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

ByteNotes

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

Long Answer

Medium difficulty • Structured explanation

Medium

Question 1

Long Form

Derive an expression for electrostatic potential at distance r from a point charge Q. Explain how potential and electric field differ in their distance dependence.

For charge Q at origin, calculate work done bringing unit positive charge from ∞ to P (distance r) along radial path. At intermediate point r', force = Q/(4πε₀r'²).
Work against this force for displacement dr': dW = −Q/(4πε₀r'²)dr'. Integrating r' = ∞ to r: W = Q/(4πε₀r).
Therefore V(r) = Q/(4πε₀r). Valid for any sign of Q. For Q > 0, V > 0; for Q < 0, V < 0.
V ∝ 1/r (slow decrease) while E ∝ 1/r² (faster decrease). Graphs show V as gentle hyperbola, E as steeper curve.
At infinity both vanish, consistent with V(∞) = 0 reference. Work to bring charge q to point P is W = qV = qQ/(4πε₀r).

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 2

Long Form

Derive the expression for potential due to an electric dipole at a general point P(r, θ). Discuss axial and equatorial special cases.

Dipole: +q and −q, separation 2a, moment p = 2qa. For P at (r, θ) with r >> a: using geometry r₁ ≈ r − a cosθ, r₂ ≈ r + a cosθ. V = (q/4πε₀)(1/r₁ − 1/r₂).
Binomial approximation: 1/r₁ ≈ (1/r)(1 + a cosθ/r), 1/r₂ ≈ (1/r)(1 − a cosθ/r). Substituting: V = q(2a cosθ)/(4πε₀r²) = p cosθ/(4πε₀r²) = p·r̂/(4πε₀r²).
Axial point (θ = 0 or π): V = ±p/(4πε₀r²). Positive at θ = 0 (near +q side); negative at θ = π.
Equatorial point (θ = π/2): cos(π/2) = 0, so V = 0. Contributions from +q and −q cancel exactly.
Contrast with single charge: V ∝ 1/r; dipole V ∝ 1/r² (faster falloff) and depends on θ (not spherically symmetric).

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 3

Long Form

Describe the six key electrostatic properties of a conductor in static equilibrium and explain the principle of electrostatic shielding.

E = 0 inside: free electrons redistribute until internal field vanishes. If E ≠ 0, charges drift — contradicting static equilibrium.
E normal to surface: any tangential component would drive surface charges — also contradicting equilibrium. Surface field E = (σ/ε₀)n̂.
Excess charge on surface only: Gauss's law applied to any internal closed surface (where E = 0 gives zero flux) shows no enclosed charge inside.
Potential constant everywhere: since E = 0 inside and no tangential E at surface, no work done moving charge — no potential difference anywhere.
Electrostatic shielding: E = 0 in any charge-free cavity regardless of external fields. Sensitive instruments enclosed in metallic shells are fully protected.
One-directional shielding: exterior not shielded from charges inside cavity. This asymmetry is important in designing Faraday cages for protection.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 4

Long Form

Analyse the effect on capacitance of a parallel plate capacitor when (a) plate separation is halved, (b) plate area doubled, (c) dielectric K inserted fully, (d) a conducting slab of thickness t is inserted.

Base: C₀ = ε₀A/d. (a) d → d/2: C = ε₀A/(d/2) = 2C₀. Halving separation doubles capacitance.
(b) A → 2A: C = ε₀(2A)/d = 2C₀. Doubling area doubles capacitance — more charge stored at same field.
(c) Dielectric K fully inserted: C = Kε₀A/d = KC₀. Polarisation reduces V for same Q; since K > 1, C always increases.
(d) Conducting slab of thickness t: field inside conductor is zero; effective gap = d − t. C = ε₀A/(d−t) > C₀. For negligible t, C ≈ C₀.
Summary: C = ε₀A/d shows C increases with A, decreases with d, multiplies by K for dielectric. These effects are independent and multiplicative.

ByteNotes

Long Answer

Medium difficulty • Structured explanation

Medium

Question 5

Long Form

Compare series and parallel combinations of capacitors — derive equivalent capacitance formulae and state practical situations where each is preferred.

Series: all capacitors carry same charge Q. V = V₁ + V₂ + ... = Q/C₁ + Q/C₂ + ... Dividing by Q: 1/C = 1/C₁ + 1/C₂ + ... + 1/Cₙ. C_series < smallest individual C.
Parallel: all at same potential V. Q = Q₁ + Q₂ + ... = C₁V + C₂V + ... Dividing by V: C = C₁ + C₂ + ... + Cₙ. C_parallel > largest individual C.
Key difference: series — same Q, V divides; parallel — same V, Q divides. In series, smallest C gets largest voltage share.
Practical series use: when higher voltage tolerance needed (voltage shared among capacitors); to obtain smaller net C from larger ones.
Practical parallel use: when larger net capacitance or charge storage is needed at given voltage, e.g., power supply filters.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 6

Long Form

Derive the expression for energy stored in a capacitor. Show that this energy can be viewed as stored in the electric field and derive the energy density expression.

Charging from 0 to Q: when charge Q' is on capacitor, potential = Q'/C. Work to transfer dQ': dW = (Q'/C)dQ'.
Integrate: W = ∫₀Q (Q'/C)dQ' = Q²/(2C). This is the stored energy U = Q²/(2C) = (1/2)CV² = (1/2)QV.
Energy resides in the field between plates. For parallel plate capacitor (area A, sep d): U = (σ²/2ε₀) × Ad.
Using E = σ/ε₀: U = (ε₀E²/2) × Ad = (1/2)ε₀E² × Volume.
Energy density u = (1/2)ε₀E². This general result holds for any electric field — not just parallel plate capacitors.

