Equipotential Surfaces and Relation Between Field and Potential

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

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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)

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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.

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Equipotential Surfaces and Relation Between Field and Potential

Two charges \( 14 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are at \( (3, 0, 0) \) and \( (-3, 0, 0) \, \text{cm} \)

**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. Distance to midpoint = 0.03 m. V = 9 × 10⁹ ( (14 × 10⁻⁶/0.03) + (-6 × 10⁻⁶/0.03) ) = 9 × 10⁹ × (8 × 10⁻⁶/0.03) . V = 9 × 10⁹ × (8 × 10⁻⁶/0.03) = 2.4 × 10⁶ V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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Equipotential Surfaces and Relation Between Field and Potential

A charge of \( 2 \, \mu\text{C} \) is moved from infinity to a point with potential \( 500 \, \text{V} \). What is the w

**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. Work done = Potential energy = q V . W = 2 × 10⁻⁶ × 500 = 10⁻³ J = 1 mJ . 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 mJ follows, reflecting potential-capacitance relations.

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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.

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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

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Equipotential Surfaces and Relation Between Field and Potential

A spherical conductor of radius 8 cm has a charge of \( 4 \times 10^{-8} \, \text{C} \). What is the electric field at 1

**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. For r = 0.12 m > R = 0.08 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (4 × 10⁻⁸/(0.12)²) = 9 × 10⁹ × (4 × 10⁻⁸/0.0144) = 2.5 × 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

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Equipotential Surfaces and Relation Between Field and Potential

A parallel plate capacitor has plates of area \( 0.06 \, \text{m}^2 \) and separation 0.3 mm in air. What is its capacit

**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. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.06/0.3 × 10⁻³) = 1.77 × 10⁻⁹ F = 1770 pF . 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 1770 pF follows, reflecting potential-capacitance relations.

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Equipotential Surfaces and Relation Between Field and Potential

Two charges \( 6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 12 cm apart. What is the potential energy of the syst

**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. U = (1/4 π ε₀) (q₁ q₂/r) = 9 × 10⁹ × (6 × 10⁻⁶ × (-3 × 10⁻⁶)/0.12) = 9 × 10⁹ × (-18 × 10⁻¹²/0.12) = -1.35 J . 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 -1.35 J follows, reflecting potential-capacitance relations.

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Equipotential Surfaces and Relation Between Field and Potential

A parallel plate capacitor has plates of area \( 0.09 \, \text{m}^2 \) and separation 0.5 mm in air. What is its capacit

**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. C = (ε₀ A/d) = (8.85 × 10⁻¹² × 0.09/0.5 × 10⁻³) = 1.593 × 10⁻⁹ F = 1593 pF . 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 1593 pF follows, reflecting potential-capacitance relations.

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Equipotential Surfaces and Relation Between Field and Potential

Three charges \( +9 \, \mu\text{C} \), \( -6 \, \mu\text{C} \), and \( +4 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (

**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. Distances: r₁ = √(9² + 9²) = 9√(2) m , r₂ = 9 m , r₃ = 9 m . V = 9 × 10⁹ ( (9 × 10⁻⁶/9√(2)) + (-6 × 10⁻⁶/9) + (4 × 10⁻⁶/9) ) . V = 9 × 10⁹ ( (9 × 10⁻⁶/12.728) - (6 × 10⁻⁶/9) + (4 × 10⁻⁶/9) ) . V

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