NCERT Solutions Class 12 Physics Chapter 1 Electric Charges and Fields

What the question is asking. We have two tiny spheres, each carrying a known positive charge, placed in air a fixed distance apart. We need the force one feels from the other.

Given.

• Charge on first sphere, q₁ = 210^-7 C
• Charge on second sphere, q₂ = 310^-7 C
• Distance between them, r = 30 cm = 0.30 m
• Medium: air. The relative permittivity of air is _r ≈ 1, so we can treat it exactly like vacuum.

Concept used — Coulomb's law. The electrostatic force between two stationary point charges separated by a distance r is F = 14π₀ q₁ q₂r². The number 14π₀ is the Coulomb constant. In SI units, 14π₀ = k = 910⁹ N m² C^-2. Same-sign charges repel; opposite-sign charges attract. Both charges here are positive, so we expect a repulsive force.

Step 1 — write the formula with symbols. F = k q₁ q₂r².

Step 2 — substitute the numbers. F = (910⁹) (210^-7)(310^-7)(0.30)².

Step 3 — simplify the numerator. Multiply the two charges: q₁ q₂ = (2× 3)×(10^-7× 10^-7) = 610^-14 C².

Step 4 — simplify the denominator. r² = (0.30)² = 0.09 m².

Step 5 — combine. F = 910⁹ × 610^-140.09. First compute 610^-140.09. Since 0.09 = 910^-2, 610^-14910^-2 = 69× 10^-14+2 = 0.667× 10^-12.

Step 6 — multiply by the Coulomb constant. F = 910⁹ × 0.66710^-12 = 610^-3 N.

Step 7 — interpret the sign / direction. Both charges are positive, so the force is repulsive: each sphere is pushed away from the other along the line joining them. The magnitude is the same on both spheres (Newton's third law).

Final answer. F = 610^-3 N (repulsive).

Given.

• q₁ = 0.4 = 0.410^-6 C
• q₂ = -0.8 = -0.810^-6 C
• Magnitude of force, F = 0.2 N
• Medium: air _r ≈ 1.

(a) Distance between the spheres.

Step 1 — set up Coulomb's law. F = 14π₀ |q₁ q₂|r². We take the absolute value of q₁ q₂ because F is a magnitude (always positive). The negative sign on q₂ only tells us the force is attractive, not the size.

Step 2 — rearrange for r. r² = k |q₁ q₂|F, r = √k |q₁ q₂|F.

Step 3 — compute |q₁ q₂|. |q₁ q₂| = (0.410^-6)(0.810^-6) = 0.3210^-12 C² = 3.210^-13 C².

Step 4 — substitute everything. r² = (910⁹)(3.210^-13)0.2 = 28.810^-40.2 = 14410^-4 = 1.4410^-2 m².

Step 5 — take the square root. r = √1.4410^-2 = 0.12 m = 12 cm.

(b) Force on the second sphere due to the first.

By Newton's third law, the two spheres push or pull each other with equal and opposite forces. So the magnitude of the force on the second sphere is also 0.2 N, directed toward the first sphere (since the charges are opposite, the force is attractive).

Final answer.

• (a) r = 0.12 m = 12 cm.
• (b) F₂₁ = 0.2 N, directed from the second sphere toward the first (attraction).

What the ratio is. The numerator k e² is the Coulomb-form numerator for the electric force between an electron and a proton (each carrying charge e) at unit separation. The denominator G m_e m_p is the corresponding numerator for the gravitational force between them. So the ratio compares electric vs gravitational attraction between an electron–proton pair.

Step 1 — list the SI units.

• k = 14π₀ has units N m² C^-2.
• e (electron charge) has units C.
• G (gravitational constant) has units N m² kg^-2.
• m_e, m_p (electron and proton masses) have units kg.

Step 2 — substitute units in the ratio. k e²G m_e m_p ⟶ (N m²C^-2)(C²)(N m²kg^-2)(kg)(kg).

Step 3 — cancel units. The C² in the numerator cancels the C^-2 factor. The two kg terms in the denominator cancel the kg^-2 factor. The N m² factors cancel between numerator and denominator. What remains: 11 = dimensionless.

Step 4 — plug in numerical values.

