Chapter 2: Electrostatic Potential and Capacitance Application Questions | physics Class 12 CBSE

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Chapter 2: Electrostatic Potential and Capacitance Application Questions | physics Class 12 CBSE | ByteNotes.in

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A charged comb run through dry hair attracts small bits of paper. Explain using polarisation. What changes on a rainy day?

The comb gets charged by friction with dry hair. Bringing it near paper, the comb's field induces dipole moments in paper molecules — positive charges shift toward the comb, negative charges away.
The non-uniform field of the comb exerts a net attractive force on the polarised paper (force on nearer induced charges > force on farther ones), pulling paper toward the comb.
On a rainy day: moisture reduces friction, so comb gains little charge. Also, humid air conducts slightly, leaking away any charge gained. Therefore the comb cannot attract paper on rainy days.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

Aircraft tyres are made slightly conducting. A bird on a high-voltage wire is safe; a person touching it is not. Explain both using electrostatics.

Aircraft tyres: friction with air during flight generates static charge on the aircraft. Slightly conducting tyres drain this charge to ground upon landing, preventing dangerous static build-up that could spark and ignite fuel.
Bird on wire: both legs are at nearly the same high potential. Current flows only with a potential difference — there is essentially zero potential difference across the bird's body, so no current flows.
Person touching wire while standing on ground: body forms circuit from high-potential wire to ground (V = 0). Large potential difference drives fatal current through the body via the path of least resistance.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

With reference to C = ε₀A/d and C = KC₀, explain how capacitors used in portable electronics are made compact yet have large capacitance.

From C = ε₀A/d: to increase C, increase plate area A or decrease separation d. Modern capacitors use extremely thin dielectric films (d ~ micrometres) and folded/rolled plate geometry to maximise A/d in small volume.
Inserting dielectric of high K further multiplies C by factor K. Ceramic capacitors use BaTiO₃ (K ~ 1000–10000); electrolytic capacitors use oxide layers (K ~ 8–30) with very small d.
Combined effect: C = Kε₀A/d. Both small d and large K are exploited. A 1 μF capacitor that would require 30 km² plates in air at 1 cm separation can be made coin-sized using high-K thin-film dielectrics.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A parallel plate capacitor (air, C = 8 pF) has its plate separation halved and is filled with dielectric K = 6. Calculate the new capacitance.

Initial: C₀ = ε₀A/d = 8 pF. Plate separation halved: d → d/2. New C due to separation only = ε₀A/(d/2) = 2C₀ = 16 pF.
Dielectric K = 6 inserted: further multiplies C by 6. Final C = 6 × 2C₀ = 12C₀ = 12 × 8 = 96 pF.
Physical explanation: halving d doubles C (factor 2) because the same charge produces smaller potential difference. Inserting K = 6 dielectric multiplies C by 6 (polarisation reduces field). Both effects are independent and multiplicative: 2 × 6 = 12.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

Three 9 pF capacitors are connected in series to 120 V. Find total capacitance, charge, and voltage across each.

Series combination of three identical C = 9 pF: 1/C_total = 1/9 + 1/9 + 1/9 = 3/9 = 1/3, so C_total = 3 pF.
Total charge: Q = C_total × V = 3 × 10⁻¹² × 120 = 3.6 × 10⁻¹⁰ C. In series, each capacitor carries this same charge Q.
Voltage across each: V₁ = V₂ = V₃ = Q/C = 3.6 × 10⁻¹⁰/(9 × 10⁻¹²) = 40 V. Check: 3 × 40 = 120 V ✓. Equal voltages because all three are identical capacitors sharing the same charge.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

Explain why the potential at the equatorial plane of a dipole is zero and describe the equipotential surfaces of a dipole qualitatively.

At equatorial plane, θ = 90° for any point. From V = p cosθ/(4πε₀r²): cos90° = 0, so V = 0. Alternatively, +q and −q are equidistant from any equatorial point, giving equal and opposite contributions that cancel.
The equatorial plane (entire infinite plane through dipole midpoint, perpendicular to axis) is the V = 0 equipotential surface. Points on the dipole axis have the largest |V| at a given r (cosθ = ±1).
Other equipotential surfaces of a dipole are closed curves that are bulged toward the + charge (high positive V) on one side and toward the − charge (high negative V) on the other. The electric field lines are perpendicular to all these surfaces.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

A 12 pF capacitor is connected to 50 V battery. Find energy stored. It is then disconnected and connected to another uncharged 12 pF capacitor. Find energy lost.

