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#conductors

19 public questions tagged with this topic.

Combination of Capacitors - Series and Parallel

In a system of two identical conductors initially charged differently and then connected by a wire, why does the final e

**Series combination** of capacitors has same charge Q on each because connected end-to-end, single path for charge flow, induced charges equal, total voltage V = Σ V_i = Q Σ 1/C_i, so 1/C_eq = Σ 1/C_i. Different potential differences V_i = Q/C_i inversely proportional to C_i, smaller C gets larger V. Initially, the conductors have charges Q₁ and Q₂ , with energy U_i = (Q₁²/2C) + (Q₂²/2C) . Upon connection, charge redistributes to equal potentials, total charge Q₁ + Q₂ splits equally ( (Q₁ + Q₂/2) each), so final energy U_f = 2 × (((Q₁ + Q₂/2))²/2C) = ((Q₁ + Q₂)²/4C) . Typically, (Q₁ +

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Combination of Capacitors - Series and Parallel

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Equipotential Surfaces and Relation Between Field and Potential

A conductor has a surface charge density of \( 4.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just ou

**Equipotential through midpoint** of +q and -q is perpendicular bisector, V=0 everywhere on it because contributions k q/r and k(-q)/r cancel. For two opposite charges, potential at midpoint zero, but field non-zero, pointing from positive to negative, illustrating vector vs scalar nature. E = (sigma/ε₀) = (4.5 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 5.085 × 10⁵ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 5.085 × 10⁵ N/C follows, reflecting potential-capacitance relations.

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

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Equipotential Surfaces and Relation Between Field and Potential

A conductor has a surface charge density of \( 5.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just ou

**Equipotential surface** is surface where V constant, no work done moving charge along it because W = q ΔV =0 when ΔV=0. Electric field E always perpendicular to equipotential surface, direction from higher to lower potential, magnitude E = -dV/dr, steeper potential gradient means stronger field. E = (sigma/ε₀) = (5.5 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 6.215 × 10⁵ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 6.215 × 10⁵ N/C follows, reflecting potential-capacitance relations.

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

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Equipotential Surfaces and Relation Between Field and Potential

A conductor has a surface charge density of \( 4 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outs

**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. E = (sigma/ε₀) = (4 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 4.52 × 10⁵ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U = ½ C V² and common potential V = Q_total/C_total, result 4.52 × 10⁵ N/C follows, reflecting potential-capacitance relations.

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

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Conductors, Electrostatic Shielding and Dielectrics

Why can't the electric field inside a conductor in electrostatic equilibrium have a tangential component along its surfa

**Conductor in electrostatic equilibrium** has E=0 inside, charges reside on surface, potential constant throughout conductor. Hollow shell with no internal charge has zero field inside cavity, even if external field present, charges on outer surface screen interior, principle used in Faraday cage. In electrostatic equilibrium, the electric field inside a conductor is zero, and charges reside on the surface. If the electric field had a tangential component along the surface, it would exert a force on the free charges, causing them to move along the surface. This movement would contradict the condition of equilibrium where no net motion of charges occurs. Therefore, the field must

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

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Potentiometer, Conductivity and Special Cases

Why does the current density in a conductor remain uniform across its cross-section under steady-state conditions?

**Potential difference drop** across resistor when current flows because charges lose potential energy qV = I² R t as heat, field E = V/l drives drift, maintaining current. At very high E, velocity saturation or heating changes τ, causing non-ohmic behaviour. Current density ( j = I / A ) is uniform if the current distributes evenly. In steady state, charge conservation (Kirchhoff’s junction rule) ensures a constant current through a uniform conductor, making j consistent across the area. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε - I r and

Ref: NCERT > Physics Book > Current Electricity > Potentiometer, Conductivity and Special Cases

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Electric Current, Drift Velocity and Mobility

A conductor has a resistivity of \( 1.0 \times 10^{-7} \, \Omega \text{m} \) and \( \alpha = 4 \times 10^{-3} \, ^\circ\

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Use: rho_t = rho₀ [1 + α (T - T₀)] . Substitute: rho_t = 1.0 × 10⁻⁷ [1 + 4 × 10⁻³ (85 - 25)] . Calculate: rho_t = 1.0 × 10⁻⁷ [1 + 0.24] = 1.0 × 10⁻⁷ × 1.24 = 1.24 × 10⁻⁷ Ω m . Applying I = n e A v_d, R = ρ l/A, R_t

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Electric Current, Drift Velocity and Mobility

What is the significance of the relaxation time in the context of electron drift in a conductor?

**Mobility** μ = v_d/E = e τ/m, τ relaxation time (s), measures ease of electron drift under field E (V/m). Conductivity σ = n e μ = 1/ρ, linking microscopic τ to macroscopic resistivity, explaining why metals conduct well due to large n and τ. Relaxation time ( tau ) is the average time between electron collisions with lattice ions. It determines drift velocity ( v_d = e E tau / m ), affecting how quickly electrons respond to the field and thus the conductor’s conductivity. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel,

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Electric Current, Drift Velocity and Mobility

Why does the drift velocity of electrons in a conductor remain constant despite continuous acceleration by an electric f

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Electrons accelerate due to the electric field but collide with lattice ions, losing momentum. These collisions occur at random intervals, and the average time between collisions ( tau ) stabilizes the drift velocity ( v_d = e E tau / m ), balancing acceleration with energy loss. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Electric Current, Drift Velocity and Mobility

Why does the drift velocity of electrons in a conductor remain much smaller than their thermal velocity?

**Current and drift relation** I = n e A v_d shows current proportional to drift velocity and area. For A=6×10⁻⁷ m², I=1.8 A, n=8.5×10²⁸ m⁻³, v_d =1.8/(8.5×10²⁸×1.6×10⁻¹⁹×6×10⁻⁷)=2.2×10⁻⁴ m/s, illustrating small drift speed even for ampere currents. Drift velocity ( v_d = e E tau / m ) is small because E is typically weak and tau is short due to frequent collisions, whereas thermal velocity arises from random motion at high speeds (proportional to √(k T / m) ), unaffected by the field. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0, ΣV=0, R_eq series/parallel, V = ε

Ref: NCERT > Physics Book > Current Electricity > Electric Current, Drift Velocity and Mobility

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Electric Dipole - Moment, Field and Torque

Why can’t the electric field inside a charged insulator be zero, unlike in a conductor?

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. In insulators, charges are fixed and cannot move to cancel an internal field. If charges are present inside, they generate a field that persists, as there are no free charges to redistribute and neutralize it, unlike in conductors. Substituting values gives Lack of free charges, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Dipole - Moment, Field and Torque

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