NCERT Exemplar Class 12 Physics Chapter 1 Solutions PDF

Symmetry-first reading. The original configuration is mirror-symmetric about the x-axis: q₂ at +y, q₃ at -y. Symmetry forces the net force on q₁ to lie along the x-axis; the problem says this is +x.

1. Add +Q at (x,0). This point is on the x-axis, so the new force on q₁ from Q is along the x-axis (no y component to disturb the cancellation).
2. Direction of the new force: +Q on q₁, both placed on the x-axis, gives a Coulomb force along the x-axis. The Exemplar geometry (with Q between origin and the existing resultant +x direction) makes this contribution add to, not subtract from, the existing +x force.
3. Magnitude grows; direction unchanged.

Alternative method, explicit vector sum. Write the positions: q₁ at (a,0) on the +x-axis, q₂ at (0,b), q₃ at (0,-b). The Coulomb force on q₁ from q₂: F⃗₁₂ = kq₁ q₂r² r⃗₁₂|r⃗₁₂|, r⃗₁₂ = (a-0, 0-b) = (a, -b). Similarly for q₃. The sum F⃗₁₂+F⃗₁₃ has only an x-component (the y-components cancel by symmetry), giving the existing +x resultant. Adding F⃗_{Q→ q1}, also along x̂, simply scales the magnitude.

Vector-superposition principle. Coulomb's superposition says that the force from many charges on a single test charge is the vector sum of individual Coulomb forces. There's no "shielding" or "interference", each pair acts independently. So adding +Q to the system just appends one more term to the sum without modifying the others.

Why this matters. Recognising that adding a collinear charge along an existing-force direction strictly augments magnitude is a recurring MCQ pattern, and the underlying superposition principle underpins all of electrostatics, from charged spheres to molecular forces.

Option (a): force increases, direction stays +x.

Two-line reading. Any field-line picture near a conductor must satisfy three rules: (i) lines are perpendicular to the conductor surface, (ii) no lines exist inside the bulk, (iii) lines start on + charge and terminate on - charge.

1. Scan each option for those rules. Only Fig. (i) has every line meeting the sphere at 90^∘ and no lines inside.
2. Fig. (ii)/(iii)/(iv) show oblique or tangent lines, or lines passing through the body of the sphere, all unphysical.
3. Charge bookkeeping: lines from the external +q terminate on the induced - density on the near face; new lines emerge from the induced + density on the far face and run to infinity. The induced charges sum to zero (sphere stays neutral overall).

Alternative reading, boundary-condition method. Inside the metal, E⃗=0, hence the tangential component of E⃗ just outside must also vanish (continuity of E_t across a surface that has no surface current). Only Fig. (i) respects this constraint at every point. The normal component E_n = _ind/₀ varies around the sphere, which is why the line density is higher on the near hemisphere.

Why this matters. The "perpendicular to conductor" rule is the fingerprint to spot the right field-line sketch in any conductor-MCQ, and it underpins how a Faraday cage screens the interior of a conductor from external fields.

Option (a).

Gauss-law shortcut. When the same charge is enclosed, the flux is identical. No calculation needed.

1. Confirm each panel encloses +q. Yes, (i)–(iv) all enclose a single +q inside S.
2. Apply Gauss: Φ = q_enc/₀ is constant.
3. Options (a),(b),(c) all assert shape-dependent flux, which contradicts Gauss's law.

Direct integration as a cross-check. For panel (i) where the surface is a sphere of radius r centred on +q: E = q/(4π₀ r²), A = 4π r², so Φ = E· A = q/₀, independent of r. For an irregular surface (panel (iv)), integration is harder, but Gauss's law guarantees the same answer without doing the integral. This is the power of a global theorem.

Topology vs geometry. Flux is a topological invariant: stretching, squishing, or wrinkling the surface (without crossing the charge) doesn't change Φ. Only one property of the surface matters: how many times it "winds around" the charge. For a simple closed surface enclosing +q once, Φ = q/₀ always.

Common misreading of the options. Students often pick (c) because it sounds plausible: "same flux through (ii) and (iii)". But (c) also says "smaller than (iv)", which mixes a true statement (same flux for (ii), (iii), (iv)) with a false one (one of them is "smaller"). Eliminating (a)/(b)/(c) leaves (d).

Why this matters. Flux is a topological quantity: it counts enclosed charge, not field strength on the surface. This is the foundation of all electrostatic shielding arguments and the basis of how Maxwell's equations decouple "source-counting" from "geometry-counting".

Option (d): Φ = q/₀ in every case.

Two-bin reading. Sort each side of Gauss's law into "what it sees".

1. LHS sees the field at S, generated by every charge in space.
2. RHS sees the charge inside S only.
3. In Fig. 1.4: q₂,q₄ inside; q₁,q₃,q₅ outside.
4. LHS has contributions from all five; RHS from q₂,q₄.

Why outside charges contribute to E⃗ but not to flux. Outside charges produce field lines that thread through S, they enter and leave the surface. Each line that enters once must leave once (continuity), so each line contributes zero net flux. But while threading through S, the line passes through each point on S with a definite, non-zero E⃗. So the integrand E⃗· dS⃗ at each point includes outside-charge contributions; only the sum cancels.

Concrete check. Place a closed surface in empty space and move a charge q_out far away. The flux through S is zero (Gauss says so), but the field at every point on S is non-zero (Coulomb's law from q_out). Both statements are simultaneously true.

Why the equation is still useful. Gauss's law is so powerful precisely because E⃗ in the LHS integrand is the actual field (including all sources), yet the RHS only involves enclosed charge. In high-symmetry problems (sphere, cylinder, sheet), this lets us compute E⃗ from q_enc directly, the outside-charge complication is present in the integrand but absorbed by the global integral.

Why this matters. Misreading Gauss's law as "E⃗ on S comes only from enclosed charges" is the single most common error on flux problems. The LHS-RHS asymmetry is a recurring source of trick questions in JEE/NEET.

Option (b).

Field-density rule. A dipole in a non-uniform field migrates toward denser (stronger) field lines when oriented with the heavier-pulled end on the strong-field side.

1. Identify field gradient: lines crowd on the left (strong), spread on the right (weak).
2. Position of ± q: -q on the strong-field side (left), +q on the weak-field side (right).
3. |F_-| = qE_left > qE_right = |F₊|. The bigger force is on -q and points opposite to the local field at the left end (which is rightward), so F⃗_- is leftward.
4. Net direction: leftward.

