#conductor

Energy Stored in Capacitor and Energy Density

When a conductor is placed in an external electric field, why does the potential throughout its volume become constant i

**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. In electrostatic equilibrium, the electric field inside a conductor is zero because free charges rearrange to cancel any internal field. Since the electric field is the negative gradient of potential ( E = -(dV/dr) ), if E = 0 , the potential gradient must be zero, implying the potential V is constant

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Energy Stored in Capacitor and Energy Density

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Equipotential Surfaces and Relation Between Field and Potential

When a charged conductor is placed in contact with an uncharged conductor of smaller size, why does the smaller conducto

**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. When two conductors reach equilibrium, they share charge and attain the same potential. Potential on a conductor's surface is V = (Q/4 π ε₀ R) for a sphere (or similar for other shapes). For equal V , (Q₁/R₁) = (Q₂/R₂) , so Q ∝ R . Surface charge density sigma = (Q/4 π R²) , so sigma ∝ (Q/R²)

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 \( 1.5 \times 10^{-6} \, \text{C/m}^2 \). 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) = (1.5 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 1.695 × 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 1.695 × 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 does the electric field inside a conductor remain zero when a non-symmetric external field is applied?

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. In electrostatic equilibrium, the electric field inside a conductor must be zero, regardless of the external field’s symmetry. Free charges redistribute on the conductor’s surface to cancel the external field inside. For a non-symmetric field, the surface charge distribution adjusts accordingly

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

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Conductors, Electrostatic Shielding and Dielectrics

In an electrostatic shielding scenario, a cavity inside a conductor contains a charge. What can be said about the electr

**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. In electrostatic equilibrium, a conductor with a cavity shields the external region from charges inside the cavity. The charge inside induces an equal and opposite charge on the inner surface of the cavity, and the conductor adjusts its outer surface charge distribution to ensure the field inside the conductor (outs

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

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a conductor vanish in electrostatic equilibrium, even if the conductor is irregularly

**Electrostatic shielding** inside hollow conducting shell field zero when no charges inside, regardless of external field, because free charges redistribute on outer surface to cancel external field inside conductor, E=0 inside material in equilibrium, consequence of Gauss's law and conductor property. In electrostatic equilibrium, free charges in a conductor redistribute to cancel any internal electric field. If there were a field inside, it would exert a force on the free charges, causing them to move until the field becomes zero everywhere inside. This principle holds regardless of shape b

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

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Electric Potential and Potential Difference

A conductor has a surface charge density of \( 3 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outs

**Potential due to point charge** V = (1/4π ε₀)Q/r, positive Q gives positive V, negative gives negative. At r=0.3 m from Q=3×10⁻⁸ C, V=9×10⁹×3×10⁻⁸/0.3=900 V. Work done moving charge q from infinity (V=0) to point where V=40 V is W = q V =9×10⁻⁶×40=3.6×10⁻⁴ J. E = (sigma/ε₀) = (3 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 3.39 × 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 3.39 × 10⁵ N/C follows, reflecting potential-capacitance relations.

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

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Electric Potential and Potential Difference

A conductor has a surface charge density of \( 5 \times 10^{-7} \, \text{C/m}^2 \). What is the electric field just outs

**Spherical conductor** behaves as point charge outside, potential V = k Q/R at surface, field E = k Q/r² for r>R, zero inside for rR, so E = k Q/r² =9×10⁹×6×10⁻⁸/0.16=3375 N/C. E = (sigma/ε₀) = (5 × 10⁻⁷/8.85 × 10⁻¹²) ≈ 5.65 × 10⁴ 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 5.65 × 10⁴ N/C follows, reflecting potential-capacitance relations.

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

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Electric Potential and Potential Difference

A conductor has a surface charge density of \( 2.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just ou

**Potential due to point charge** V = (1/4π ε₀)Q/r, positive Q gives positive V, negative gives negative. At r=0.3 m from Q=3×10⁻⁸ C, V=9×10⁹×3×10⁻⁸/0.3=900 V. Work done moving charge q from infinity (V=0) to point where V=40 V is W = q V =9×10⁻⁶×40=3.6×10⁻⁴ J. E = (sigma/ε₀) = (2.5 × 10⁻⁶/8.85 × 10⁻¹²) ≈ 2.82 × 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 2.82 × 10⁵ N/C follows, reflecting potential-capacitance relations.

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

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Wheatstone Bridge and Meter Bridge

A conductor of length \( 1 \, \text{m} \) and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance

**Unbalanced bridge** has potential difference between galvanometer nodes, current direction determined by which node higher potential, i.e., if R₁/R₂ > R₃/R₄, left node higher, current flows one way, else opposite. Galvanometer deflection indicates imbalance magnitude. Resistance is given by R = (rho l/A) . Rearrange: rho = (R A/l) . Given: R = 2 Ω , A = 1 × 10⁻⁶ m² , l = 1 m . Substitute: rho = (2 × 1 × 10⁻⁶/1) = 2 × 10⁻⁶ Ω m . 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 > Wheatstone Bridge and Meter Bridge

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Temperature Dependence of Resistance and Resistivity

In a conductor, if the electric field is suddenly doubled while keeping the conductor's properties unchanged, what happe

**Temperature dependence** of resistance R_t = R₀[1+α(T-T₀)], α temperature coefficient (per °C), R₀ resistance at T₀ (Ω). For metals α positive ≈10⁻³ /°C, resistance increases with temperature because τ decreases due to increased phonon scattering, n nearly constant. Drift velocity ( v_d ) is given by v_d = (e E tau/m) , where E is the electric field, e is the electron charge, tau is the relaxation time, and m is the electron mass. If E is doubled, v_d becomes 2v_d , assuming tau and other properties remain constant. Applying I = n e A v_d, R = ρ l/A, R_t = R₀[1+αΔT], Kirchhoff's ΣI=0,

Ref: NCERT > Physics Book > Current Electricity > Temperature Dependence of Resistance and Resistivity

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Electric Current, Drift Velocity and Mobility

Why does a conductor exhibit zero net current in the absence of an electric field?

**Drift velocity** v_d = I/(n e A), I current (A), n number density of conduction electrons (m⁻³) ≈8.5×10²⁸ m⁻³ for copper, e =1.6×10⁻¹⁹ C, A cross-sectional area (m²). Typical v_d ≈10⁻⁴ m/s for 1 A in mm² wire, slow despite fast signal propagation due to electric field establishment. Without an electric field, electrons move randomly due to thermal energy, with no preferred direction. The average velocity of electrons cancels out, resulting in zero net current. 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 P = I²R, evaluation

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

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