That's why high-voltage transmission lines use thick conductors, to push the corona discharge threshold up.

_{smaller sphere} > _{larger sphere}.

Strategic angle. Sign of the charge flips the direction of motion relative to E⃗. That's the entire content of the question.

1. E⃗ points from high V to low V.
2. Electron force F⃗ = -eE⃗ is opposite to E⃗.
3. Electrons therefore move from low V to high V, i.e. into the higher-potential region.

Alternative method, energy argument. The PE of a charge q at potential V is U = qV. For an electron, q = -e < 0, so U = -eV. A free electron seeks lower PE (equilibrium tendency), which means lower U = -eV, which in turn means larger V. So electrons drift toward higher-V regions to minimise their PE.

Common pitfall, confusing electron flow with current. Conventional current is defined as the direction of positive charge flow, which is from high V to low V. The physical carriers in metals (electrons) drift the opposite way, from low V to high V. A battery's positive terminal is at higher V; electrons flow into the positive terminal externally and out of the negative terminal. Both descriptions describe the same current I.

Concept linkage, semiconductors. In a p-type semiconductor, the majority carriers are holes (effective positive charges). Holes drift from high V to low V, like positive charges. In n-type, electrons are the majority and drift toward high V. Identifying ``which direction the carriers go'' depends on the sign of the charge carrier, purely a chapter-2 fact applied to chapter-3 material.

Towards the higher-potential region.

Strategic angle. The relation V = Q/C separates ``charge'' from ``geometry''. Same Q, different C ⇒ different V.

1. Take a sphere of radius R₁ and another of radius R₂, each carrying Q. Their potentials at the surface are kQ/R₁ and kQ/R₂, which differ for R₁ ≠ R₂.
2. So yes, equal charges can sit on conductors at different potentials.

Numerical example. A 1 cm radius sphere and a 10 cm radius sphere each carrying 1 nC: V₁ = (9× 10⁹)(10^-9)/(0.01) = 900 V, V₂ = (9× 10⁹)(10^-9)/(0.1) = 90 V. A Δ V = 810 V exists between them even though they carry equal charge.

Common pitfall, confusing ``same charge'' with ``same potential''. Equal charge equal potential. Equal potential equal charge. The two are coupled through V = Q/C, and C is purely geometric. Two conductors with the same C at the same V do have equal Q, but that's a special case.

Concept linkage, what equalises in contact. If the two conductors are now connected by a wire, charge flows until their potentials are equal, not until their charges are equal. Equalising V usually means unequal final charges (Q₁/Q₂ = C₁/C₂ = R₁/R₂ for spheres). This is the basis of many ``two-sphere connection'' problems in this chapter (Q 2.27, for instance).

Yes, a potential difference can exist.

Strategic angle. The single fact ∇² V = 0 in charge-free space, together with its mean-value property, settles the question.

1. In free space, ρ = 0, so ∇² V = 0 (Laplace).
2. Harmonic functions satisfy the mean-value property and therefore cannot have strict interior maxima or minima (maximum principle).
3. Consequently V achieves its extreme values only on the boundary of the region (typically: on conductors at the edge of the region).

Alternative method, physical proof by contradiction. Suppose V had a local maximum at some interior point P in free space. Then V decreases in every direction away from P, so ∇ V points inward from any nearby point, and E⃗ = -∇ V points outward from P on every side. The flux of E⃗ through a small sphere around P is then positive, ∮E⃗· dA⃗ > 0. But by Gauss's law, this equals q_enclosed/₀. Free space means no enclosed charge, so the flux must be zero, a contradiction. Hence no interior maximum.

Concept linkage, Earnshaw's theorem and stable levitation. A positive test charge would need V to have a local minimum at the equilibrium point to be stably trapped (so that any displacement raises U = qV). Laplace's equation forbids this in free space, so static electrostatic levitation is impossible. Real-world ``electrostatic traps'' (like ion traps used in atomic physics and quantum computing) work only with time-varying fields, which sidestep Earnshaw's theorem by averaging over an oscillation cycle.

Common pitfall. Students sometimes argue V has a maximum at a positive point charge and a minimum at a negative point charge. True, but the point charge itself is not ``free space'', it is a source. The question asks about a region outside all sources, where ρ = 0. There, Laplace's equation holds, and no extrema are allowed.