Vector-calculus derivation. For a dipole p⃗=qd⃗ in a non-uniform field, F⃗ = (p⃗·∇)E⃗. Take p⃗=p x̂ (pointing -q→+q). Then F_x = p ∂ E_x/∂ x. From Fig. 1.5, E_x is large at left and small at right, so ∂ E_x/∂ x < 0, giving F_x < 0, leftward, consistent with the line-density argument.

Quick numerical check. If E_left=2E₀ and E_right=E₀ at the two charge positions with separation d, then F_net = q(2E₀)-q(E₀) = qE₀ leftward, a finite, non-zero pull confirming the direction.

Why this matters. This pattern (dipole pulled toward stronger field) underlies dielectric attraction (why an uncharged piece of paper jumps to a charged comb), the dielectrophoretic trapping of biomolecules, and the alignment force on polar molecules inside a capacitor's fringing field.

Option (c).

Boundary-condition reading. Two universal rules at a conductor surface in electrostatics: (i) E⃗ just outside is normal to the surface; (ii) it points outward from positive surface charge.

1. Isolated plate with +q on one side: near face gets -q, far face gets +q. Field at P (far side) is set by the +q on the far face.
2. By rule (i), E⃗ at P is perpendicular to the plane. By rule (ii), it points away from the far face.
3. This rules out (b), (c), (d).

Alternative method, image charges. Place a virtual image charge -q at the mirror position across the plane. On the same side as the real +q, the field is the Coulomb superposition of the real +q and the image -q. On the far side, however, the field inside and beyond the conductor is set by induced surface charges only. The conductor shields any point on the far side from a direct +q-style field; what remains is the contribution from the +σ on the far face, which behaves locally like an infinite sheet (uniform, normal, outward). This confirms (a).

Why the field is not radial. A common trap is to draw "radial E⃗ from +q" reaching the far side. That picture ignores the conductor entirely. The metal bulk has E⃗=0 inside; the only physical source of field at P is the surface charge on the far face.

Why this matters. The "isolated conducting plane" is a JEE mainstay: recognise that the back face is a uniformly charged sheet sourcing the only field on the far side. This is also the working principle of an electrostatic shield.

Option (a).

Mirror-symmetry shortcut. Pair charge patches across the mirror plane through P perpendicular to the diameter; tangential components cancel.

1. Place P on the diameter.
2. Mirror plane: perpendicular to the diameter through P. Charges pair up.
3. Each pair: equal and opposite components along the diameter (⇒ cancel), equal components perpendicular (⇒ add).

Formal symmetry argument. Let the hemisphere be the upper half of a sphere of radius a, oriented so its diameter lies along the x-axis. The hemisphere has reflection symmetry about the yz-plane (the mirror plane through the centre). Any field point P on the diameter at (x₀,0,0) has its mirror image (-x₀,0,0), but after reflecting the hemisphere, the charge distribution is identical, and the field at (-x₀,0,0) must equal the original field at (x₀,0,0) with x-component sign-flipped. For this to be self-consistent, the x-component of E⃗ at (x₀,0,0) must satisfy E_x = -E_x, i.e. E_x = 0. The field must be perpendicular to the diameter.

Equivalent integration approach. Parametrise the hemisphere in polar coordinates (θ from the axis, φ around). A charge element at (asinφ, asinφ, acosθ) contributes a Coulomb field at P with a component along the diameter that, when integrated over φ∈[0,2π], gives zero by the orthogonality of cosφ and 1 over a full period. So ∫ E_x dq = 0 explicitly, but the symmetry argument bypasses the messy integral.

Direction perpendicular to diameter, which way? The perpendicular component points away from the hemisphere (in the direction the field would push a + test charge). For an "upper" hemisphere with P on the diameter, the field at P points downward (toward the open side of the hemisphere).

Why this matters. Symmetry arguments save heavy integration on Exemplar MCQs about half-rings, hemispheres, half-discs. Always check for reflection or rotational symmetry before reaching for the integral sign.

Option (a): perpendicular to the diameter.

Strategic angle. Translate "zero flux" into geometric and algebraic statements separately, then test each option against both.

1. Geometric: lines in = lines out ⇒ (c) true.
2. Algebraic: q_enc = 0 ⇒ (d) consistent (charges outside or balanced inside; option (d) names one valid scenario).
3. Counter-example for (a): a uniform external field passing through a closed surface gives Φ = 0 but E ≠ 0.
4. Counter-example for (b): an external non-uniform field with no enclosed charge gives non-uniform E⃗ inside.

Three explicit configurations that give Φ=0. (1) Empty surface in vacuum, E⃗=0 everywhere: trivially zero flux. (2) Empty surface in a uniform external field: E⃗ is constant and non-zero, but flux entering one side cancels flux leaving the other. (3) Surface enclosing +q and -q together: net enclosed charge is zero, so total flux vanishes, yet the local field on S is highly non-uniform. All three give Φ=0 but only the first has E⃗=0 on S, destroying option (a).

Why this matters. Many students mistakenly read Φ = 0 as "E = 0"; this question is designed to break that habit. The distinction between a global integral and a pointwise statement recurs in every Maxwell equation.

(c), (d).

Strategic angle. The continuity of E⃗ is set by Coulomb's 1/r² behaviour: smooth except at the source.

1. Coulomb field E ∝ 1/r² diverges at r = 0 (the charge location). So E⃗ is discontinuous there ⇒ (d) true.
2. For any point with no local charge, the field is a sum of smooth Coulomb fields from distant sources, hence continuous ⇒ (b) true.
3. Charge sign doesn't change continuity behaviour ⇒ (c) false.
4. (a) is false because of the divergence at point-charge locations.

Two types of discontinuity in electrostatics. (1) Singularity at a point charge: |E⃗|→∞ as r→ 0; the field has no finite limit. (2) Jump across a surface-charge sheet: E_⊥ above - E_⊥ below =σ/₀. The tangential component, however, remains continuous. Both kinds count as "discontinuity at a charge" in this MCQ, only signs of charge are irrelevant.

Why this matters. The discontinuity at sources is what makes Gauss's law useful, flux gets contributions only from enclosed sources. The surface-jump rule, Δ E_⊥ = σ/₀, is also the working tool for all boundary-value problems with charged sheets and conductors.

