#charged particle motion

Conductors, Electrostatic Shielding and Dielectrics

In a system where a charged particle moves along a curved path between two points with different potentials, what can be

**Dielectric polarization** when slab inserted, bound charges appear reducing effective field, capacitance increases by factor K, potential difference for constant charge V = Q/C decreases, for constant voltage charge increases. Dielectric constant K = ε/ε₀ >1, e.g., K≈5 for glass. The work done by the electric field on a charge q moving between two points is W = q (Vfiₙₐl - Viₙitiₐl) , where V is the potential. Since the electrostatic field is conservative, this work depends only on the potential difference between the points and not on the path taken (straight or curved). Thus, the work done is path-independent and determined solely by the

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Conductors, Electrostatic Shielding and Dielectrics

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Equipotential Surfaces and Relation Between Field and Potential

A charged particle moves along a path between two points in an electric field where the potential difference is zero. Wh

**Relation E = -∇V** shows field points down potential gradient. For system of opposite charges close together, equipotential near midpoint between them has V≈0, but shape distorted, not spherical, reflecting superposition of potentials V = k q₁/r₁ + k q₂/r₂. If the potential difference between two points is zero, they lie on the same equipotential surface. The path taken by the particle must lie entirely on this equipotential surface (where V is constant), as any deviation would imply a potential difference. Equipotential surfaces are perpendicular to the electric field, so the particle moves perpendicular to the field direction, doing no work against the field along

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Equipotential Surfaces and Relation Between Field and Potential

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Electric Potential and Potential Difference

A positively charged particle moves from a point of higher potential to a point of lower potential in an electrostatic f

**Electric potential** V = k Q/r, k = 1/(4π ε₀)=9×10⁹ N·m²/C², Q charge (C), r distance (m), scalar, potential at surface of spherical conductor radius R, V = k Q/R. Potential difference ΔV = V_B - V_A = -∫ E·dl, work per unit charge to move charge without acceleration. The electric field does work on a charge as W = q Δ V . For a positive charge moving from higher potential ( V_h ) to lower potential ( V_l ), Δ V = V_l - V_h < 0 . Since q > 0 , the work done by the field W = q (V_l

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Electric Potential and Potential Difference

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Solenoid, Toroid and Ampere's Law

A proton moves at \( 4 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.05 \, \text{T} \). What is the radi

**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 4 × 10⁷/1.6 × 10⁻¹⁹ × 0.05) = (6.68 × 10⁻²⁰/8 × 10⁻²¹) = 8.35 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I A B

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Solenoid, Toroid and Ampere's Law

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Magnetic Force on Moving Charge - Lorentz Force and Motion

A proton moves at \( 3.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.08 \, \text{T} \). What is the ra

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 3.5 × 10⁷/1.6 × 10⁻¹⁹ × 0.08) = (5.845 × 10⁻²⁰/1.28 × 10⁻²⁰) = 4.5664 ≈ 4.57 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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PHYSICS

An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 ×

Given: An electron moves at 8 × 10⁶ m/s perpendicular to a field of 0.15 T . What is the radius of its path? (Mass = 9.1 × 10⁻³¹ kg, charge = 1.6 × 10⁻¹⁹ C ) These values define the system as per NCERT data. Formula: r = mv/qB. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: r = frac9.1 × 10⁻³¹ × 8 × 10⁶¹.6 × 10⁻¹⁹ × 0.15 = frac7.28 × 10⁻²⁴².4 × 10⁻²⁰= 3.033 × 10⁻⁴ m approx 0.0303 cm . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

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