No, V has no interior maximum or minimum in a charge-free region.

Strategic angle. The shape of the closed loop is a red herring. Conservative-field ⇒ zero work on any closed loop.

1. E⃗ of a point charge is conservative: ∮E⃗· dl⃗ = 0.
2. Therefore W₁ = q_C1E⃗· dl⃗ = 0.
3. Therefore W₂ = q_C2E⃗· dl⃗ = 0.
4. W₁ = W₂ = 0.

Alternative method, energy bookkeeping. The test charge q returns to its starting point in both loops. Since electrostatic PE depends only on position (U = qV), the PE at the start equals the PE at the end: Δ U = 0. By the work-energy theorem applied to the conservative force, W_elec = -Δ U = 0. The shape of the path is again irrelevant.

Common pitfall, ``area of the loop matters''. For magnetic loops (Ampere's law) and Faraday's law, the enclosed area matters. The Exemplar deliberately points this out because students who have studied magnetism instinctively expect ``equal area ⇒ equal something''. For electrostatic work, area is a red herring; only the conservative-field property ∮E⃗· dl⃗ = 0 counts.

Concept linkage, conservative force families. Gravity, electrostatic force, and the spring force are all conservative. For each, work around a closed loop is zero, and PE depends only on position. Friction is non-conservative; its loop integral is non-zero (negative, it always opposes motion). The Exemplar question is reinforcing the conservative-field signature.

Both are zero; equal.

Strategic angle. Laplace's equation plus the maximum principle pin V to its boundary value.

1. Inside the closed surface, ρ = 0, so ∇² V = 0 , i.e. V is harmonic.
2. Maximum-principle theorem for harmonic functions: maxima (and minima) of V on a closed bounded region occur on the boundary.
3. Boundary has V = V₀ (constant); hence both maximum and minimum of V on the boundary equal V₀.
4. Therefore V₀ ≤ V(r) ≤ V₀ for all interior points, giving V(r) = V₀.

Alternative method, uniqueness theorem. The interior satisfies ∇² V = 0 with boundary condition V = V₀ on the closed surface S. The uniqueness theorem for Laplace's equation guarantees exactly one solution for given Dirichlet boundary conditions. But V(r) ≡ V₀ is obviously a solution (trivially harmonic and matches the boundary). By uniqueness, it is the solution. Hence V = V₀ everywhere inside.

Concept linkage, Faraday cage rigorous form. A hollow conductor at potential V₀ has V = V₀ on its inner surface (the inner surface is just another part of the same equipotential conductor). The result just proved says the entire enclosed charge-free volume sits at V = V₀, with E⃗ = 0 inside. This is the precise mathematical statement of why a Faraday cage shields its interior from external fields.

Exam tip. CBSE doesn't usually ask the maximum-principle proof directly, but it tests the corollary: ``Prove that the electric field inside a hollow charged conductor is zero.'' Use this Q 2.19 result: interior is at the surface potential, so ∇ V = 0 inside, so E⃗ = 0 inside. Two lines, full marks.

V ≡ V₀ throughout the enclosed region.

Strategic angle. Battery-off means Q is the conserved quantity. Track each variable via Q and C.

1. C falls from KC₀ to C₀ when the dielectric leaves.
2. Q is fixed (no current path).
3. V = Q/C rises by factor K.
4. E = V/d rises by factor K.
5. U = Q²/(2C) rises by factor K.

Where does the extra energy come from? The dielectric is attracted into the capacitor's fringing field (a polarised slab gets pulled into a region of stronger field). Pulling it out against that attraction requires external work. With the battery off, no energy can flow back through the wires, so all the external work piles up as field-energy in the capacitor: W_external = U_f - U_i = (K - 1) U_i. This is consistent with conservation of energy and explicitly shows why U grows by factor K, not by some other factor.

Compare with battery-on case. If the battery had stayed connected during dielectric removal:

• V stays at V₀; C drops to C₀; Q drops by factor K;
• E = V₀/d stays the same; U = 12C V² drops by factor K.