(b), (d).

Strategic angle. Gauss's law is sign-blind: it only counts algebraic charge enclosed. With one sign, the "algebraic" sum is just the arithmetic sum.

1. No charge inside ⇒ Φ = 0 regardless of external configuration. (b) true.
2. Charge q inside ⇒ Φ = q/₀. (d) true.
3. (a) false: the surface could be drawn to enclose no charge.
4. (c) false: ∮E⃗· dS⃗ is always defined.

One-sign field-line geometry. If the universe contained only positive charges, every field line would start on a +q and run to infinity (no - charges to terminate on). A closed surface that encloses none of these + sources is pierced by lines that enter once and leave once ⇒ net flux is zero. Enclose one +q: now the field lines from that q exit S without returning, giving net outward flux q/₀.

Why this matters. The "one sign of charge" thought experiment is sometimes used to motivate the cosmological hypothesis of a slight e_p/e_e imbalance driving cosmic expansion (cf. Q1.26). Gauss's law's sign-blindness is what makes it powerful even when the absolute signs of the unbalanced sources are uncertain.

(b), (d).

Strategic angle. Multipole expansion + conservative-field property.

1. Q = 0 kills monopole. Leading non-zero term is dipole (1/r³) in general.
2. (c) true: at large r, dominant scaling is 1/r³.
3. (d) true: electric field is conservative, line integral over closed path is zero.
4. (a) false: only 1/r² vanishes; higher multipoles survive.
5. (b) false: higher-order moments (quadrupole, octupole) also contribute beyond the dipole.

Explicit dipole field for orientation. Choose p⃗=pẑ. The far-field is E⃗_dip(r,θ) = p4π₀ r³ (2cosθ r̂ + sinθ θ̂), which scales as 1/r³, confirming (c). Note the angular structure: the field is strongest along the dipole axis (θ = 0,π) and weakest in the equatorial plane.

Conservative-field check for (d). For any static charge distribution, E⃗ = -∇ V, so ∮ E⃗· dl⃗ = -∮ ∇ V· dl⃗ = 0 by the fundamental theorem of calculus on closed loops. Work done on a charged particle along a closed loop is W = q∮E⃗· dl⃗ = 0, regardless of the charge distribution's internal complexity.

Why this matters. The multipole expansion is the foundation of antenna theory, molecular electrostatics (computing forces between water molecules), and stellar potential calculations. The 1/r³ dipole tail is also why van der Waals forces decay so quickly with separation.

(c), (d).

Strategic angle. Decompose flux question and field question separately.

1. Flux is global: only enclosed charges count. q_enc = Q + (-2Q) = -Q, so Φ = -Q/₀. (a) holds.
2. Outside charges contribute zero flux ⇒ (c) holds.
3. Field on surface is pointwise: -2Q off-centre makes the field non-uniform on the Gaussian sphere ⇒ (b) and (d) fail.

Why (b) is a trap. The formula E = Q/(4π₀ R²) for the field on a Gaussian sphere is valid only when the enclosed charge sits at the centre and there are no other charges. Here -2Q is off-centre at r=R/2, so the spherically symmetric form breaks. Use Gauss's law to compute total flux, but never to claim a uniform-magnitude field unless the symmetry is genuinely spherical about the surface's centre.

Order-of-magnitude check on (a). The total flux is |Φ|=Q/₀, the same as if a single -Q sat anywhere inside. The position of -2Q, the size of R, and the values of 5Q outside all drop out. This invariance is the topological content of Gauss's law.

Why this matters. Distinguishing global flux from pointwise field is the hardest skill in this chapter and is repeatedly tested in JEE/NEET via "off-centre charge in a sphere" geometries.

(a), (c).

Strategic angle. Examine stability separately along the axis and in the plane.

1. At centre, E⃗ = 0 by symmetry.
2. In-plane stability for q > 0: the Exemplar reads this as a pseudo-restoring scenario in which the closer arc of the ring pushes the test charge toward the far side of the centre. (a) marked correct.
3. For q < 0 in plane: the attractive force toward the closer arc pulls the charge to the ring; (b) correct.
4. Axial SHM for q < 0: the axial field of a ring on a -q test charge is repulsive at small displacements (not SHM); (c) false.
5. Plane equilibrium for q > 0 is unstable in the standard analysis; (d) correct.

Axial-field formula for the ring. For a ring of charge +Q and radius R, the on-axis field at distance z is E_axis(z) = 14π₀ Qz(R²+z²)^3/2, pointing away from the centre. A test charge +q on the axis feels force F = qE along +z, pushing it further away, not restoring. So no axial SHM for like-sign test charge either; (c) is correctly false. For unlike-sign (-q on the axis), the same field attracts the charge back toward the centre, that case does give SHM (cf. Q1.31).

Earnshaw consistency check. Earnshaw's theorem forbids stable static equilibrium in pure electrostatics: there is always some direction along which the equilibrium is unstable. For the ring centre: stable axially for -q but unstable in-plane for -q; stable in-plane (pseudo) for +q but unstable axially for +q. Either way, full stability is impossible.

Why this matters. Ring-and-axial-charge stability problems are a JEE staple. The deeper lesson, Earnshaw's theorem, is why ion traps (Paul, Penning) require time-varying fields, not static ones.

(a), (b), (d).

Quick reading. A dipole encloses zero net charge, hence zero flux through any surface enclosing it.

1. q_enc = +q + (-q) = 0.
2. Gauss: Φ = 0/₀ = 0.

Geometric picture. Every field line that starts on +q either terminates on -q inside the same surface, or escapes to infinity. By the dipole's bound geometry (the two charges are close together compared to "infinity"), every escaping line that exits S must eventually loop back through S to reach -q. So "line crossings out" equal "line crossings in", and net flux is zero regardless of how the surface is shaped.

Caveat. The surface must enclose both charges. If the surface threads between the two charges (enclosing only +q or only -q), the flux is ± q/₀.

Why this matters. Total flux is sign-additive, equal + and - charges cancel even when the local field is highly non-trivial. The same logic explains why a perfectly neutral spherical body produces no monopole field outside, even though internally it may have complex charge structure.

Φ = 0.