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 7

Long Form

Explain polarisation of a dielectric and contrast the behaviour of a conductor and a dielectric in an external electric field.

In dielectric (non-conductor), charges are bound. External field E₀ displaces charges in non-polar molecules (induced dipole) or partially aligns polar molecules' permanent dipoles.
Net dipole moment per unit volume = polarisation P = ε₀χeE. This creates induced surface charges ±σp on dielectric surfaces.
Opposing field E_in from σp partially cancels E₀. Net field inside dielectric: E = E₀ − E_in = E₀/K, where K > 1.
In conductor: free charges redistribute until E_inside = 0 completely (E₀ fully cancelled by induced charges).
Key difference: conductor — E_inside = 0 (complete cancellation); dielectric — E_inside = E₀/K > 0 (partial reduction). Hence conductors shield; dielectrics merely reduce the field.

ByteNotes

Long Answer

Medium difficulty • Structured explanation

Medium

Question 8

Long Form

Derive the expression for potential energy of a system of two charges in an external electric field and explain the physical significance of each term.

Work bringing q₁ to r₁ in external field: W₁ = q₁V(r₁). This is stored as q₁V(r₁).
Work bringing q₂ to r₂: against external field gives q₂V(r₂); against q₁'s field gives q₁q₂/(4πε₀r₁₂).
Total potential energy: U = q₁V(r₁) + q₂V(r₂) + q₁q₂/(4πε₀r₁₂).
First term q₁V(r₁): interaction energy of q₁ with external field. Second term q₂V(r₂): interaction of q₂ with external field.
Third term q₁q₂/(4πε₀r₁₂): mutual interaction between q₁ and q₂, independent of external field. Assembly order doesn't matter (conservative force).

ByteNotes

Long Answer

Medium difficulty • Structured explanation

Medium

Question 9

Long Form

Explain equipotential surfaces with examples for (a) single charge, (b) uniform field, (c) dipole. State two important properties with proof.

Single charge Q: V = Q/(4πε₀r) = constant when r = constant. Equipotential surfaces are concentric spheres centred at Q. Field lines are radial and perpendicular to spheres.
Uniform field E along x-axis: V = −Ex + const. Constant V means constant x — surfaces are planes parallel to y-z plane (⊥ to E).
Dipole: V = 0 equatorial plane; other surfaces are complex closed curves. Near charges, surfaces resemble single-charge spheres; farther away they distort.
Property 1 (proof): E always perpendicular to equipotential. If E had a tangential component, it would do work moving a charge along the surface — contradicting V = constant.
Property 2: Closer-spaced surfaces indicate stronger field (E = |dV/dl| is large where dl is small for same dV).

ByteNotes

Long Answer

Hard difficulty • Structured explanation

Hard

Question 10

Long Form

A charged capacitor is disconnected from the battery and a dielectric K is inserted. Analyse what happens to Q, V, C, E, and U. Explain why energy decreases.

Disconnected → Q constant = C₀V₀ (no charge path). Q does not change.
C increases: C = KC₀ (dielectric polarises, reduces net field, increases capacitance by K).
V decreases: V = Q/C = Q/(KC₀) = V₀/K (same Q, larger C).
E decreases: E = V/d = V₀/(Kd) = E₀/K. The polarisation charges create opposing field E_in = E₀(1 − 1/K).
U decreases: U = Q²/(2C) = Q²/(2KC₀) = U₀/K. The energy reduction U₀(1 − 1/K) represents work done by the electrostatic force drawing the dielectric into the capacitor — the system spontaneously lowers its energy.

ByteNotes

Premium Access

Unlock all Long Questions

You are viewing 10 free long questions from this chapter. Upgrade to Premium to unlock the remaining 10 questions.

Upgrade to Premium10 questions locked

Chapter sections

Chapter 2: Electrostatic Potential and Capacitance | Long Questions

Derive an expression for electrostatic potential at distance r from a point charge Q. Explain how potential and electric field differ in their distance dependence.

Derive the expression for potential due to an electric dipole at a general point P(r, θ). Discuss axial and equatorial special cases.

Describe the six key electrostatic properties of a conductor in static equilibrium and explain the principle of electrostatic shielding.

Analyse the effect on capacitance of a parallel plate capacitor when (a) plate separation is halved, (b) plate area doubled, (c) dielectric K inserted fully, (d) a conducting slab of thickness t is inserted.

Compare series and parallel combinations of capacitors — derive equivalent capacitance formulae and state practical situations where each is preferred.

Derive the expression for energy stored in a capacitor. Show that this energy can be viewed as stored in the electric field and derive the energy density expression.

Explain polarisation of a dielectric and contrast the behaviour of a conductor and a dielectric in an external electric field.

Derive the expression for potential energy of a system of two charges in an external electric field and explain the physical significance of each term.

Explain equipotential surfaces with examples for (a) single charge, (b) uniform field, (c) dipole. State two important properties with proof.

A charged capacitor is disconnected from the battery and a dielectric K is inserted. Analyse what happens to Q, V, C, E, and U. Explain why energy decreases.

Unlock all Long Questions