#charge

Applications of EM Waves in Communication and Medicine

What type of charges cannot produce electromagnetic waves?

**Microwaves in ovens** cause water molecules to rotate at 2.45 GHz, friction heats food, penetration depth few cm, efficient heating, also radar uses reflection of microwaves from objects, Doppler shift measures speed, medical diathermy uses microwaves for tissue heating. The document states that neither stationary charges nor charges in uniform motion (steady currents) can produce electromagnetic waves; only accelerated charges can. Using c = fλ, E₀/B₀ = c, I_d = ε₀ dΦ_E/dt, and spectrum classification λ = c/f, evaluation yields Stationary charges, illustrating EM wave transverse nature and

Ref: NCERT > Physics Book > Electromagnetic Waves > Applications of EM Waves in Communication and Medicine

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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