Quick reading. Inside conductor E⃗ = 0 forces the total enclosed by any Gaussian surface inside the bulk to be zero, fixing the induced inner-surface charge to -Q. Charge conservation then puts +Q on the outer surface.

1. Inner surface: -Q over area 4π R₁² ⇒ _in = -Q/(4π R₁²).
2. Outer surface: +Q over area 4π R₂² ⇒ _out = +Q/(4π R₂²).

Why the inner and outer densities differ in magnitude. Both surfaces carry |Q| total, but they have different areas. Since R₁ < R₂, the inner-surface area 4π R₁² is smaller, so the inner density is larger in magnitude: |_in||_out| = R₂²R₁² > 1. This is the static-shell version of the "pointiness ⇒ higher σ" rule.

Generalisation: shell with net charge Q₀. If the shell already carries an extra net charge Q₀ (instead of starting neutral), the inner surface still locks at -Q (set by the cavity charge), but the outer surface adjusts to Q + Q₀ (so the total on the shell is -Q + (Q+Q₀) = Q₀).

Position of the cavity charge. If the inner charge is not at the centre, the inner-surface density becomes non-uniform, but the total induced charge is still -Q. The outer-surface density, however, remains uniformly +Q/(4π R₂²) because the conductor screens the asymmetry of the cavity charge.

Why this matters. The "induced inner -Q + outer +Q" pattern is the template for every spherical-cavity electrostatics problem and underpins capacitor design.

_in = -Q/(4π R₁²), _out = +Q/(4π R₂²).

Strategic angle. Distinguish microscopic and macroscopic fields, they are two different averaging scales.

1. Microscopic field at Å scale: ∼ 10¹¹ V/m between proton and electron. Real.
2. Macroscopic field = spatial average over many atoms. Internal atomic fields cancel because atoms are neutral.
3. Free electrons in a conductor screen external macroscopic fields by redistributing until E= 0 inside.

Order-of-magnitude reasoning. Atomic dimension ∼ 10^-10 m; charge e = 1.6× 10^-19 C; Coulomb's k ≈ 9× 10⁹ N m2/C2. So E_atom ∼ ker² = (9× 10⁹)(1.6× 10^-19)(10^-10)² ≈ 1.4× 10¹¹ V/m. For comparison, a 1 V battery across a 1 mm gap gives only 10³ V/m , eight orders of magnitude smaller. Atomic-scale fields are simply on a different planet.

Screening time-scale. In a metal, free-electron density is ∼ 10²⁸/m³ and the Drude relaxation time is ∼ 10^-14 s. So when an external field is applied, the electrons re-equilibrate to cancel the macroscopic interior field in roughly a femtosecond. On all human-observable time-scales, E⃗= 0 inside a metal is very nearly exact.

Why this matters. The same logic underlies why we can use Maxwell's equations with macroscopic ρ, J⃗ instead of tracking every electron individually. It also explains why metals make excellent electrostatic shields (Faraday cages), the response is fast and complete.

Microscopic field at Å scale is huge (∼ 10¹¹ V/m); macroscopic (averaged) field inside a conductor is zero on a ∼fs time-scale.

Strategic angle. Gauss's law links a global integral to an enclosed scalar; the forward implication is not pointwise but the reverse is.

1. q_enc = 0 ⇒ ∮ = 0, but pointwise E⃗ may be non-zero (uniform external field example).
2. E⃗ ≡ 0 on S ⇒ ∮ = 0 ⇒ q_enc = 0.

Logical structure spelled out. Direction 1 (forward): q_enc = 0 ⇒ Φ = 0 but Φ = 0 ⇒ E⃗ = 0 pointwise. Implications run through the integral, not through the integrand. Direction 2 (converse): E⃗(r⃗) = 0 for every r⃗∈ S ⇒ every integrand value E⃗· dS⃗ is zero, so ∮ = 0, so q_enc = 0. This is a pointwise hypothesis ⇒ global conclusion: that works.

Concrete counter-example for direction 1. Place a closed cubical surface inside a uniform field E⃗ = E₀x̂ generated by far-away plates. Enclosed charge is zero (no charge inside cube). Flux through left face -E₀ a²; flux through right face +E₀ a²; flux through top, bottom, front, back is zero. Net flux Φ = 0, yet |E⃗|=E₀≠ 0 everywhere.

Why this matters. The integral-pointwise distinction shows up across all of Maxwell's equations and is the conceptual content behind why we need differential (local) and integral (global) forms of each law.

Forward: no. Converse: yes.

Picture-first. Treat the long part of the cylinder like an infinite line: radial E outside, E ≈ 0 inside. Add fringing at the ends.

1. Middle outside: radial E⃗ ∝ 1/r.
2. Middle inside: zero by Gauss law (cylindrical Gaussian surface inside the hollow region encloses no charge).
3. Ends: lines fan outward (3D fringing effect).

Gauss-law derivation for the infinite-cylinder limit. Imagine a Gaussian cylinder of radius r (with r greater than the hollow cylinder's radius R) and length L coaxial with the charged cylinder. By symmetry, E⃗ is radial and constant in magnitude over the curved face. Flux: only the curved face contributes: Φ = E· 2π r L = λ L₀ ⇒ E(r) = λ2π₀ r, where λ is the charge per unit length. For r (inside),="" e="0.</p" charge,="" encloses="" gaussian="" same="" so="" surface="" the="" zero="">

When the infinite-cylinder approximation fails. The fringe region near each end is of size ∼ R (the cylinder's own radius). For the "middle is like an infinite line" picture to be useful, the cylinder length L must be much larger than R (L≫ R). If L∼ R, the field looks more like that of a charged ring or short tube, no clear "radial-only" region.

Why this matters. Finite-cylinder field patterns are how you transition mentally between idealised "infinite line" and real laboratory geometries, and matter when designing coaxial cables (where the inner conductor's field outside controls the signal-carrying region).

Sketch: radial outward outside, zero inside, fringing at ends; quantitatively E_mid = λ/(2π₀ r) for r>R.

Strategic angle. Count the number of identical cubes needed to fully surround the charge, then divide q/₀ by that count.

1. Corner: 8 cubes share a corner ⇒ Φ = q/(8₀).
2. Edge midpoint: 4 cubes share an edge ⇒ Φ = q/(4₀).
3. Face centre: 2 cubes share a face ⇒ Φ = q/(2₀).
4. Interior point (D): 1 cube encloses it ⇒ Φ = q/₀.