• k = 910⁹
• e = 1.610^-19 C ⇒ e² = 2.5610^-38 C²
• G = 6.6710^-11
• m_e = 9.1110^-31 kg
• m_p = 1.6710^-27 kg

Step 5 — compute the numerator. k e² = 910⁹× 2.5610^-38 = 23.0410^-29 = 2.30410^-28.

Step 6 — compute the denominator. G m_e m_p = 6.6710^-11× 9.1110^-31× 1.6710^-27 ≈ 1.010^-67.

Step 7 — divide. k e²G m_e m_p ≈ 2.30410^-281.010^-67 ≈ 2.310³⁹.

Interpretation. The electric force between an electron and a proton is roughly 2.310³⁹ times stronger than the gravitational force between them. That's why gravity is utterly negligible inside atoms — chemistry is electromagnetic, not gravitational.

(a) What 'quantised' means for electric charge.

Every observable electric charge in nature is an integer multiple of one fundamental "smallest" unit — the elementary charge, e = 1.610^-19 C. That is, any charge q that exists in nature satisfies q = n e, n ∈ 0, ± 1, ± 2, ± 3, … . You can have a body carrying +1e, +2e, -3e, and so on, but never +1.5 e, +π e, or any non-integer fraction. Charge comes in discrete "lumps" — like grains of sand, not like water flowing continuously.

What about quarks? The fundamental particles inside protons and neutrons (quarks) carry fractional charges ± e/3 and ± 2e/3. But quarks are confined — they never appear alone. They always combine in groups of three (forming protons, neutrons) or in quark–antiquark pairs, and the combinations always have integer charge. So at the level of free observable particles, the rule q = ne is exact.

(b) Why quantisation is invisible at large scales.

Consider a charge of just 1 = 10^-6 C. The number of elementary charges in it is n = qe = 10^-61.610^-19 ≈ 6.2510¹². That's over six trillion elementary charges. Adding or removing even a million electrons changes the count by less than one part in a million — far below any laboratory measurement's precision.

So at the macroscopic level, charge appears continuous. We can speak of a charge being "0.4 " or "2.7 " without worrying about whether the number of electrons is exactly right. The discreteness is "hidden" by the enormous size of n.

Analogy. A sand dune looks like a smooth surface from a distance, even though it's made of 10²⁰ discrete sand grains. Up close (under a microscope), the graininess shows — just as inside an atom, where charges are small, quantisation is essential.

Law of conservation of charge. The total electric charge of an isolated system stays constant. Charges can move around or transfer between objects, but the algebraic sum (positive minus negative) is fixed.

What rubbing actually does. Glass and silk are both electrically neutral before being rubbed — they contain equal numbers of protons (in nuclei) and electrons. When the two surfaces are rubbed together, friction provides enough energy for some electrons to be wrenched away from the glass and transferred to the silk. (Why glass loses electrons and silk gains them depends on each material's electron affinity — glass sits higher in the triboelectric series, so it gives up electrons easily.)

What this means for charges.

• Before rubbing: glass is neutral, silk is neutral. Total charge of the pair: 0.
• After rubbing: glass has lost some electrons → glass is now positively charged, say +q.
• After rubbing: silk has gained exactly those electrons → silk is now negatively charged, -q.
• Total charge of the pair: (+q) + (-q) = 0. Unchanged.

Consistency with conservation. No new charges were created or destroyed. Electrons were merely moved from one object to the other. The total charge of the isolated (glass + silk) system stayed at zero throughout. This is exactly what conservation of charge predicts: charges can flow around, but the net amount is preserved.

Final statement. The phenomenon (neutral) → (+q, -q) involves opposite-but-equal charges appearing on the two bodies, summing to zero — fully consistent with the law of conservation of electric charge.

Set-up. Square ABCD with side 10 cm. Label the corners going around:

• A = +2 — say top-left corner.
• B = -5 — top-right corner.
• C = +2 — bottom-right corner (diagonally opposite to A).
• D = -5 — bottom-left corner (diagonally opposite to B).

A test charge q₀ = +1 sits at the centre O of the square.

Step 1 — distance from each corner to the centre. For a square of side a, the diagonal is a√2, so the centre is at half-diagonal from each corner: r_OA = r_OB = r_OC = r_OD = a√22 = (0.10)√22 = 0.0707 m.

Step 2 — use symmetry on the diagonals. The diagonal AC connects the two +2 charges. The diagonal BD connects the two -5 charges. These diagonals are perpendicular to each other and cross at O.