Common pitfall, confusing U = 12 CV² with U = Q²/(2C). Both formulas are correct, but they make different things obvious. Use Q²/(2C) when Q is the conserved quantity (battery off); use 12 CV² when V is pinned (battery on). Picking the wrong one is the typical CBSE board-exam slip.

Q constant; C ↓ by K; V, E, U all ↑ by K.

Strategic angle. Use the monotone potential drop along field lines from the charged body to infinity.

1. Field lines flow from A (at V_A) outward, eventually to infinity (V = 0). Along each line, V decreases monotonically.
2. The uncharged conductor B is on the path: field lines hit its near side, emerge from its far side, continue to infinity. B is an equipotential.
3. Since B lies between A and infinity along the line, V_B lies between V_A and 0. Strict inequalities follow from induced charges existing on B.

Alternative method, superposition. The potential at B's position equals (potential due to A alone) plus (potential due to B's own induced charges, evaluated at B's surface). The first term is some value V_{A→ B} > 0 (since A is at V_A > 0 and B is nearer than infinity). The second term, the self-potential of B's induced charges, averages to zero on B because the total induced charge on B is zero (it's uncharged overall). So V_B is positive but less than V_A.

Common pitfall, assuming the uncharged body is at zero potential. Many students reason ``B has no charge, so it must be at V = 0''. That's wrong: the potential at a conductor depends on the field of all nearby charges, not just on the charge of the conductor itself. An uncharged conductor near a charged body is induced into a non-zero potential by the external field. ``Zero charge zero potential''.

Concept linkage, grounding. A conductor is at V = 0 only if it is connected to ground (a reservoir at the potential of infinity). An isolated uncharged conductor near a charged one floats at the induced intermediate potential, exactly what this question proves.

0 < V_B < V_A.

Strategic angle. Ring potential on axis is the familiar kQ/√R²+z²; multiply by (-q) for the PE.

1. V(z) = kQ/√R²+z² at axial distance z.
2. U(z) = -qV(z) = -kqQ/√R²+z².
3. U(0) = -kqQ/R (deepest); U→ 0^- as |z|→∞. So the centre is a PE minimum along the axis.
4. PE minimum ⇒ small displacement gives a restoring force; the charge oscillates back to centre.

Small-oscillation frequency. Expand U(z) around z=0 to second order: √R²+z² ≈ R(1 + z²/(2R²)), so U(z) ≈ -kqQR(1 - z²2R²) = -kqQR + kqQ2R³z². Comparing with the simple-harmonic form U = 12mω² z² + const: mω² = kqQR³ ω = √kqQm R³. The charge oscillates with this angular frequency for small axial displacements, a classic ``SHM near a stable equilibrium'' result.

Common pitfall, stability off the axis. The PE minimum is only along the axis. Off-axis, the charge -q is attracted toward the nearest part of the ring, which is unstable (transverse direction). The full 3D PE has a saddle at z = 0 (stable axially, unstable transversely). This is Earnshaw's theorem at work, there can be no fully-stable equilibrium of a point charge in any electrostatic field.

Concept linkage, Rutherford scattering analogue. An electron oscillating along the axis of a positively-charged ring is a toy model of an electron bound to a nucleus along one direction. The -kqQ/√R²+z² form softens the Coulomb singularity at small z to a finite value -kqQ/R, a useful regulator for some scattering problems.

U = -kqQ/√R²+z²; stable along axis.

Strategic angle. For axial points of a ring, every charge element is at the same distance, so the scalar potential integral collapses trivially.

1. All ring elements are at distance √R²+z² from the axial point (0,0,z).
2. V = k∫ dQ/r = k Q/√R²+z².

Cross-check with axial E⃗. Differentiate: E_z(z) = -dVdz = -ddz[kQ√R²+z²] = kQz(R²+z²)^3/2. This is the standard axial-E⃗ formula for a uniformly charged ring, which we know from Chapter 1. The consistency check confirms the potential calculation. At z = 0, E_z = 0 (symmetry); the field maxima are at z = ± R/√2.

Common pitfall, confusing ring with disc. For a disc, the potential at the centre is σ R/(2₀); for a ring, the potential at the centre is kQ/R = Q/(4π₀ R). These look superficially similar but differ in factors. Always keep track of whether the source is a 1D loop (ring) or a 2D sheet (disc).

