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#point charge

32 public questions tagged with this topic.

Energy Stored in Capacitor and Energy Density

A point charge \( Q = 5 \times 10^{-9} \, \text{C} \) is placed at the origin. What is the potential at a point 10 m awa

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (5 × 10⁻⁹/10) = 9 × 10⁹ × 0.5 × 10⁻⁹ = 4.5 V . 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 > Energy Stored in Capacitor and Energy Density

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Energy Stored in Capacitor and Energy Density

A point charge \( Q = 21 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 7 m a

**Work done charging capacitor** is integral ∫ V dQ = ∫ Q/C dQ = Q²/2C, stored as electrostatic energy. When capacitor discharges, energy released as heat or work, explaining spark when shorted, energy proportional to V². Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (21 × 10⁻⁹/7) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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 27 V follows, reflecting potential-capacitance relations.

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

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Energy Stored in Capacitor and Energy Density

A point charge \( Q = 6 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 2 m aw

**Energy density** in electric field u = ½ ε E², ε = K ε₀, E = V/d, total energy U = u·volume = ½ ε E²·A d =½ ε A d·(V/d)²=½ ε A V²/d=½ C V², consistent. For parallel plate, E = V/d ≈10⁶ V/m for 400 V across 0.4 mm, u≈½×8.85×10⁻¹²×10¹²≈4.4 J/m³. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (6 × 10⁻⁹/2) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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

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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

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Conductors, Electrostatic Shielding and Dielectrics

A point charge \( Q = 9 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 3 m aw

**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 due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (9 × 10⁻⁹/3) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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 =

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

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Conductors, Electrostatic Shielding and Dielectrics

A point charge \( Q = 18 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 6 m a

**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 due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (18 × 10⁻⁹/6) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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

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

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Conductors, Electrostatic Shielding and Dielectrics

Why does the electric field inside a uniformly charged spherical shell vary linearly with distance from the center if a

**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. Without the point charge, the field inside a uniformly charged spherical shell is zero (Gauss’s law). With a point charge Q at the center, the field inside the shell is due to the point charge only ( E = (1/4 π ε₀) (Q/r²) ), which varies as (1/r²) , not linearly. The question may imply a misunderstanding; in standard electrostatics (per the PDF), the shell

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 \( 5 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 300 \, \text{V} \). What

**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. Work done = Potential energy = q V . W = 5 × 10⁻⁶ × 300 = 1.5 × 10⁻³ J = 1.5 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 1.5 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 point charge \( Q = 15 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 5 m a

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (15 × 10⁻⁹/5) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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 > Equipotential Surfaces and Relation Between Field and Potential

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Equipotential Surfaces and Relation Between Field and Potential

A point charge \( Q = 12 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 6 m a

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (12 × 10⁻⁹/6) = 9 × 10⁹ × 2 × 10⁻⁹ = 18 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 > Equipotential Surfaces and Relation Between Field and Potential

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Potential Energy of System of Charges

A point charge \( Q = 24 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 8 m a

**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. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (24 × 10⁻⁹/8) = 9 × 10⁹ × 3 × 10⁻⁹ = 27 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 > Potential Energy of System of Charges

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Potential Energy of System of Charges

A point charge \( Q = 8 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 4 m aw

**System of charges** potential energy is sum over pairs U = Σ k q_i q_j/r_ij, work required to assemble charges from infinity. For 28 μC and -14 μC, 0.28 m apart, U=9×10⁹×28×(-14)×10⁻¹²/0.28= -12.6 J, negative indicates bound system. Potential due to a point charge: V = (1/4 π ε₀) (Q/r) . Substitute: V = 9 × 10⁹ × (8 × 10⁻⁹/4) = 9 × 10⁹ × 2 × 10⁻⁹ = 18 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 > Potential Energy of System of Charges

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