Step 3 — forces from the AC diagonal the two +2 corners.

• Force from q_A on q₀: repulsive (both positive). The force on q₀ points away from A, i.e., along the AC diagonal toward C.
• Force from q_C on q₀: also repulsive. Points away from C, i.e., along the AC diagonal toward A.
• Same magnitude (same charges, same distance), opposite directions. They cancel exactly.

Step 4 — forces from the BD diagonal the two -5 corners.

• Force from q_B on q₀: attractive (opposite signs). The force on q₀ points toward B, i.e., along BD toward B.
• Force from q_D on q₀: also attractive. Points toward D, i.e., along BD toward D.
• Same magnitude, opposite directions. They also cancel exactly.

Step 5 — net force. All four pairwise forces have a partner that exactly cancels it: F_net = F_A + F_B + F_C + F_D = 0.

Final answer. The net electric force on the central charge is zero.

What an electric field line represents. A field line is a curve drawn so that at every point along it, the line's tangent points in the direction of the electric field E at that point. The closer the lines are packed, the stronger the field.

(a) Why field lines have no breaks.

Suppose a field line had a sudden break at some point P. There would be two equally absurd consequences:

• Either the field is undefined at P — but in any region without a point charge sitting at P, the field is a well-defined finite vector at every point in space (you can always compute it via Coulomb's law from the source charges).
• Or a test charge moving along the line would suddenly stop being acted upon by any force at P — which violates the smooth behaviour of the electrostatic force.

In short, the field is a continuous function of position (away from source charges), so field lines drawn to follow it must also be continuous curves. They start on positive charges, end on negative charges, or extend off to infinity — but they don't just stop in the middle of empty space.

(b) Why two field lines never cross.

Suppose two field lines did cross at a point Q. At Q, the tangent to one line points in direction d₁; the tangent to the other line points in a different direction d₂. Since the field line tangent gives the direction of E at that point, this would mean E at Q has two different directions simultaneously — a logical contradiction.

At any point in space, the electric field has one and only one direction. So at most one field line can pass through any given point — hence no two field lines can cross.

One exception. Field lines meet at the locations of point charges themselves (lines start at +q, end at -q). These endpoints are mathematical singularities where the field magnitude is infinite — not "crossings" in the regular sense.

Given.

• q_A = +3 = 310^-6 C
• q_B = -3 = -310^-6 C
• Separation AB = 20 cm = 0.20 m
• Midpoint O is 0.10 m from both A and B.

(a) Electric field at the midpoint O.

Step 1 — direction of each contribution.

• E_A due to +q_A at O: a positive source produces a field that points away from itself. So at O, the field due to A points from A toward B.
• E_B due to -q_B at O: a negative source produces a field that points toward itself. At O, this means pointing from O toward B. That is also from A toward B.

Both contributions point in the same direction (from A toward B). So they add.

Step 2 — magnitude of each. |E_A| = |E_B| = 14π₀|q|r² = (910⁹)310^-6(0.10)² = (910⁹)310^-610^-2. = 9× 3× 10^9-6+2 = 2710⁵ = 2.710⁶ N/C.

Step 3 — total field at O. E_O = |E_A| + |E_B| = 2.710⁶ + 2.710⁶ = 5.410⁶ N/C, directed from A toward B.

(b) Force on the negative test charge at O.

Test charge: q_t = -1.510^-9 C (negative).

Step 1 — magnitude of force. |F| = |q_t| E_O = (1.510^-9)(5.410⁶) = 8.110^-3 N.

Step 2 — direction. Force on a charge is F = qE. Since q_t is negative, F points opposite to E. The field at O points A → B, so the force on the negative test charge points B → A.

Final answer.

• (a) E_O = 5.410⁶ N/C, directed from A to B.
• (b) |F| = 8.110^-3 N, directed from B toward A.

Given.

• q_A = +2.510^-7 C at A = 0, 0, -15 cm — i.e., 15 cm below the origin on the z-axis.
• q_B = -2.510^-7 C at B = 0, 0, +15 cm — i.e., 15 cm above the origin.
• So the positive charge sits at -z, the negative charge at +z. Separation between them is 30 cm = 0.30 m.

(a) Total charge. Q_total = q_A + q_B = (+2.510^-7) + (-2.510^-7) = 0. The system has no net charge — it is an electrically neutral arrangement, specifically a dipole.

