Chapter 1: Electric Charges and Fields Long Questions | physics Class 12 CBSE

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Chapter 1: Electric Charges and Fields Long Questions | physics Class 12 CBSE | ByteNotes.in

Long Answer

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Question 1

Long Form

Derive the expression for the electric field of an electric dipole at a point on its axial line. State how the field direction relates to the dipole moment.

Consider a dipole with charges +q at B and −q at A, separated by 2a. Let P be a point on the axis at distance r from the centre O. The field at P due to +q is E₊ = q/[4πε₀(r−a)²] directed away from B (along p̂), and due to −q is E₋ = q/[4πε₀(r+a)²] directed toward A (along −p̂).
The net axial field E = E₊ − E₋ = (q/4πε₀)[1/(r−a)² − 1/(r+a)²] along p̂, which simplifies to E = q × 4ar / [4πε₀(r²−a²)²] along p̂.
For r >> a, (r²−a²)² ≈ r⁴, and using p = 2qa, the field becomes E = 2p/[4πε₀r³] along p̂.
The field on the axial line is in the same direction as the dipole moment vector p, i.e., from −q toward +q.
The field falls off as 1/r³ (not 1/r² as for a point charge), showing that the dipole field diminishes faster with distance.

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Question 2

Long Form

Derive the expression for the electric field of an electric dipole at a point on its equatorial plane. Compare this result with the axial field.

Let P be a point on the equatorial plane at distance r from the centre O of the dipole (+q at B and −q at A). Since BP = AP = √(r²+a²), the magnitudes of fields due to +q and −q at P are equal: E₊ = E₋ = q/[4πε₀(r²+a²)].
The components of E₊ and E₋ perpendicular to the dipole axis (normal to p̂) are equal and opposite, so they cancel. The components along the dipole axis (along −p̂) add up.
Each contributes a component E cosθ along −p̂, where cosθ = a/√(r²+a²). Net equatorial field: E = −2qacosθ/[4πε₀(r²+a²)] p̂ = −p/[4πε₀(r²+a²)^(3/2)] p̂.
For r >> a: E_equatorial ≈ −p/[4πε₀r³], directed opposite to p. At the same distance, E_axial = 2p/[4πε₀r³], so E_axial = 2 × E_equatorial in magnitude.
Both fields decrease as 1/r³ unlike the 1/r² dependence of a single charge; the equatorial field is directed antiparallel to p, while the axial field is parallel to p.

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Question 3

Long Form

Using Gauss's law, derive an expression for the electric field due to an infinitely long straight wire with uniform linear charge density λ.

An infinitely long straight wire with uniform linear charge density λ has cylindrical symmetry. By symmetry, the electric field must be radial (perpendicular to the wire) and have the same magnitude at all points equidistant from the wire.
Choose a cylindrical Gaussian surface of radius r and length l coaxial with the wire. The total flux through this surface is the sum of fluxes through the two circular ends and the curved cylindrical surface.
The electric field is parallel to the two flat ends (normal to E), so flux through them is zero. On the curved surface, E is perpendicular to the surface and constant in magnitude, giving flux = E × 2πrl.
The charge enclosed by the Gaussian surface is λl. Applying Gauss's law: E × 2πrl = λl/ε₀, which gives E = λ/(2πε₀r).
In vector form, E = (λ/2πε₀r) n̂, where n̂ is the radial unit vector. The field is directed radially outward if λ > 0 and varies as 1/r with distance.

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Question 4

Long Form

Using Gauss's law, derive the electric field inside and outside a uniformly charged thin spherical shell. What is the significance of the zero field inside?

For a spherical shell of radius R with uniform surface charge density σ (total charge q = 4πR²σ), consider a Gaussian sphere of radius r centred at the shell's centre.
Outside the shell (r > R): The entire charge q is enclosed. By spherical symmetry, E is radial and constant on the Gaussian surface. Gauss's law gives E × 4πr² = q/ε₀, so E = q/(4πε₀r²) outward — identical to the field of a point charge q at the centre.
Inside the shell (r < R): No charge is enclosed by the Gaussian sphere. Gauss's law gives E × 4πr² = 0, hence E = 0 everywhere inside the shell.
The zero field inside the shell is a direct and profound consequence of the inverse-square law; experimental verification of this result confirms the 1/r² dependence in Coulomb's law.
This result has practical significance: a metallic enclosure (Faraday cage) shields its interior from external electric fields, because free charges rearrange on the surface to ensure zero field inside — directly analogous to the zero-field result for a charged spherical shell.

