#midpoint calculation

Energy Stored in Capacitor and Energy Density

Two charges \( 16 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (4, 0, 0) \) and \( (-4, 0, 0) \, \text{cm} \)

**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. Distance to midpoint = 0.04 m. V = 9 × 10⁹ ( (16 × 10⁻⁶/0.04) + (-4 × 10⁻⁶/0.04) ) = 9 × 10⁹ × (12 × 10⁻⁶/0.04) . V = 9 × 10⁹ × (12 × 10⁻⁶/0.04) = 2.7 × 10⁶ V . Using V = kQ/r, U = k q₁q₂/r,

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

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Conductors, Electrostatic Shielding and Dielectrics

Two charges \( 6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are at \( (10, 0, 0) \) and \( (-10, 0, 0) \, \text{cm} \

**Conductor in electrostatic equilibrium** has E=0 inside, charges reside on surface, potential constant throughout conductor. Hollow shell with no internal charge has zero field inside cavity, even if external field present, charges on outer surface screen interior, principle used in Faraday cage. Distance to midpoint = 0.1 m. V = 9 × 10⁹ ( (6 × 10⁻⁶/0.1) + (-3 × 10⁻⁶/0.1) ) = 9 × 10⁹ × (3 × 10⁻⁶/0.1) = 2.7 × 10⁵ 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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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

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

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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 =

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

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Electric Field and Electric Field Lines

Two charges \( +5 \, \mu\text{C} \) and \( -7 \, \mu\text{C} \) are 35 cm apart. What is the electric field magnitude at

**Electric field** defined as E = F/q₀, force per unit positive test charge, unit N/C or V/m, direction along force on positive test charge. For point charge, E = k q/r² radially outward for q>0. Field lines start on positive and end on negative, density indicates strength. Midpoint distance = 17.5 cm = 0.175 m. E₁ = 9 × 10⁹ × (5 × 10⁻⁶/(0.175)²) = 1.47 × 10⁶ N/C (towards -7 μC ). E₂ = 9 × 10⁹ × (7 × 10⁻⁶/(0.175)²) = 2.06 × 10⁶ N/C (towards -7 μC ). Net E = 1.47 × 10⁶ + 2.06 × 10⁶ = 3.53 × 10⁶

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Field and Electric Field Lines

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