(b) Electric dipole moment.

Step 1 — definition. The electric dipole moment of a pair of equal and opposite point charges +q and -q separated by displacement d (going from - to +) is p = q d, where q is the magnitude of either charge and d is the vector pointing from the negative charge to the positive charge.

Step 2 — magnitude of p. Here q = 2.510^-7 C and |d| = 0.30 m: |p| = q |d| = (2.510^-7)(0.30) = 7.510^-8 C m.

Step 3 — direction of p. Negative charge q_B is at +z. Positive charge q_A is at -z. Going from negative to positive means going from +z to -z, i.e., along -z (downward). p = 7.510^-8 C m, directed along -z.

Final answer.

• (a) Q_total = 0 (the dipole is neutral).
• (b) p = 7.510^-8 C m along the negative z-axis (from B to A).

Given.

• Dipole moment magnitude: p = 410^-9 C m.
• Electric field magnitude: E = 510⁴ N/C.
• Angle between p and E: θ = 30°.

Concept used — torque on a dipole in a uniform field. A dipole experiences no net force in a uniform field (the two equal-and-opposite forces on the +q and -q ends are equal-and-opposite and so cancel). But the two forces form a couple, producing a torque that tries to rotate the dipole into alignment with E. The torque is τ = p × E, |τ| = p E sinθ. Why sinθ? It's the component of p perpendicular to E (the lever arm) times the force qE on each end. When p is parallel to E θ = 0, no torque. When p is perpendicular θ = 90°, torque is maximum.

Step 1 — write the formula. |τ| = p E sinθ.

Step 2 — value of sin 30°. sin 30° = 0.5.

Step 3 — substitute. |τ| = (410^-9)(510⁴)(0.5).

Step 4 — multiply. |τ| = (4× 5× 0.5)× 10^-9+4 = 10× 10^-5 = 110^-4 N m.

Final answer. |τ| = 10^-4 N m. The torque tries to rotate the dipole so that p aligns with E i.e., reduce θ toward zero.

(a) Number of electrons transferred.

Step 1 — identify direction of transfer. The polythene ended up with a negative charge. Negative charge means an excess of electrons. So electrons flowed onto the polythene. The wool must have given them up.

Direction: wool → polythene.

(This is consistent with the triboelectric series — wool sits higher than polythene, so wool tends to lose electrons to polythene during friction.)

Step 2 — set up the count. The total negative charge on polythene equals the elementary charge multiplied by the number of extra electrons: q = n e ⇒ n = qe.

Step 3 — plug in. n = 310^-71.610^-19 = 31.6× 10^-7+19 = 1.87510¹² electrons.

(b) Mass transfer.

Step 1 — does mass move? Yes. Every electron has mass m_e = 9.1110^-31 kg. So when 1.87510¹² electrons leave the wool and land on the polythene, mass moves with them.

Step 2 — total mass transferred. Δ m = n m_e = (1.87510¹²)(9.1110^-31). = (1.875× 9.11)× 10^12-31 ≈ 17.1× 10^-19 = 1.7110^-18 kg.

Step 3 — interpretation. About 1.710^-18 kg — about one billion-billionth of a gram. The polythene literally weighs that much MORE after rubbing; the wool weighs that much LESS. In practice, this is utterly undetectable — even the most sensitive laboratory balance can only measure changes of order 10^-9 kg, nine orders of magnitude coarser.

Final answer.

• (a) Approximately 1.87510¹² electrons were transferred from wool to polythene.
• (b) Yes, mass is transferred — about 1.7110^-18 kg of mass moves from wool to polythene. But the change is far too small to measure.

Given (part a). Two identical insulated charged copper spheres, each with charge q₁ = q₂ = 6.510^-7 C, separated by r = 0.50 m. Sphere radii are negligible — treat as point charges.

(a) Initial force.

Step 1 — set up Coulomb's law. F = 14π₀ q₁ q₂r² = k q²r², with q = 6.510^-7.

Step 2 — compute q². q² = (6.5)²× 10^-14 = 42.25× 10^-14 C².

Step 3 — compute r². r² = (0.50)² = 0.25 m².

Step 4 — divide. q²r² = 42.2510^-140.25 = 169× 10^-14 = 1.6910^-12.