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Question 5

Long Form

Derive the electric field due to a uniformly charged infinite plane sheet with surface charge density σ using Gauss's law. Explain why the field is independent of distance.

Consider an infinite plane sheet with uniform surface charge density σ. By symmetry, the electric field E must be perpendicular to the sheet on both sides and depend only on the perpendicular distance from the sheet.
Choose a cylindrical (pillbox) Gaussian surface of cross-sectional area A with its axis perpendicular to the sheet, extending equally on both sides. The curved cylindrical surface has E parallel to it, contributing zero flux.
The two flat faces each have E perpendicular to them (E parallel to n̂). Each contributes flux EA, giving total flux = 2EA.
The charge enclosed is σA. Gauss's law gives 2EA = σA/ε₀, so E = σ/(2ε₀), directed away from the sheet if σ > 0.
The field is independent of the distance from the sheet because the Gaussian surface area (and hence enclosed charge) scales with the same area A regardless of how far the faces are placed; this is a consequence of the infinite planar extent of the sheet and holds approximately for large finite sheets far from the edges.

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Question 6

Long Form

Analyse the behaviour of an electric dipole placed in a uniform and a non-uniform external electric field. Include torque expression and its consequences.

In a uniform external field E, the force on +q is qE and on −q is −qE. The net force is zero, so the dipole undergoes no translational motion. However, the two forces act at different points, constituting a couple that produces a torque.
The magnitude of the torque is τ = qE × 2a sinθ = pE sinθ, where θ is the angle between p and E. In vector form, τ = p × E. This torque tends to align the dipole along the field.
The torque is zero when p is parallel (θ = 0, stable equilibrium) or antiparallel (θ = 180°, unstable equilibrium) to E; it is maximum when θ = 90°.
In a non-uniform field, the forces on +q and −q have different magnitudes, so the net force is non-zero. When p is parallel to E, the dipole moves toward stronger field; when antiparallel, toward weaker field.
This non-uniform field effect explains why a charged comb attracts neutral paper: the comb's non-uniform field induces a dipole in the paper and exerts a net force toward the comb, overcoming the weaker repulsion on the farther end.

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Question 7

Long Form

State and explain the superposition principle for electric forces. Derive the expression for the total force on charge q₁ in a system of n charges.

The superposition principle states that the total force on any one charge in a system of multiple charges equals the vector sum of all individual Coulomb forces exerted on it by each of the other charges, with each force computed as if the remaining charges were not present.
Consider n charges q₁, q₂, ..., qₙ. The force on q₁ due to q₂ alone is F₁₂ = (1/4πε₀)(q₁q₂/r²₁₂) r̂₁₂; similarly, F₁₃ = (1/4πε₀)(q₁q₃/r²₁₃) r̂₁₃, and so on.
By superposition, the total force F₁ = F₁₂ + F₁₃ + ... + F₁ₙ = (q₁/4πε₀) Σᵢ (qᵢ/r²₁ᵢ) r̂₁ᵢ for i = 2 to n.
Each individual force is pairwise and independent of all other charges present. No additional multi-body forces arise when more than two charges are present simultaneously.
All of electrostatics — from the electric field to Gauss's law — is ultimately a consequence of Coulomb's law combined with the superposition principle, demonstrating the foundational importance of this principle.

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Question 8

Long Form

Explain the physical significance of the concept of electric field. Why is it indispensable in time-varying situations even though it may seem unnecessary in electrostatics?