Concept linkage. Off-axis points have a more complex expression involving elliptic integrals, not Class 12 material. The reason the axial point is so clean is the exact symmetry: every charge element is equidistant. Any departure from the axis breaks this and forces a non-trivial integral.

V(z) = kQ√R²+z².

Strategic angle. Cylindrical symmetry ⇒ V depends only on r. Build V(r) from E(r) by integration.

1. Gauss's law (cylindrical surface, radius r, length L): E· 2π r L = λ L/₀, so E = λ/(2π₀ r).
2. Integrate from r₀ (reference) to r: V(r) = -_r0^r E dr' = -(λ/2π₀)ln(r/r₀).
3. Equipotential V = const ⇒ r = const ⇒ coaxial cylinder.

Why reference can't be at infinity. For a line charge, the potential integral V(r) = -_∞^r λ/(2π₀ r') dr' diverges logarithmically because ∫ dr'/r' blows up at large r'. The physical origin: an infinite line has infinite total charge, so the standard ``V → 0 at infinity'' convention cannot be enforced. We must pick a finite reference radius, here, the surface r = r₀. (Compare with a point charge, where V at infinity is naturally zero because the source is bounded.)

Alternative method, cylindrical capacitor. A coaxial cable is two cylinders of radii r₀ and R with line charge densities +λ and -λ. The potential difference between them is Δ V = λ2π₀lnRr₀. This is the standard formula for capacitance per unit length of a coaxial cable: C/L = 2π₀/ln(R/r₀). The Exemplar question is the building block for that capacitor formula, and for the structure of coaxial transmission lines used in cable TV and CRO probes.

Exam tip. CBSE problems on cylinders almost always specify ``infinite''. That single word legitimises using Gauss's law with the symmetric cylindrical surface and ignoring end-cap flux. Real cables are finite, but the infinite-line approximation is excellent for the region far from the ends.

Coaxial cylinders, V(r) = (λ/2π₀)ln(r₀/r).

Structural observation. Equal-magnitude opposite charges ⇒ the equation V = 0 collapses to ``equidistant from both charges''.

1. V(P) = k(q/r₊ - q/r_-) = 0 requires r₊ = r_-.
2. Locus of points equidistant from two given points is the perpendicular bisector plane of the segment joining them.
3. Segment endpoints (± d/2, 0, 0) ⇒ bisector plane is x = 0.

Field on the zero-potential plane. Although V = 0 on the perpendicular bisector plane, E⃗ is not zero there. By symmetry, E_x and E_z both vanish on the plane, but E_y points from +q toward -q (i.e. in the +x direction) with magnitude E_y = -2kp/(p² + y² + z²)^3/2· (d/2) at distance r = √y²+z² from the dipole axis. The plane is equipotential but not field-free.

Common pitfall. Some students conclude ``V = 0 ⇒ no force on a test charge''. Wrong. V = 0 means ``no work to bring a charge from infinity to this point'', not ``no force on a charge at this point''. The force is governed by E⃗ = -∇ V, which is the rate of change of V, not its value.

Concept linkage, dipole far-field. On the dipole's perpendicular bisector plane (the ``equatorial'' plane in spherical language), the standard dipole formula gives V = kpcosθ/r² with θ = π/2, so V = 0. This is true to all orders in the dipole expansion, not just an approximation. The Exemplar derivation confirms it from first principles.

x = 0 (the yz-plane).

Strategic angle. Charge conservation with a non-linear capacitor gives a quadratic in the final voltage.

1. Pre-connection: Q₀ = C₀ U₀.
2. Post-connection (parallel): vacuum cap holds C₀ U, dielectric cap holds α U · C₀ · U = α C₀ U².
3. C₀ U₀ = C₀ U + α C₀ U² ⇒ α U² + U - U₀ = 0.
4. Discriminant √1 + 4α U₀ = √1+4· 2· 78 = √625 = 25, and U = (-1+25)/(2α) = 24/4 = 6 V.