Step 5 — multiply by k. F = (910⁹)(1.6910^-12) = 15.2110^-3 ≈ 1.510^-2 N.

So the initial repulsive force is about 1.510^-2 N or 0.015 N.

(b) Each charge is doubled; distance is halved.

Step 1 — track how each factor scales.

• New charges: each is now 2q. So the product q₁ q₂ = (2q)(2q) = 4q². The numerator quadruples (×4).
• New distance: r/2. So r² becomes (r/2)² = r²/4. Dividing by r²/4 is the same as multiplying by 4 compared to dividing by r². The denominator effect quadruples the force (×4).

Step 2 — combine the scalings. F_new = 4 × 4 × F_old = 16 F_old.

Step 3 — compute. F_new = 16× 1.510^-2 = 0.24 N.

Final answer.

• (a) Repulsive force: 1.510^-2 N.
• (b) After doubling each charge and halving the distance: 0.24 N — sixteen times larger.

Starting state. From Exercise 1.12: q_A = q_B = q = 6.510^-7 C, separation r = 0.50 m. Initial force F₀ = 1.510^-2 N.

Key rule for identical conductors in contact. When two identical conducting spheres touch, the total charge on the pair gets shared equally between them (by symmetry: each sphere has the same capacitance, and at equilibrium they have the same potential, hence the same charge).

Step 1 — sphere C (uncharged) touches A (charge q).

Total charge on the (A, C) pair before contact: q + 0 = q. At equilibrium after contact, both spheres share equally: each ends with q/2.

So after this step: q_A = q/2, q_C = q/2, q_B = q (unchanged — wasn't touched).

Step 2 — now C (carrying q/2) touches B (carrying q).

Total charge on the (B, C) pair before contact: q/2 + q = 3q/2. Shared equally: each ends with 3q/22 = 3q4.

So after this step: q_B = 3q/4, q_C = 3q/4, q_A = q/2 (unchanged — wasn't touched again).

Step 3 — C is removed. Spheres A and B remain. Their charges are now: q_A = q2, q_B = 3q4.

Step 4 — compute the new force between A and B. F_new = k q_A q_Br² = k (q/2)(3q/4)r² = 38 k q²r² = 38 F₀.

Step 5 — numerical value. F_new = 38× 1.510^-2 = 0.375× 1.510^-2 = 5.62510^-3 N ≈ 5.710^-3 N.

Final answer. F_new ≈ 5.710^-3 N, still repulsive. This is smaller than the original 1.510^-2 N because both spheres now carry less charge.

What the figure shows. Three charged particles enter a region with a uniform electric field and follow curved tracks. We need to read off (1) the sign of each charge and (2) compare their charge-to-mass ratios.

How charge interacts with a uniform field. A charge q in a uniform field E feels a constant force F = q E. This produces a constant acceleration a = Fm = q Em. The deflection in any given time interval depends on a, which is proportional to the charge-to-mass ratio q/m. Particles with higher q/m curve more.

Step 1 — determine the sign of each particle.

The field direction in the figure can be inferred from the deflection: tracks 1 and 2 curve in one direction (say upward), and track 3 curves in the opposite direction (downward).

• If the field points downward and a particle deflects upward, then F must point upward. Since F = qE and E is downward, the particle must have q < 0 — negative charge.
• If a particle deflects downward (along the field), then F is along the field, meaning q > 0 — positive charge.

From the standard NCERT figure orientation:

• Particles 1 and 2 deflect upward (against the field) — they are negative.
• Particle 3 deflects downward (along the field) — it is positive.

Step 2 — compare charge-to-mass ratios.

Two particles passing through the same field over the same length of region differ in deflection by exactly the ratio of their q/m values. Greater deflection over the same horizontal distance means greater q/m.

From the figure, particle 3 has the largest deflection. So particle 3 has the highest charge-to-mass ratio.

Final answer.

• Particle 1: negative.
• Particle 2: negative.
• Particle 3: positive.
• Highest |q/m|: particle 3.

Concept used — electric flux. For a flat surface of area A in a uniform field E, the flux is Φ = E · A = E A cosθ, where θ is the angle between E and the area's outward normal n. Flux is a scalar (a signed number): positive if E and n are on the same side, negative if they oppose, zero if perpendicular.

Given. E = 310³ N/C (along +x). Square side 0.10 m, so area A = (0.10)² = 10^-2 m².