In electrostatics, the electric field at a point is a convenient intermediate quantity that characterises the electrical environment created by a source charge configuration, independent of any test charge placed there.
In static situations, the electric field is not strictly necessary: one could compute forces directly using Coulomb's law and superposition. The field merely provides an elegant way to describe the effect of a source charge at every point in space.
However, in time-dependent situations involving accelerated charges, a fundamental problem arises: electromagnetic effects travel at the finite speed of light c. There is a time delay between the cause (motion of source charge) and the effect (force on distant charge).
The field concept elegantly resolves this: the accelerated source charge creates time-varying electromagnetic fields (waves), which propagate at speed c and carry energy, ultimately reaching and exerting a force on a distant charge.
Electric (and magnetic) fields are thus not merely mathematical tools but physical entities with their own dynamics and energy content, capable of existing and propagating even without charges. The concept was first introduced by Faraday and is now central to all of modern physics.

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Question 9

Long Form

Describe the three types of continuous charge distributions. Write expressions for the electric field contribution from each type of charge element.

Linear charge density λ = ΔQ/Δl (C/m): applicable to thin wires or rods, where ΔQ is the charge on a small length element Δl. The element contributes a field dE = (1/4πε₀)(λΔl/r'²) r̂' at a field point.
Surface charge density σ = ΔQ/ΔS (C/m²): applicable to charged surfaces, where ΔQ is the charge on small area element ΔS. The element contributes dE = (1/4πε₀)(σΔS/r'²) r̂' at a field point.
Volume charge density ρ = ΔQ/ΔV (C/m³): applicable to charged bulk material, where ΔQ = ρΔV is the charge in a small volume element ΔV. The element contributes ΔE = (1/4πε₀)(ρΔV/r'²) r̂' at a field point.
In all cases, r' is the distance from the charge element to the field point and r̂' is the unit vector from the element to the point. The total field is obtained by summing (integrating) over all elements using the superposition principle.
These macroscopic densities smooth out the discrete microscopic nature of charge (quantisation) and treat charge as continuously distributed — analogous to how bulk mass density treats matter, ignoring its atomic structure.

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Question 10

Long Form

Discuss the important properties of Gauss's law. Explain how a violation of Gauss's law would indicate a departure from Coulomb's inverse-square law.

Gauss's law (φ = q/ε₀) is true for any closed surface regardless of its shape, size, or position in space, as long as the enclosed charge q is correctly accounted for.
The electric field E on the left side of Gauss's law is due to all charges — both inside and outside the Gaussian surface — but only the charges inside contribute to the right side (q/ε₀).
Gauss's law is most useful when the charge distribution has sufficient symmetry (spherical, cylindrical, planar) allowing a Gaussian surface to be chosen where E is constant and perpendicular to (or parallel to) every part of the surface.
The Gaussian surface must not pass through any discrete point charge (because E is undefined at the charge location), but it may pass through a continuous charge distribution.
Gauss's law is fundamentally based on the inverse-square dependence of Coulomb's law. If E followed a different power law (e.g., 1/r^n with n ≠ 2), the total flux through a sphere enclosing a charge would depend on the sphere's radius, violating Gauss's law. Any experimental violation of Gauss's law would therefore signal a breakdown of Coulomb's inverse-square law.

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Chapter 1: Electric Charges and Fields | Long Questions

Derive the expression for the electric field of an electric dipole at a point on its axial line. State how the field direction relates to the dipole moment.

Derive the expression for the electric field of an electric dipole at a point on its equatorial plane. Compare this result with the axial field.

Using Gauss's law, derive an expression for the electric field due to an infinitely long straight wire with uniform linear charge density λ.

Using Gauss's law, derive the electric field inside and outside a uniformly charged thin spherical shell. What is the significance of the zero field inside?

Derive the electric field due to a uniformly charged infinite plane sheet with surface charge density σ using Gauss's law. Explain why the field is independent of distance.

Analyse the behaviour of an electric dipole placed in a uniform and a non-uniform external electric field. Include torque expression and its consequences.

State and explain the superposition principle for electric forces. Derive the expression for the total force on charge q₁ in a system of n charges.

Explain the physical significance of the concept of electric field. Why is it indispensable in time-varying situations even though it may seem unnecessary in electrostatics?

Describe the three types of continuous charge distributions. Write expressions for the electric field contribution from each type of charge element.

Discuss the important properties of Gauss's law. Explain how a violation of Gauss's law would indicate a departure from Coulomb's inverse-square law.

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