Class 12 Physics Chapter 1: Electric Charges and Fields worksheet with answer key
A 20-mark worksheet on Electric Charges and Fields for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.
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Answer: (c) 1.0 μC
q² = Fr²/k = 0.9 × (0.10)² / (9 × 10⁹) = 1.0 × 10⁻¹² C², so q = 1.0 × 10⁻⁶ C = 1.0 μC.
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Answer: (d) −2 μC
Charge is a scalar and adds algebraically: (+4) + (−7) + (+2) + (−1) = −2 μC. Adding magnitudes gives 14 μC, which ignores the signs.
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Answer: (a) electrons move from the rod to the silk
In solids only some loosely bound electrons are transferred. The rod loses electrons to the silk, so the rod is left positive and the silk negative; no charge is created or destroyed.
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Answer: (a) F/3, repulsive
Total charge is conserved: +12 − 4 = +8 μC, shared equally by identical spheres, so each has +4 μC. F ∝ |q₁q₂|: before 12 × 4 = 48, after 4 × 4 = 16. New force = 16F/48 = F/3, and like charges repel.
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Answer: (d) 4 cm
The two forces cancel when 2/x² = 8/(12 − x)², i.e. (12 − x)/x = 2, so x = 4 cm, nearer the smaller charge. 2.4 cm comes from dividing in the ratio of the charges instead of the square root of that ratio.
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E = F/q: unit N C⁻¹; dimensions [MLT⁻²]/[AT] = [MLT⁻³A⁻¹]. Flux φ = E·ΔS: unit N C⁻¹ m² (equivalently V m); dimensions [MLT⁻³A⁻¹][L²] = [ML³T⁻³A⁻¹].
- E: N C⁻¹, [MLT⁻³A⁻¹] (1)
- flux: N C⁻¹ m², [ML³T⁻³A⁻¹] (1)
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E = λ/(2πε₀r) = 2 × (9 × 10⁹) × (2.0 × 10⁻⁶)/0.10 = 3.6 × 10⁵ N C⁻¹, radially outward. F = eE = 1.6 × 10⁻¹⁹ × 3.6 × 10⁵ = 5.76 × 10⁻¹⁴ N ≈ 5.8 × 10⁻¹⁴ N. The electron is negative, so the force is towards the wire.
- E = λ/(2πε₀r) = 3.6 × 10⁵ N C⁻¹ (1)
- F = eE ≈ 5.8 × 10⁻¹⁴ N, directed towards the wire (1)
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The charged comb produces a field that polarises the paper: it induces a dipole moment in the paper along the field. The comb's field is non-uniform, stronger nearer the comb, so the induced dipole feels a net force towards the stronger field, and the paper moves towards the comb.
- the comb's field polarises the paper, inducing a dipole along the field (1)
- field is non-uniform, so a net force acts towards the comb (1)
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Q must lie between the two charges (outside, the two forces on it point the same way). Let it be x from +q. Equilibrium of Q: kqQ/x² = k(4q)Q/(L − x)², so (L − x) = 2x and x = L/3. Equilibrium of +q: the repulsion from +4q, k·4q²/L², must be balanced by an attraction towards Q, so Q is negative with kq|Q|/(L/3)² = 4kq²/L², giving |Q| = 4q/9. So Q = −4q/9 at L/3 from +q. (The charge +4q is then also in equilibrium, since the net force on the system is zero.)
- Q between the charges; x = L/3 from +q (1)
- balance on +q gives |Q| = 4q/9 (1)
- Q negative: Q = −4q/9 (1)
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(a) Charge enclosed = +3.0 − 3.0 + 4.0 = +4.0 μC. φ = q/ε₀ = 4.0 × 10⁻⁶/8.854 × 10⁻¹² = 4.5 × 10⁵ N m² C⁻¹. (b) The field at each point of the surface is due to all charges, inside and outside, so the −6.0 μC charge changes it; but it encloses no part of the surface, so it adds no net flux: the flux counts only the charge inside.
- enclosed charge +4.0 μC (dipole contributes zero) (1)
- φ = q/ε₀ = 4.5 × 10⁵ N m² C⁻¹ (1)
- outside charge changes E but not the net flux (1)
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Each sheet gives a field of (its σ)/2ε₀, away from a positive sheet and towards a negative one. A: 2σ/2ε₀ = σ/ε₀, away from A. B: σ/2ε₀, towards B. (a) Left of A: A's field to the left, σ/ε₀; B's field to the right, σ/2ε₀. Net σ/2ε₀ to the left. (b) Between: A's field to the right, σ/ε₀; B's field to the right, σ/2ε₀. Net 3σ/2ε₀ to the right. (c) Right of B: A's field to the right, σ/ε₀; B's to the left, σ/2ε₀. Net σ/2ε₀ to the right.
- left of A: σ/2ε₀, to the left (1)
- between: 3σ/2ε₀, from A to B (1)
- right of B: σ/2ε₀, to the right (1)