(a) Square parallel to the yz-plane.

Step 1 — determine the area's normal. A square in the yz-plane has its normal along the x-axis. Take n = (along +x).

Step 2 — angle between E and n. Both are along +, so θ = 0° and cosθ = 1.

Step 3 — compute flux. Φ = E A cosθ = (310³)(10^-2)(1) = 30 N m²/C.

(b) Same square, but normal makes 60° with the x-axis.

Step 1 — the angle. Now θ = 60°, so cosθ = 0.5.

Step 2 — compute flux. Φ = E A cos 60° = (310³)(10^-2)(0.5) = 15 N m²/C.

Final answer.

• (a) Φ = 30 N m²/C.
• (b) Φ = 15 N m²/C.

Reasoning step by step.

Step 1 — set up. E = 310³ N/C, cube of side L = 0.20 m. The cube's six faces are pairs of three: two perpendicular to (the field direction), two perpendicular to , two perpendicular to k.

Step 2 — identify each face's outward normal.

• Face at +x: outward normal = +.
• Face at -x: outward normal = -.
• Faces at ± y: outward normals = ± both perpendicular to E.
• Faces at ± z: outward normals = ± k both perpendicular to E.

Step 3 — flux through the +x face. _+x = E· A = E L² (+·) = +E L². Plugging E = 310³ and L² = 0.04: _+x = (310³)(0.04) = 120 N m²/C.

Step 4 — flux through the -x face. _-x = E· A = E L² (·(-)) = -E L² = -120 N m²/C. (Field lines entering this face cross it inward — negative flux by convention.)

Step 5 — flux through the ± y and ± z faces. The field has no or k component, so E·= 0 and E·k = 0. All four of these faces contribute zero flux.

Step 6 — net flux. _net = _+x + _-x + 0 + 0 + 0 + 0 = (+120) + (-120) = 0.

Final answer. The net flux through the cube is zero, regardless of the field strength or cube size.

Why this makes intuitive sense. Every field line that enters one face of the cube exits through the opposite face — what goes in must come out, because there are no charges inside the cube to "create" or "absorb" field lines.

Concept used — Gauss's law. _net = ∮ E· dA = q_enclosed0. The total electric flux through any closed surface equals the enclosed charge divided by ₀.

(a) Net charge enclosed.

Step 1 — rearrange. q_enc = _{0 net}.

Step 2 — substitute. ₀ = 8.85410^-12 C²/N m², Φ = 8.010³ N m²/C: q_enc = (8.85410^-12)(8.010³).

Step 3 — multiply. q_enc = (8.854× 8.0)× 10^-12+3 ≈ 70.83× 10^-9 ≈ 7.08× 10^-8 C. Equivalently 70.8 nC or about 0.071 .

The sign is positive: outward flux means field lines emerge from the surface, which by Gauss requires positive net charge inside.

(b) If the net flux were zero, can we conclude no charges inside?

No. Zero net flux only tells us that the algebraic sum of enclosed charges is zero, not that there are no charges at all.

Example: place a +5 charge and a -5 charge anywhere inside the box. The enclosed charges sum to zero, so _net = 0. But the box obviously DOES contain charges. The field lines from the positive charge emerge through some part of the box's surface, only to re-enter through some other part (terminating on the negative charge). Net flux is zero, but the individual fluxes through pieces of the surface are not.

So a zero net flux gives the much weaker conclusion: net enclosed charge is zero. It says nothing about whether individual charges exist inside.

Final answer.

• (a) q_enc ≈ +7.0810^-8 C (positive, since outward flux is positive).
• (b) No — a zero net flux implies the net enclosed charge is zero, but not that there are no charges. A dipole (or any neutral combination) inside the box would give zero net flux despite having internal charges.

Set-up. A point charge q = +10 = 1010^-6 C sits exactly 5 cm above the centre of a 10 cm× 10 cm square. We want the flux through that square.

Direct attack (hard). Computing the flux directly would require integrating E· dA over the square — a messy double integral because the distance from the charge to each point on the square varies. Avoid this.

The clever trick — embed the square in a cube. Imagine the square as the top face of a cube of side 10 cm, with the charge sitting at the geometric centre of the cube 5 cm below the top face.

Step 1 — total flux from the charge through the cube. By Gauss's law, the total flux from +q through ANY closed surface around it is _total = q₀. Substituting: _total = 10× 10^-68.85410^-12 ≈ 1.1310⁶ N m²/C.