Initial energy: U_i = (1/2)CV² = (1/2)(12 × 10⁻¹²)(50²) = (1/2)(12 × 10⁻¹²)(2500) = 1.5 × 10⁻⁸ J = 15 nJ. Charge: Q = 12 × 10⁻¹² × 50 = 6 × 10⁻¹⁰ C.
After disconnection and connection to identical uncharged capacitor: charge shares equally. Each gets Q/2 = 3 × 10⁻¹⁰ C. Common voltage V' = Q/(2C) = 25 V. Final energy: U_f = 2 × (1/2)CV'² = 12 × 10⁻¹² × 625 = 7.5 × 10⁻⁹ J = 7.5 nJ.
Energy lost = 15 − 7.5 = 7.5 nJ, exactly half the initial energy. It is dissipated as heat and electromagnetic radiation during transient current flow and cannot be recovered.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

For a uniformly charged spherical shell, explain why potential inside equals the surface potential, and why outside it behaves like a point charge.

Outside shell (r ≥ R): by Gauss's law, E = Q/(4πε₀r²) — same as a point charge. Integrating E from ∞ to r gives V = Q/(4πε₀r), identical to a point charge potential.
Inside shell (r < R): by Gauss's law, E = 0 (no enclosed charge). Since E = −dV/dl and E = 0, V is constant throughout the interior.
By continuity, interior V must equal surface V: V_inside = V_surface = Q/(4πε₀R). No work is done moving any charge inside the shell — consistent with E = 0 inside.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A mole of a dipolar substance (dipole moment per molecule = 10⁻²⁹ C m) is polarised by field 10⁶ V/m. Field suddenly rotates by 60°. Estimate heat released.

Total dipole moment: p = 6 × 10²³ × 10⁻²⁹ = 6 × 10⁻⁶ C m. Initial (θ = 0°): U_i = −pE cos0° = −6 × 10⁻⁶ × 10⁶ = −6 J.
Final (θ = 60°): U_f = −pE cos60° = −6 × 10⁻⁶ × 10⁶ × 0.5 = −3 J. Change in PE = U_f − U_i = −3 − (−6) = +3 J.
PE increased by 3 J but no external work done — this energy comes from thermal (molecular kinetic) energy as dipoles partially align with new field direction. Heat released = 3 J.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

Two charges 3 × 10⁻⁸ C and −2 × 10⁻⁸ C are 15 cm apart. Find the point(s) on the joining line where potential is zero.

Let positive charge at origin O, negative at A (x = 15 cm). For point P at distance x from O between them: kq₁/x + kq₂/(15−x) = 0 → 3/x = 2/(15−x) → 3(15−x) = 2x → x = 9 cm. First zero at x = 9 cm.
For point Q beyond A at distance x from O (x > 15 cm): 3/x − 2/(x−15) = 0 → 3(x−15) = 2x → x = 45 cm. Second zero at x = 45 cm.
Physically: at 9 cm (between charges), larger positive and smaller negative cancel. At 45 cm (beyond negative charge), farther positive charge's potential still exactly cancels nearer negative charge's — both points are zero-potential equilibria.

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Chapter sections

Chapter 2: Electrostatic Potential and Capacitance | Application Questions

A charged comb run through dry hair attracts small bits of paper. Explain using polarisation. What changes on a rainy day?

Aircraft tyres are made slightly conducting. A bird on a high-voltage wire is safe; a person touching it is not. Explain both using electrostatics.

With reference to C = ε₀A/d and C = KC₀, explain how capacitors used in portable electronics are made compact yet have large capacitance.

A parallel plate capacitor (air, C = 8 pF) has its plate separation halved and is filled with dielectric K = 6. Calculate the new capacitance.

Three 9 pF capacitors are connected in series to 120 V. Find total capacitance, charge, and voltage across each.

Explain why the potential at the equatorial plane of a dipole is zero and describe the equipotential surfaces of a dipole qualitatively.

A 12 pF capacitor is connected to 50 V battery. Find energy stored. It is then disconnected and connected to another uncharged 12 pF capacitor. Find energy lost.

For a uniformly charged spherical shell, explain why potential inside equals the surface potential, and why outside it behaves like a point charge.

A mole of a dipolar substance (dipole moment per molecule = 10⁻²⁹ C m) is polarised by field 10⁶ V/m. Field suddenly rotates by 60°. Estimate heat released.

Two charges 3 × 10⁻⁸ C and −2 × 10⁻⁸ C are 15 cm apart. Find the point(s) on the joining line where potential is zero.

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