Electric Potential and Potential Difference

Electric Potential and Potential Difference

A spherical conductor of radius 25 cm has a charge of \( 10 \times 10^{-8} \, \text{C} \). What is the electric field at

**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. For r = 0.6 m > R = 0.25 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (10 × 10⁻⁸/(0.6)²) = 9 × 10⁹ × (10 × 10⁻⁸/0.36) = 2.5 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq = ΣC

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Electric Potential and Potential Difference

Two charges \( 8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (6, 0, 0) \) and \( (-6, 0, 0) \, \text{cm} \).

**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. Distance to midpoint = 0.06 m. V = 9 × 10⁹ ( (8 × 10⁻⁶/0.06) + (-4 × 10⁻⁶/0.06) ) = 9 × 10⁹ × (4 × 10⁻⁶/0.06) . V = 9 × 10⁹ × (4 × 10⁻⁶/0.06) = 6 × 10⁵ V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series,

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Electric Potential and Potential Difference

A spherical conductor of radius 30 cm has a charge of \( 12 \times 10^{-8} \, \text{C} \). What is the electric field at

**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. For r = 0.7 m > R = 0.3 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (12 × 10⁻⁸/(0.7)²) = 9 × 10⁹ × (12 × 10⁻⁸/0.49) ≈ 2.204 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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Electric Potential and Potential Difference

Three charges \( +5 \, \mu\text{C} \), \( -2 \, \mu\text{C} \), and \( +3 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (

**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. Distances: r₁ = √(5² + 5²) = 5√(2) m , r₂ = 5 m , r₃ = 5 m . V = 9 × 10⁹ ( (5 × 10⁻⁶/5√(2)) + (-2 × 10⁻⁶/5) + (3 × 10⁻⁶/5) ) . V = 9 × 10⁹ ( (5 × 10⁻⁶/7.07) - (2 × 10⁻⁶/5) + (3 × 10⁻⁶/5) ) . V =

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Electric Potential and Potential Difference

A charge of \( 8 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 50 \, \text{V} \). What i

**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. Work done = Potential energy = q V . W = 8 × 10⁻⁶ × 50 = 4 × 10⁻⁴ J = 0.4 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,

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Electric Potential and Potential Difference

A charged conductor is surrounded by a thin concentric hollow conducting shell. If the shell is grounded, what happens t

**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. When the outer shell is grounded (potential V = 0 ), the potential on the inner conductor adjusts due to charge redistribution. If the inner conductor has charge Q , the inner surface of the shell induces -Q , and since the shell's potential is zero, the outer surface of the shell acquires +Q . The

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Electric Potential and Potential Difference

A spherical conductor of radius 6 cm has a charge of \( 6 \times 10^{-8} \, \text{C} \). What is the potential at its su

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (6 × 10⁻⁸/0.06) = 9 × 10⁹ × 10⁻⁶ = 9000 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

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Electric Potential and Potential Difference

Two charges \( 10 \, \mu\text{C} \) and \( -2 \, \mu\text{C} \) are at \( (8, 0, 0) \) and \( (-8, 0, 0) \, \text{cm} \)

**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. Distance to midpoint = 0.08 m. V = 9 × 10⁹ ( (10 × 10⁻⁶/0.08) + (-2 × 10⁻⁶/0.08) ) = 9 × 10⁹ × (8 × 10⁻⁶/0.08) . V = 9 × 10⁹ × (8 × 10⁻⁶/0.08) = 9 × 10⁵ V . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq series, C_eq =

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Electric Potential and Potential Difference

Three charges \( +8 \, \mu\text{C} \), \( -5 \, \mu\text{C} \), and \( +3 \, \mu\text{C} \) are at \( (0, 0, 0) \), \( (

**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. Distances: r₁ = √(8² + 8²) = 8√(2) m , r₂ = 8 m , r₃ = 8 m . V = 9 × 10⁹ ( (8 × 10⁻⁶/8√(2)) + (-5 × 10⁻⁶/8) + (3 × 10⁻⁶/8) ) . V = 9 × 10⁹ ( (8 × 10⁻⁶/11.314) - (5 × 10⁻⁶/8) + (3 × 10⁻⁶/8)

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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.

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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.

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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.

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