Step 2 — use symmetry to share among 6 faces. Because the charge sits at the cube's centre, by symmetry the total flux is divided equally among the six faces of the cube: _{each face} = _total6.

Step 3 — compute. _{each face} = 1.1310⁶6 ≈ 1.8810⁵ N m²/C.

Our specific square (the top face) is one of these six identical faces by symmetry, so:

Final answer. _square ≈ 1.88× 10⁵ N m²/C.

Concept used — Gauss's law. _net = q_enclosed0. The flux depends only on the enclosed charge — not on the surface's shape, size, or where the charge sits within the surface.

Given. q = 2.0 = 2.010^-6 C. Surface: a cube of edge 9.0 cm with the charge at its centre.

Step 1 — write the formula. _net = q₀.

Step 2 — substitute values. ₀ = 8.85410^-12 C²/N m²: _net = 2.010^-68.85410^-12.

Step 3 — compute. _net = 2.08.854× 10^-6+12 ≈ 0.226× 10⁶ = 2.2610⁵ N m²/C.

Final answer. _net ≈ 2.2610⁵ N m²/C.

Important — the cube's size doesn't matter. If we'd used a cube of edge 50 cm or 1 cm, the flux would still be the same. Doubling the edge multiplies each face's area by 4, but the field at the surface drops by 4 (the charge is now twice as far on average), so the integral is unchanged. Gauss's law captures this elegantly: only q_enc matters.

Given. Spherical Gaussian surface of radius R = 10.0 cm centred on a point charge q (sign unknown initially). Net flux through it: Φ = -1.010³ N m²/C. The negative sign means net flux is inward (field lines enter the surface).

(a) Flux when the radius is doubled.

Step 1 — apply Gauss's law. Φ = q₀. This depends ONLY on the enclosed charge, not on the radius. So doubling the radius from 10 cm to 20 cm doesn't change q_enc (the charge is still inside) and doesn't change the flux.

Step 2 — answer. _new = _old = -1.010³ N m²/C.

Why this is initially counterintuitive. When you double the radius, the field at the surface drops by 4 since E ∝ 1/r². But the surface area increases by 4 since A = 4π r². These exactly cancel, leaving the total flux unchanged.

(b) Value of the point charge.

Step 1 — rearrange Gauss's law. q = ₀ Φ.

Step 2 — substitute. q = (8.85410^-12)(-1.010³) = -8.854× 10^-9 C.

Step 3 — report. q ≈ -8.85 nC (negative). The negative sign confirms what the negative flux told us: a net inward flux means a negative source (field lines terminate on negative charges).

Final answer.

• (a) Flux stays the same: -1.010³ N m²/C, independent of the Gaussian radius.
• (b) The point charge is q ≈ -8.85 nC about -8.8510^-9 C.

Given.

• Conducting sphere of radius R = 0.10 m.
• Observation point at distance r = 0.20 m from the centre (i.e., outside the sphere).
• Electric field at this point: E = 1.510³ N/C, pointing radially inward.

Key fact about charged conducting spheres. Outside a uniformly charged conducting sphere, the electric field is exactly the same as if all the charge were concentrated at the centre as a point charge. So we can treat the sphere as a point charge q at its centre for the purpose of computing fields at external points.

Concept used — Coulomb's law (outside the sphere). E = 14π₀ |q|r² = k |q|r².

Step 1 — set up. We know E, r; want |q|. Rearrange: |q| = E r²k.

Step 2 — compute r². r² = (0.20)² = 0.04 m².

Step 3 — substitute. |q| = (1.510³)(0.04)910⁹ = 60910⁹.

Step 4 — compute. |q| = 6.6710^-9 C.

Step 5 — determine the sign. The field at the external point points radially inward (toward the sphere). A field that points toward a source means the source is negative — field lines terminate on negative charges. So the sphere carries a NEGATIVE charge: q = -6.6710^-9 C = -6.67 nC.

Final answer. q ≈ -6.67 nC (negative).

Given.

• Diameter D = 2.4 m, so radius R = D/2 = 1.2 m.
• Surface charge density σ = 80.0 /m² = 8010^-6 C/m².

(a) Total charge on the sphere.

Step 1 — surface area of the sphere. A = 4π R² = 4π(1.2)² = 4π (1.44) = 5.76π m² ≈ 18.10 m².

