#charge

Combination of Capacitors - Series and Parallel

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the potential at its su

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (5 × 10⁻⁸/0.05) = 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

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

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

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (8 × 10⁻⁸/0.04) = 9 × 10⁹ × 2 × 10⁻⁶ = 18000 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

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

View Details

Conductors, Electrostatic Shielding and Dielectrics

A spherical conductor of radius 3 cm has a charge of \( 3 \times 10^{-8} \, \text{C} \). What is the potential at its su

**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. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (3 × 10⁻⁸/0.03) = 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 9000 V follows, reflecting potential-capacitance relations.

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

View Details

Equipotential Surfaces and Relation Between Field and Potential

A charge of \( 6 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 150 \, \text{V} \). What

**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 = 6 × 10⁻⁶ × 150 = 9 × 10⁻⁴ J = 0.9 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 0.9 mJ follows, reflecting potential-capacitance relations.

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

A charge of \( 3 \, \mu\text{C} \) is brought from infinity to a point where the potential is \( 400 \, \text{V} \). 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. Work done = Potential energy = q V . W = 3 × 10⁻⁶ × 400 = 1.2 × 10⁻³ J = 1.2 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 1.2 mJ follows, reflecting potential-capacitance relations.

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

A spherical conductor of radius 15 cm has a charge of \( 9 \times 10^{-8} \, \text{C} \). What is the electric field at

**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 = 0.2 m > R = 0.15 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (9 × 10⁻⁸/(0.2)²) = 9 × 10⁹ × (9 × 10⁻⁸/0.04) = 2.025 × 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

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

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A spherical conductor of radius 10 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the electric field at

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. For r = 0.25 m > R = 0.1 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (5 × 10⁻⁸/(0.25)²) = 9 × 10⁹ × (5 × 10⁻⁸/0.0625) = 7.2 × 10³ 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

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

View Details

Capacitance of Parallel Plate and Spherical Capacitor

A spherical conductor of radius 5 cm has a charge of \( 5 \times 10^{-8} \, \text{C} \). What is the potential at its su

**Parallel plate capacitor** capacitance C = ε₀ A/d, ε₀=8.85×10⁻¹² F/m, A plate area (m²), d separation (m), for air, with dielectric C = K ε₀ A/d. For A=0.08 m², d=0.4 mm=4×10⁻⁴ m, C=8.85×10⁻¹²×0.08/4×10⁻⁴=1.77×10⁻⁹ F=1.77 nF, illustrating small capacitance for cm separation. Potential at the surface: V = (1/4 π ε₀) (Q/R) . V = 9 × 10⁹ × (5 × 10⁻⁸/0.05) = 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 9000 V follows, reflecting potential-capacitance relations.

Ref: NCERT > Physics Book > Electrostatic Potential and Capacitance > Capacitance of Parallel Plate and Spherical Capacitor

View Details

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

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

View Details

Electric Potential and Potential Difference

What is the work done in moving a \( 5 \, \mu\text{C} \) charge from infinity to a point where the potential is \( 2000

**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. Work done = Potential energy = q V . W = 5 × 10⁻⁶ × 2000 = 10⁻² J = 0.01 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 0.01 J follows, reflecting potential-capacitance relations.

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

View Details

Electric Potential and Potential Difference

A spherical conductor of radius 35 cm has a charge of \( 14 \times 10^{-8} \, \text{C} \). 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 = 0.8 m > R = 0.35 m , E = (1/4 π ε₀) (Q/r²) . E = 9 × 10⁹ × (14 × 10⁻⁸/(0.8)²) = 9 × 10⁹ × (14 × 10⁻⁸/0.64) ≈ 1.969 × 10³ N/C . Using V = kQ/r, U = k q₁q₂/r, E = -dV/dr, C = ε₀A/d, 1/C_eq

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

View Details

Electric Potential and Potential Difference

A charge of \( 9 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 40 \, \text{V} \). What i

**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. Work done = Potential energy = q V . W = 9 × 10⁻⁶ × 40 = 3.6 × 10⁻⁴ J = 0.36 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 0.36 mJ follows, reflecting potential-capacitance relations.

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

View Details