Why discard the negative root. The quadratic α U² + U - U₀ = 0 has roots U = (-1± 25)/4, giving U = 6 or U = -6.5 V. The negative root is physically unacceptable because the capacitor cannot have U < 0 once charged to a positive U₀, connecting two capacitors in parallel cannot reverse the polarity of the source capacitor. So U = +6 V is the only physical answer.

Final charges check. With U = 6 V at equilibrium: Q₁ = C₀ U = 6 C₀, Q₂ = α C₀ U² = 2 · C₀ · 36 = 72 C₀. Total: 78 C₀. Compare with initial Q₀ = C₀ U₀ = 78 C₀. Charge conservation holds. The non-linear capacitor takes the lion's share of the charge (≈ 92%) because its effective capacitance grows with U.

Concept linkage, non-linear ``ferroelectric'' capacitors. Real ferroelectric materials show ε that depends on the applied field, a strongly non-linear capacitor. This Exemplar problem is a toy version. Solving non-linear charge-balance equations (here, quadratic) is the standard technique, and the discriminant must always be checked to lie within physical bounds.

U = 6 V.

Strategic angle. Compute the electrostatic pull per unit area on the disc; balance it against gravity.

1. E = V/d between the plates; σ = ₀ V/d on either plate (and on the disc).
2. Force per unit area pulling the disc up: P = σ²/(2₀) = ₀ V²/(2d²).
3. Total upward force: F = P· π r² = π r² ₀ V²/(2d²).
4. Set F = mg and solve: V_min = d√2mg/(π ₀ r²).

Unit check. Inside the radical: [mg]/[₀ r²] = (kg/s²)/(C²/(N· m²)· m²) = N/(C²/N) = N²/C² = (V/m)². Multiply by d² (in m²): (V/m)² · m² = V². Square root: V. The formula is dimensionally correct.

Numerical illustration. For m = 1 mg = 10^-6 kg, r = 1 cm = 0.01 m, d = 1 mm = 10^-3 m: V_min² = 2(10^-6)(9.8)(10^-3)²π (8.854× 10^-12)(0.01)² ≈ 1.96× 10^-112.78× 10^-15 ≈ 7.05× 10³, so V_min ≈ 84 V. A small disc inside a millimetre gap lifts at a few tens of volts, within the range of a laboratory high-voltage supply.

Common pitfall, the factor of 1/2. A frequent error is to use F = QE = σ A · σ/₀ = σ² A/₀ instead of the correct σ² A/(2₀). The factor 1/2 comes from the fact that, at the surface, half the field is due to the conductor's own surface charge (which cannot exert a force on itself), so the effective field felt by the surface charges is half the total. Forgetting this 1/2 predicts a lifting voltage smaller by 2, wrong by ∼ 30%.

V_min = d√2mgπ₀ r².

Strategic angle. The whole problem reduces to summing q_i q_j over the three pairs and dividing by the common side length.

1. Neutron: pairs give (-2 - 2 + 1)/9 = -1/3 in units of e². So U_n = -e²/(3· 4π₀ L) ≈ -0.48 MeV. Ratio to m_n c²: ≈ 5× 10^-4.
2. Proton: pairs give (4 - 2 - 2)/9 = 0. So U_p = 0 identically. Ratio to m_p c²: 0.

Numerical cross-check (neutron). Using SI units directly: e²4π₀ L = (1.6× 10^-19)²4π(8.854× 10^-12)(10^-15) = 2.56× 10^-381.113× 10^-25 ≈ 2.30× 10^-13 J. Convert to MeV (divide by 1.6× 10^-13 J/MeV): ≈ 1.44 MeV. So U_n = -1.44/3 ≈ -0.48 MeV. |U_n|/m_n c² = 0.48/939 ≈ 5.1× 10^-4.

Why the proton cancellation is exact. The proton's three charges are +23, +23, -13 (in units of e). Sum of pair products: 49 - 29 - 29 = 0. This is algebraic and depends only on the ratios of the charges, not on the geometry, so the result U_p = 0 would hold even if the triangle were not equilateral, as long as all three pair distances were equal. But change the side ratios (non-equilateral), and generally U_p ≠ 0.

