#electric dipole

Energy Stored in Capacitor and Energy Density

An electric dipole with moment \( p = 8 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at

**Energy stored in capacitor** U = ½ C V² = ½ Q V = Q²/(2C) (J), C capacitance (F), V voltage (V), Q charge (C). For 4 μF charged to 250 V, U=0.5×4×10⁻⁶×62500=0.125 J. Energy resides in electric field, energy density u = ½ ε₀ E² (J/m³), E field between plates. Along the dipole axis ( θ = 180° ): V = -(1/4 π ε₀) (p/r²) . V = -9 × 10⁹ × (8 × 10⁻⁹/4²) = -9 × 10⁹ × (8 × 10⁻⁹/16) = -4.5 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq

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

View Details

Combination of Capacitors - Series and Parallel

Why does the potential due to an electric dipole fall off as \( 1/r^2 \) at large distances, unlike the potential of a p

**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. The potential due to a point charge falls as 1/r because it behaves as a single source of charge. An electric dipole consists of two equal and opposite charges separated by a small distance, so their potentials partially cancel out at large distances. The net potential depends on the dipole moment and the angle

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

An electric dipole with moment \( p = 7 \times 10^{-9} \, \text{C m} \) lies along the x-axis. What is the potential at

**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. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (7 × 10⁻⁹/4²) = 9 × 10⁹ × (7 × 10⁻⁹/16) = 3.9375 V . 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

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

An electric dipole with moment \( p = 3 \times 10^{-9} \, \text{C m} \) lies along the x-axis. What is the potential at

**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 the equatorial plane ( θ = 90° ): V = (1/4 π ε₀) (p cos θ/r²) . Since cos 90° = 0 , V = 0 V . 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 0 V follows, reflecting potential-capacitance relations.

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 9 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 9 × 10⁻⁹ × 5 × 10⁴ × (cos 90° - cos 180°) . W = 9 × 10⁻⁹ × 5 × 10⁴ × (0 - (-1)) = 9 × 10⁻⁹ × 5 × 10⁴ × 1 = 4.5 × 10⁻⁴ J . Using V = kQ/r, U = k q₁q₂/r,

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 5 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 180^\circ \) in a field

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 5 × 10⁻⁹ × 2 × 10⁵ × (cos 90° - cos 180°) . W = 5 × 10⁻⁹ × 2 × 10⁵ × (0 - (-1)) = 5 × 10⁻⁹ × 2 × 10⁵ × 1 = 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr,

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

A dipole \( p = 10 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \

**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. Work done: W = p E (cos θ₀ - cos θ₁) = 10 × 10⁻⁹ × 4 × 10⁵ × (cos 0° - cos 90°) . W = 10 × 10⁻⁹ × 4 × 10⁵ × (1 - 0) = 4 × 10⁻³ J . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC

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

View Details

Equipotential Surfaces and Relation Between Field and Potential

An electric dipole with moment \( p = 2 \times 10^{-10} \, \text{C m} \) is at the origin, aligned along the x-axis. Wha

**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. Position vector r = (0, 2, 0) , r = 2 m , hatr = (0, 1, 0) . Dipole moment p = (2 × 10⁻¹⁰, 0, 0) . V = (1/4 π ε₀) (p · hatr/r²) = 9 × 10⁹ × ((2 × 10⁻¹⁰) · (0)/2²) = 0 V (since cos θ = 0 , equatorial plane). Using V = kQ/r, U

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

View Details

Equipotential Surfaces and Relation Between Field and Potential

An electric dipole with moment \( p = 2 \times 10^{-9} \, \text{C m} \) lies along the z-axis. What is the potential at

**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. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (2 × 10⁻⁹/4²) = 9 × 10⁹ × (2 × 10⁻⁹/16) = 1.125 V . 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

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

View Details

Potential Energy of System of Charges

A dipole \( p = 3 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 180^\circ \) in a field \

**Sign of potential energy** indicates nature of interaction, like charges U>0, opposite U

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

View Details

Potential Energy of System of Charges

An electric dipole with moment \( p = 5 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. Along the dipole axis ( θ = 0° ): V = (1/4 π ε₀) (p/r²) . V = 9 × 10⁹ × (5 × 10⁻⁹/3²) = 9 × 10⁹ × (5 × 10⁻⁹/9) = 5 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

View Details

Potential Energy of System of Charges

An electric dipole with moment \( p = 6 \times 10^{-9} \, \text{C m} \) lies along the z-axis. What is the potential at

**Potential energy of two charges** U = k q₁ q₂/r, k=9×10⁹ N·m²/C², q₁,q₂ in coulombs, r separation (m), positive for like charges (repulsive, work needed to bring together), negative for opposite (attractive, work released). For 20 μC and -8 μC, 0.2 m apart, U=9×10⁹×20×10⁻⁶×(-8×10⁻⁶)/0.2= -7.2 J. Along the dipole axis ( θ = 180° ): V = -(1/4 π ε₀) (p/r²) . V = -9 × 10⁹ × (6 × 10⁻⁹/2²) = -9 × 10⁹ × (6 × 10⁻⁹/4) = -13.5 V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC parallel, U =

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Potential Energy of System of Charges

View Details