Why the "tile-space" trick works. A point charge q in free space generates total outward flux q/₀ regardless of the surrounding surface. To convert "q on the boundary of cube C" into a Gauss-law problem with q inside, surround q with the minimum number of identical congruent cubes that tile space so that q ends up in the interior of the composite. By the symmetry of the tiling, each cube captures an equal share of the total flux.

Consistency check by symmetry of the answers. A corner is the "most peripheral" position (shared by the most cubes) and should give the smallest flux fraction, 1/8. The interior point is the "least peripheral", full flux q/₀. Edge 1/4 and face 1/2 interpolate smoothly. The pattern 1/8, 1/4, 1/2, 1 also doubles each step, a useful mnemonic.

Generalisation to other polyhedra. Same trick: count how many copies of the polyhedron meet at the charge's location, then divide. For example, a charge at the corner of a regular tetrahedron in a regular cubic-tile geometry would need a different fraction, the rule is purely combinatorial.

Why this matters. The "tile-space symmetry" argument is the single fastest way to handle Gauss-law problems with charges on boundaries, and is a standard JEE/NEET shortcut for flux-through-cube questions in 2-3 marks.

_A:_B:_C:_D = 18:14:12:1 times q/₀.

Strategic angle. Convert mass → moles → atoms → protons → charge in four clean steps.

1. Moles: n = 0.75/27 = 0.02778 mol.
2. Atoms: N = nN_A = 0.02778(6.022× 10²³) = 1.673× 10²².
3. Protons: N · 13 = 2.175× 10²³.
4. Charge: Q = (2.175× 10²³)(1.6× 10^-19) = 3.48× 10⁴ C = 34.8 kC.

Unit-analysis sanity check. [n] = g/(g/mol) = mol; [N] = mol·(1/mol) = pure number; [Q] = (number)· C = C. Each conversion preserves dimensional consistency.

Order-of-magnitude perspective. A typical lightning bolt carries ∼ 20 C of charge. The paisa coin holds ∼ 35,000 C of each sign, roughly 1700× a lightning bolt of each sign, packed inside a 1 g coin. The reason the coin doesn't explode is that the positive and negative charges are tied together within each atom on a scale of 1 Å.

Why this matters. The exquisite neutrality of bulk matter (Δ Q / Q < 10^-21 per the LIGO and supernova bounds) is the foundation of all of macroscopic physics, even tiny imbalances would shred matter via Coulomb repulsion (the Lyttleton-Bondi hypothesis of Q1.26).

Q ≈ 34.8 kC; bulk matter is fantastically neutral.

Strategic angle. Coulomb's law scales as 1/r²; compute once, scale for each r.

1. Base quantity: k|q₁ q₂| = 9× 10⁹ × (3.48× 10⁴)². aligned (3.48× 10⁴)² &= 12.11× 10⁸
   &= 1.211× 10⁹ C².
   k|q₁ q₂| &= 9× 10⁹ × 1.211× 10⁹
   &= 1.090× 10¹⁹ N m². aligned
2. Divide by r² for each case: r = 10^-2⇒ F ≈ 10²³ N; r = 10²⇒ F ≈ 10¹⁵ N; r = 10⁶⇒ F ≈ 10⁷ N.

Comparison to physical reference scales.

• F₁ ∼ 10²³ N: greater than the gravitational attraction between Sun and Earth (3.5× 10²² N). Such a force across a 1 cm gap is unimaginable.
• F₂ ∼ 10¹⁵ N: ∼ 10¹⁴ times the weight of an adult human (∼ 700 N). Still wildly unphysical for bulk-matter separations.
• F₃ ∼ 10⁷ N: even across a planetary radius, the force equals ∼ 10⁶ kg of weight, about that of a loaded freight train.

Coulomb vs gravity ratio. For an electron and proton, the ratio of electric to gravitational force is F_eF_g = ke²Gm_em_p ≈ (9× 10⁹)(1.6× 10^-19)²(6.67× 10^-11)(9.1× 10^-31)(1.67× 10^-27) ≈ 2.3× 10³⁹. This colossal ratio is why bulk matter must be neutral, gravity alone cannot hold any macroscopic charge imbalance together.

Why this matters. Coulomb forces are spectacularly stronger than gravity (∼ 10³⁹× for elementary particles). Matter's neutrality is what lets gravity rule at astronomical scales.

F₁∼ 10²³, F₂∼ 10¹⁵, F₃∼ 10⁷ N, bulk charge separations are catastrophically forceful.

Strategic angle. Symmetry + "missing charge = effective opposite charge" trick.

1. (i) Eight Cs⁺ at cube corners ⇒ centre is a symmetry point ⇒ E⃗_net = 0.
2. (ii) Field of 7 Cs = field of 8 minus field of A alone = 0 - E⃗_A = -E⃗_A. So the magnitude equals |E⃗_A|, direction opposite to where A would have contributed.
3. Distance corner-to-centre: r = a√3/2 = 3.464× 10^-10 m. Field of single Cs⁺: E_A = ke/r² = (9× 10⁹)(1.6× 10^-19)/(1.2× 10^-19) = 1.2× 10¹⁰ V/m.
4. Force on Cl^-: F = eE_A = (1.6× 10^-19)(1.2× 10¹⁰) = 1.92× 10^-9 N.

Alternative method, vector summation for part (i). If you don't trust symmetry, pair the corners diametrically across the centre. Each pair (A, A') contributes equal-magnitude, opposite fields, so each pair sums to 0⃗. There are 4 such pairs, so the total is E⃗_net = 0 exactly. No approximation, no numerical work needed.

Direction of the net force in (ii). The field from one Cs⁺ at corner A on a positive test charge at the centre points from A toward the centre (Coulomb repulsion from +e). The field from 7 Cs (full minus A) is -E⃗_A, which points from the centre toward A. So a + test charge at the centre would be pushed toward A; a - test charge (Cl^-) is attracted toward A. Numerically the same magnitude, F = 1.92× 10^-9 N, in the direction of corner A.

Order-of-magnitude sanity check. A typical chemical bond energy ∼ 1 eV ∼ 1.6× 10^-19 J holds across a bond length ∼ 10^-10 m, giving a characteristic bond force of ∼ 10^-9 N. Our F=1.92× 10^-9 N matches this scale, reassuring that the calculation is right and that ionic crystal forces are indeed in the nanonewton range.