Step 2 — total charge = density × area. q = σ A = (8010^-6)(18.10).

Step 3 — compute. q = 1448× 10^-6 ≈ 1.4510^-3 C = 1.45 mC.

(b) Total electric flux leaving the surface.

Step 1 — apply Gauss's law. Φ = q_enc0 = 1.4510^-38.85410^-12.

Step 2 — compute. Φ = 1.458.854× 10^-3+12 ≈ 0.164× 10⁹ = 1.64× 10⁸ N m²/C.

Final answer.

• (a) q ≈ 1.4510^-3 C or 1.45 mC.
• (b) Φ ≈ 1.6410⁸ N m²/C.

Setup. An infinitely long straight line of charge with uniform linear density λ (charge per unit length, in C/m). We're asked to find λ given the field at a perpendicular distance r from the line.

Concept used — field of an infinite line charge. By symmetry (cylindrical), the field at perpendicular distance r is radial (pointing outward if λ > 0, inward if λ < 0) with magnitude E = λ2π₀ r. The factor 1/2π₀ = 2k = 1.810¹⁰ (since k = 1/4π₀).

Where this comes from (sketch). Apply Gauss's law to a cylindrical surface of radius r and length L coaxial with the line. By symmetry, E is radial and constant in magnitude on this surface. Flux through the curved side: E2π r L. Flux through the flat end caps: zero (field perpendicular to them is zero). Charge enclosed: λ L. Gauss's law gives E2π r L = λ L/₀, so E = λ/2π₀ r.

Given. E = 910⁴ N/C, r = 2 cm = 0.02 m.

Step 1 — rearrange for λ. λ = 2π₀ r E.

Step 2 — substitute and compute. Use 2π₀ = 1/(2k) = 1/1.810¹⁰ = 5.5610^-11: λ = (5.5610^-11)(0.02)(910⁴).

Step 3 — multiply. λ = (5.56)(0.02)(9)× 10^-11+4 = 1.0008× 10^-7 ≈ 1× 10^-7 C/m.

Final answer. λ ≈ 10^-7 C/m = 0.1 /m.

Setup. Two large parallel metal plates, very close together. The inner face of plate 1 carries surface charge density +σ; the inner face of plate 2 carries -σ, with |σ| = 17.010^-22 C/m². This is the standard parallel-plate capacitor configuration.

Concept used — field of an infinite plane sheet. An infinite plane with surface charge density σ produces a uniform field of magnitude E_{single sheet} = σ2₀ on EACH side of the sheet, perpendicular to it. For +σ, the field points away from the sheet on both sides. For -σ, it points toward the sheet on both sides.

For two parallel sheets close together, the total field at any point is the vector sum of contributions from each sheet.

Step 1 — three regions to analyse. Label them: outer region of plate 1 (left of both plates), between the plates (in the gap), outer region of plate 2 (right of both plates). Take "rightward" as positive.

Step 2 — region (a): left of both plates (outer region of plate 1).

• Field from +σ plate (plate 1): points away from the plate on its left side → leftward (negative): -σ/2₀.
• Field from -σ plate (plate 2): points toward the plate from its left side → rightward (positive): +σ/2₀.
• Sum: -σ/2₀ + σ/2₀ = 0.

So E_a = 0 outside plate 1.

Step 3 — region (b): right of both plates (outer region of plate 2).

• Field from +σ plate: points away on its right side → rightward: +σ/2₀.
• Field from -σ plate: points toward the plate from its right side → leftward: -σ/2₀.
• Sum: zero.

So E_b = 0 outside plate 2.

Step 4 — region (c): between the plates.

• Field from +σ plate: points rightward (away from the plate to the right) → +σ/2₀.
• Field from -σ plate: points rightward (toward the plate from the left) → +σ/2₀.
• Sum: σ/2₀ + σ/2₀ = σ/₀.

So E_c = σ/₀, directed from the positive plate to the negative plate.

Step 5 — plug in numbers for region (c). E_c = 17.010^-228.85410^-12 ≈ 1.9210^-10 N/C.

Final answer.

• (a) E = 0 (outside the first plate).
• (b) E = 0 (outside the second plate).
• (c) E = σ/₀ ≈ 1.9210^-10 N/C, directed from the positive plate to the negative plate.