Concept linkage, strong force scale. The ∼ 0.5 MeV electrostatic contribution is dwarfed by the ∼ 939 MeV mass of the nucleon. The bulk of the nucleon mass comes from QCD (quark confinement and gluon-field energy), not electrostatics. The Exemplar question is reinforcing that the electrostatic Coulomb interaction is a contribution to nuclear physics but not the dominant one, by three orders of magnitude.

U_n ≈ -0.48 MeV; U_p = 0.

Strategic angle. Equal potential + charge conservation. Two unknowns, two equations.

1. Initial Q₁ = 4π R²σ; Q₂ = 16π R²σ. Total = 20π R²σ.
2. Equal V: Q₁'/R = Q₂'/(2R) ⇒ Q₂' = 2Q₁'.
3. Sum: 3Q₁' = 20π R²σ ⇒ Q₁' = 20π R²σ/3, Q₂' = 40π R²σ/3.
4. Divide by areas 4π R² and 16π R² respectively: ₁' = 5σ/3, ₂' = 5σ/6.

Ratio check. ₁'/₂' = (5/3)/(5/6) = 2 = R₂/R₁. This matches the general result σ ∝ 1/R at fixed potential. The smaller sphere indeed has twice the surface charge density of the larger, consistent with Q 2.13.

Energy released. Initial energy U_i = Q₁²/(8π₀ R) + Q₂²/(8π₀(2R)). Final energy uses redistributed charges Q₁' = 20π R²σ/3, Q₂' = 40π R²σ/3. Computing the ratio, the final state has lower total energy than the initial. The difference is dissipated as heat during the brief contact transient, yet another instance of unequal-potential conductors losing energy on connection.

Concept linkage, Van de Graaff generator. A small charged sphere brought into contact (via an internal brush) with the inside of a larger spherical shell transfers all its charge to the outer shell, because at the inner contact, the inner sphere is at the same potential as the outer shell's interior (which equals the shell's surface potential), and once detached, the small sphere has zero residual potential (no internal charges left). Repeating builds up huge potentials on the outer shell. The Exemplar contact problem is the first step in understanding that machine.

_R' = 5σ/3; _2R' = 5σ/6.

Strategic angle. Phase 1: C₁, C₂ in series, same charge C_serE. Phase 2: C₁ frozen, C₂ and C₃ in parallel sharing C₂'s previous charge.

1. Phase 1: C_ser = (6C· 3C)/(6C+3C) = 2C. Q = 2C · 9 = 18C = 18 μC. Both C₁, C₂ hold 18 μC; C₃ holds 0.
2. Phase 2: K₁ open isolates C₁ at 18 μC. K₂ closed puts C₂ and C₃ in parallel, total charge 18 μC, equal V': V' = 18/(C₂ + C₃) = 18/6 = 3 V. Q₂' = Q₃' = 3· 3 = 9 μC.

Voltage check (Phase 2). After redistribution, V_C2 = Q₂'/C₂ = 9/3 = 3 V and V_C3 = Q₃'/C₃ = 9/3 = 3 V. Same voltage, confirming the parallel condition. Meanwhile C₁ keeps its previous voltage V_C1 = Q₁'/C₁ = 18/6 = 3 V too , a coincidence here because C₁ = 2(C₂ + C₃), but in general V_C1 during Phase 2 is just frozen.

Energy dissipated. Initial energy: U_i = 12(Q₂)²/C₂ + 12(Q₁)²/C₁ = 12(18²)(1/3 + 1/6) = 12· 324· 0.5 = 81 μJ. Final energy: same C₁ contribution, plus 12(9)²/3 + 12(9)²/3 = 27 μJ. The C₁ part is unchanged at 12(18)²/6 = 27 μJ; the C₂-C₃ pair drops from 12(18)²/3 = 54 μJ to 27 μJ. So 27 μJ is dissipated in the wires during the Phase 2 transient.

Common pitfall, phase-2 charge on C₁. Many students also apply ``charge conservation'' to C₁ during Phase 2 and end up with all three capacitors redistributing. Wrong, K₁ is open, so C₁ is electrically disconnected from C₂, C₃. Always identify which plates are actually connected by a wire before writing redistribution equations.

Q₁ = Q₂ = 18 μC, Q₃ = 0;   Q₁' = 18 μC, Q₂' = Q₃' = 9 μC.