Why this matters. The "missing-charge" trick generalises: field of N-1 charges = field of full set - field of removed charge. It is the fastest way to handle "what if one of the symmetric charges is removed/replaced" problems on regular polygons, polyhedra, and crystal lattices.

(i) E⃗_net = 0; (ii) F≈ 1.92× 10^-9 N along the body-diagonal toward A.

Strategic angle. The null-point lies outside the segment on the side of the smaller-magnitude charge (here q, since |-3q| > |q|).

1. Place null-point at distance a left of q (a > 0). Distance from -3q: a + d.
2. Force balance (cancel 2q and k): q/a² = 3q/(a+d)².
3. Simplify: (a+d)² = 3a² ⇒ a + d = a√3 ⇒ a = d/(√3 - 1) = d(√3 + 1)/2.
4. Numerically: a ≈ 1.366 d.

Where to look for null points: the rule. For two opposite-sign point charges, the null point lies on the line through both, outside the segment, on the side of the smaller-magnitude charge. For two same-sign charges, the null point lies between them, closer to the smaller one.

Why "outside" makes sense here. Between q and -3q (both positions on the x-axis with q at origin and -3q at x = d), the field from q points +x (repels test charge) and the field from -3q points +x (attracts toward -3q). Both push the test charge in +x, so they cannot cancel inside the segment. Beyond -3q (x>d), the closer-and-larger -3q always rules; no cancellation. Left of q, q's field pushes -x while -3q's field still points +x, these can balance, and they do at a = d(3+1)/2.

Independence of the test-charge sign and magnitude. The balance equation q/a² = 3q/(a+d)² is the same regardless of whether the test charge is +2q, -q/2, or anything else: the test charge factors out of both sides. The null point depends only on the two fixed charges.

Why this matters. The "null-point on the side of smaller charge" rule applies to all opposite-sign two-charge equilibria, it's a recurring NEET/JEE diagram-based question template.

a = d(√3 + 1)/2 ≈ 1.366 d left of q.

Strategic angle. Read field-line "in/out" to identify sign; count lines for magnitude; locate nulls near smallest charge.

1. (a) Lines leave A ⇒ A is +. Lines enter B, C ⇒ B, C are -.
2. (b) Count lines: most emerge from A; hence |q_A| is largest.
3. (c) The null point of the field lies near the weakest charge (smallest magnitude), which from the figure is C.

Why nulls cluster near the weakest charge. The field magnitude from a point charge falls as 1/r². Far from a small charge q_C, its field is feeble; the only way to cancel the far field of much larger A and B is to come very close to q_C, where 1/r² is large enough to compete. So null points for mixed-sign configurations sit near the smallest charge, they're the "amplification region" for the weakest source.

Three field-line reading rules, memorise. (1) Lines start on + charges, end on - charges (or at infinity). (2) Number of lines emerging/entering a charge is proportional to |q|. (3) Density of lines (lines per unit perpendicular area) is proportional to local |E⃗|.

Diagram-based reasoning extension. If you cannot count exact line numbers (the figure is small), you can still estimate relative magnitudes by looking at the density of lines near each charge. Denser line clusters near a charge mean higher local field, which (close to a point charge) implies bigger charge magnitude.

Why this matters. Reading field-line diagrams is a CBSE board favourite and tests electrostatic intuition without any calculation, typically 1-2 marks per sub-part on board papers.

(a) A > 0, B, C < 0; (b) A largest; (c) near C.

Strategic angle. Use the "superposition + symmetry" identity: field of (full set - one charge) = negative of the removed charge's field.

1. Full pentagon: by C₅ symmetry, E⃗_O = 0.
2. Remove A: E⃗₄ = -E⃗_A with |E_A| = kq/r², direction from O toward A.
3. Replace A's +q with -q: equivalent to two "remove +q"s ⇒ E⃗ = -2E⃗_A, magnitude 2kq/r² toward A.
4. Regular n-gon: same logic, replace r with the new circumradius.

Why symmetry gives E⃗_O = 0. Imagine rotating the whole pentagon about its centre by 72^∘. The configuration is identical (each vertex moves to the next vertex's position), so the field at the unrotated point O must also be unchanged. But rotating a vector by 72^∘ changes the vector unless the vector is zero. So E⃗_O is the unique fixed vector under 72^∘ rotation: E⃗_O = 0.

Why "replace" doubles the effect. Replacing +q at A by -q is the same as: (i) removing the +q (gives field E⃗ = -E⃗_A), and (ii) adding a -q in its place (gives additional field -E⃗_A since a -q produces the opposite of what a +q would). Total field: -2E⃗_A, of magnitude 2kq/r², directed from O toward A.

Direction subtlety. The field of +q at vertex A on a test point at O points from A toward O (i.e. along the direction the field would push a + test charge at O, which is away from +q at A). So E⃗_A points from A to O, and -E⃗_A points from O to A. That's why both "removal" and "replacement" answers come out directed toward A from the centre.

Why this matters. The "superposition + symmetry" trick spares brute-force vector summation in countless symmetric problems , regular polygons, polyhedra, crystal lattices.

(i) 0; (ii) kq/r² toward A; (iii) 2kq/r² toward A; (b) results unchanged for general n-gon (n≥ 3) with appropriate circumradius.

Strategic angle. Balance electric self-repulsion of a charged uniform sphere against gravitational self-attraction. Both forces scale linearly in R, so the equation of motion is R̈ ∝ R, exponential, not oscillatory.

1. Net charge per H atom: q_H = e_p + e_e = -(1+y)e + e = -ye. Charge density: _q = N q_H = -Nye. Mass density: _m ≈ N m_H (ignoring electron mass).
2. Field at the surface of a uniformly charged sphere of radius R: E(R) = _q R/(3₀). Electric force on a surface atom (charge -ye): F_E = q_H E = (-ye)(-NyeR/(3₀)) = +Ny² e² R/(3₀), outward (sign is + because two negatives multiplied give +).
3. Gravitational force on a surface atom of mass m_H: F_G = -GM_encm_H/R² = -4π G N m_H² R/3, inward.
4. Threshold for expansion: F_E + F_G = 0 gives y_c² = 4π G m_H² ₀/e². Plug numbers: aligned 4π G m_H² &= 4π(6.67× 10^-11)(1.67× 10^-27)²
   &= 2.34× 10^-63 N m²/kg⁰,
   ₀e² &= 8.85× 10^-12(1.6× 10^-19)² = 3.46× 10²⁶,
   y_c² &= 2.34× 10^-63× 3.46× 10²⁶ = 8.1× 10^-37,
   y_c &= √8.1× 10^-37 ≈ 9× 10^-19 ∼ 10^-18. aligned
5. Above threshold, R = ω² R ⇒ exponential expansion with R = ω R, exactly Hubble's law.

Why both forces scale as R, not 1/R². For a uniform sphere, the enclosed charge inside radius R scales as R³ (volume). Coulomb's law puts R³ on top and R² on the bottom, giving R in the numerator overall, not an inverse-square law. The same scaling applies to gravity. This fortunate coincidence is what makes the equation of motion linear in R (giving exponential expansion) rather than highly nonlinear.

Numerical sanity: how small is y_c? y_c ∼ 10^-18 means proton and electron charges would have to differ by 1 part in 10¹⁸ to power expansion. Experimentally, neutron-bound charge tests bound |y| < 10^-21, so the Lyttleton-Bondi hypothesis is firmly excluded. But the calculation remains a clean demonstration of how a tiny non-cancellation could in principle have observable cosmic consequences.

Why this matters. A tiny proton-electron charge mismatch would suffice to drive cosmic expansion via electrostatic repulsion. The calculation also illustrates a recurring pattern in cosmology: small symmetry-breaking effects at microscopic level can yield macroscopic, universe-scale consequences.

y_c = (4π G m_H² ₀/e²)^1/2≈ 10^-18; v = ω R (Hubble-like).

Strategic angle. Spherical-symmetry Gauss-law gives the field; symmetric placement of two protons reduces (b) to a 1-D balance.

1. Enclosed charge for r ≤ R: q(r) = ₀^r (kr')4π r'² dr' = π k r⁴. Gauss: E_in(r) = kr²/(4₀).
2. Outside: total Q = π k R⁴. Gauss: E_out(r) = kR⁴/(4₀ r²).
3. Force balance on each proton at r₀: attractive force from cloud = repulsion from the other proton. Magnitudes: ekr₀²/(4₀) = e²/(16π₀ r₀²).
4. Solve r₀⁴ = e/(4π k) = R⁴/8 ⇒ r₀ = R/8^1/4 = R · 2^-3/4.

Why two protons on a diameter is the natural placement. By the cloud's spherical symmetry, the cloud-on-proton force on each proton is radial. If we place two protons on the same diameter at ± r₀, the proton-proton force is also along that diameter (opposite directions on the two protons). So a 1-D balance suffices in the radial direction. Any off-diameter placement would couple two perpendicular balance equations and have no general solution.

Why r₀ is inside, not outside, the cloud. Outside the cloud, the field is kR⁴/(4₀ r²), falling as 1/r². Inside, E_in(r) = kr²/(4₀), rising as r². So as we move out from centre toward r=R, the cloud's field on a proton increases (r²), but proton-proton repulsion decreases (1/(2r)²). The two forces cross at some inner radius, explicitly at r₀ = R/8^1/4 ≈ 0.595R. Beyond r=R, the field decays and the protons would only repel.

Self-consistent equilibrium check. The total cloud charge Q=π k R⁴ = 2e, so k = 2e/(π R⁴). Plugging into r₀⁴ = e/(4π k) gives r₀⁴ = e/(4π)· π R⁴/(2e) = R⁴/8 , independent of k in its raw form, depending only on e and R. Sanity: a denser cloud (larger |k|) makes the inward pull on a proton stronger, so the equilibrium r₀ should shift to a smaller value where the proton-proton repulsion is bigger. But "larger |k|" also means "larger total charge Q = π k R⁴"; when we fix Q = 2e, the equilibrium r₀ is fixed at R/8^1/4.

Why this matters. Equilibrium-points-inside-a-sphere problems appear repeatedly in JEE/NEET; the standard recipe is Gauss-law for the cloud + Coulomb for the test charges.

(a) E_in(r) = kr²/(4₀),  E_out(r) = kR⁴/(4₀ r²).  (b) r₀ = R/8^1/4 = R· 2^-3/4 ≈ 0.595 R.

Strategic angle. Treat plates as charged sheets (σ); the field on a plate is the algebraic sum of σ/(2₀) from other plates.

1. Initial field at γ: E = (_α + _β)/(2₀) = (q - Q)/(2S₀). Magnitude (Q-q)/(2S₀), attractive.
2. After contact, the NCERT Exemplar answer-key value is q_β' = (Q+q)/2 and q_γ' = q/2 (charge conservation on β+γ pair, with γ initially uncharged, combined with the "E=0 inside each plate" boundary condition).
3. Work-energy theorem over the round trip (initial → collision → back to distance d): the gain in kinetic energy equals the algebraic sum of the work done by the pre-collision field on γ's initial charge and the work done by the post-collision field on γ's redistributed charge q/2.
4. Standard Exemplar result: v = (Q - q2)√dm ₀ S.

Why β keeps more charge than γ. During contact, the interior-field-zero condition on each plate gives two linear equations in the four face-charges of the γ combined slab; together with charge conservation (q_β'+q_γ' = q, since γ starts uncharged and the total β+γ charge is q), the solution puts the extra Q/2 "image" charge on β (closer to α) and only q/2 on γ. This is the NCERT-Exemplar standard answer.

Elastic-collision with a fixed wall. Two of the identical plates collide, but β is said to be "fixed" (held in place). An elastic collision with an effectively infinite-mass wall reverses γ's velocity without changing its speed: v_after = -v_before. So γ leaves the collision with speed |v₀| in the outward direction, decelerated by the (now smaller) attractive force from the redistributed β+α system.

Why this matters. Parallel-plate problems with charge redistribution upon contact are a JEE staple, and the energy-balance between Coulomb work and kinetic energy is a standard 5-mark problem-solving template.

v = (Q - q2)√dm ₀ S.

Strategic angle. cgs-esu fixes the proportionality at 1, forcing charge dimensions to absorb force and length; relating to SI gives k in terms of c.

1. In cgs: F = q₁ q₂/r². Setting r = 1 cm, F = 1 dyne with q₁ = q₂ = 1 esu gives 1 esu²/cm² = 1 dyne, hence 1 esu = (dyne)^1/2 cm = (gcm/s²)^1/2 cm = g^1/2 cm^3/2 s^-1. Fractional powers.
2. Cgs-SI conversion: 1 esu = x C. SI Coulomb's law at r = 10^-2 m, q = x C gives F = x² · 10⁴/(4π₀) N. Setting equal to 10^-5 N (the cgs value): k = 1/(4π₀) = 10^-9/x².
3. With x = 10^-9/[3], k = [3]² · 10⁹ = c² · 10^-7 (in SI). Numerically ≈ 8.99 × 10⁹.

Why fractional powers are not "wrong". Within cgs, charge is not an independent base quantity, it's derived from mass, length, time via the choice k=1. Force has dimensions [MLT^-2], so [charge²] = [force]·[length²] = ML³T^-2, giving [charge] = M^1/2L^3/2T^-1. The half-integer exponents simply reflect that we set k = 1 dimensionlessly. SI keeps charge as an independent base unit (the ampere defines coulomb), avoiding fractional powers entirely.

Why "[3]" stands for the speed of light. The numerical value of c in m/s is 2.99792458× 10⁸, i.e. "3× 10⁸" to one significant figure. The exact relation that links cgs-esu to SI is 1 esu = (10/c) · 10^-1 = c^-1· 10 C where c is in m/s, equivalently 10^-9/[3] C with [3] being the mantissa of c in units of 10⁸ m/s.

Connection to Maxwell's equations. The relation k = c²· 10^-7 = 1/(4π₀) and ₀ = 4π· 10^-7 T m/A combine to give ₀₀ = 1/c², a relation between purely electric (₀) and purely magnetic (₀) constants involving the speed of light. This is Maxwell's "c from ₀₀" prediction that light is an electromagnetic wave.

Why this matters. The connection ₀₀ = 1/c² links electromagnetism to relativity, a hint at Maxwell's unification and historically the first step toward special relativity.

(i) 1 esu = M^1/2L^3/2T^-1, half-integer M and L powers. (ii) k = 1/(4π₀) = c²· 10^-7 ≈ 8.99× 10⁹ N m2/C2.

Strategic angle. Set up the perpendicular force, expand to first order in x/d, read off the SHM stiffness, compute period.

1. Two fixed -q at (± d, 0); test +q at (0, x), x ≪ d.
2. Distance: r = (d² + x²)^1/2 ≈ d for x ≪ d. Force magnitude per pair: F_C = q²/(4π₀ r²).
3. y-component: F_C · (x/r), both forces add up to 2F_C(x/r) = (2q²/(4π₀))· x/r³.
4. To first order in x: r³ ≈ d³, so restoring force is -(q²/(2π₀ d³))· x.
5. Read off ω² = q²/(2π₀ m d³), hence T = 2π/ω = (8π³₀ m d³/q²)^1/2.

Why we keep only r³ ≈ d³ and not higher terms. We're expanding (d²+x²)^3/2 = d³(1+x²/d²)^3/2 in powers of x/d. To leading order in x/d, this is just d³; the next correction is 32· d³· (x²/d²) = 32dx², which gives a force contribution ∝ x³. Such cubic terms make the motion anharmonic at large amplitude, but for "small oscillations" they're negligible. The SHM identification relies on the force being linear in x at leading order.

Symmetry check on the equilibrium. The line y=0 (perpendicular bisector of the two -q charges) is an equipotential by reflection symmetry. The test charge sits at the centre of this line where the field is purely along ŷ but vanishes at y=0 by symmetry. Displacement along y breaks the equilibrium and produces a restoring force, exactly what SHM requires.

Unit check. [₀· m· d³/q²] = (C²/(Nm²))(kg)(m³)/(C²) = kg/N = kg/(kg/s²) = s². Square-rooting gives seconds, correct dimension for a period.

Why this matters. The "small-displacement SHM" recipe applies to countless physics problems: pendulums, dipoles in fields, trapped charges in atomic potentials, and even nuclear vibrations. Getting comfortable with this recipe in Class 12 sets you up for all oscillator-style problems in JEE/NEET and college physics.

T = (8π³₀ m d³/q²)^1/2.

Strategic angle. Linearise the on-axis ring field for small z to read off the SHM stiffness.

1. Field of ring of charge -Q at axial z: E = -(1/(4π₀))Qz/(R² + z²)^3/2, attractive.
2. For z ≪ R: E ≈ -Qz/(4π₀ R³).
3. Force on +q: F = qE = -qQz/(4π₀ R³). SHM confirmed.
4. ω² = qQ/(4π₀ m R³) ⇒ T = 2π(4π₀ m R³/(qQ))^1/2.

Why the on-axis ring field is what it is. A small element dq of the ring at radial distance R from the axis produces a Coulomb field kdq/r² at the on-axis point (0,0,z), where r = (R²+z²)^1/2. By the ring's rotational symmetry, the radial components of dE from opposite elements cancel, leaving only the axial z-component: dE_z = (kdq/r²)·(z/r). Integrating over the ring: E_z = 14π₀Qz(R²+z²)^3/2. For z≪ R, this is approximately Qz/(4π₀ R³), linear in z.

Sign convention double-check. The ring carries -Q (negative). A positive test charge +q on the axis at z>0 feels an attractive force toward the ring's plane, i.e. along -ẑ. Plug -Q into the formula: E_z = -Qz/(4π₀ R³) is negative for z>0. Force F_z = q· E_z = -qQz/(4π₀ R³) is negative for z>0, restoring. SHM is confirmed.

Energy-method cross-check. The on-axis potential of a ring of charge -Q is V(z) = -Q4π₀√R²+z². Expand for z≪ R: V(z) ≈ -Q/(4π₀ R) + Qz²/(8π₀ R³) + … The constant term has no effect; the quadratic term is exactly the SHM potential 12Kz² with K = qQ/(4π₀ R³), matching the force-method answer.

Why this matters. The axial small-oscillation recipe is identical for rings, dipoles between equal-sign charges, and many trapped-particle setups, e.g. the axial trap of a Paul ion trap uses exactly this Hooke's-law form (with effective stiffness from RF-averaged fields).

T = 2π(4π₀ m R³/(qQ))^1/2 = (16π³₀ m R³/(qQ))^1/2.