#midpoint

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,

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

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Equipotential Surfaces and Relation Between Field and Potential

In a system of two point charges of opposite signs placed close to each other, what can be said about the equipotential

**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 two point charges of opposite signs ( +q and -q ), the potential at the midpoint is zero (since V = (q/4 π ε₀ r) - (q/4 π ε₀ r) = 0 , where r is the distance from each charge to the midpoint). The equipotential surface where V = 0 is the plane perpendicular to the line joining the charges, passing through the

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

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Electric Charge, Quantization and Conservation

Two charges \( +6 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 50 cm apart. What is the electric field magnitude at

**Quantization of charge** states observable charge is integer multiple of elementary charge e = 1.6×10⁻¹⁹ C, q = n·e, and total charge is conserved in isolated systems. Loss of electrons produces positive charge, and number of transferred electrons follows n = q/e, linking macroscopic charge measurement to microscopic carriers. Midpoint distance = 25 cm = 0.25 m. E₁ = 9 × 10⁹ × (6 × 10⁻⁶/(0.25)²) = 8.64 × 10⁵ N/C (towards -3 μC ). E₂ = 9 × 10⁹ × (3 × 10⁻⁶/(0.25)²) = 4.32 × 10⁵ N/C (towards -3 μC ). Net E = 8.64 × 10⁵ + 4.32 × 10⁵ = 1.296

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Charge, Quantization and Conservation

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Electric Charge, Quantization and Conservation

Two charges \( +3 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 15 cm apart. What is the force on a \( 2 \, \mu\text

**Charge conservation and quantization** govern rubbing processes where electrons transfer without creation. Total charge before and after remains equal, and any measured charge corresponds to n = q/e electrons, allowing counting of carriers from coulomb value. Electric field at midpoint: E₁ = 9 × 10⁹ × (3 × 10⁻⁶/(0.075)²) = 4.8 × 10⁶ N/C (towards -3 μC ). E₂ = 4.8 × 10⁶ N/C (towards -3 μC ). Net E = 4.8 × 10⁶ + 4.8 × 10⁶ = 9.6 × 10⁶ N/C . Force: F = q E = 2 × 10⁻⁶ × 9.6 × 10⁶ = 19.2 N . Substituting values gives 19.2

Ref: NCERT > Physics Book > Electric Charges and Fields > Electric Charge, Quantization and Conservation

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Electric Field and Electric Field Lines

Two charges \( +9 \, \mu\text{C} \) and \( -3 \, \mu\text{C} \) are 70 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 = 35 cm = 0.35 m. E₁ = 9 × 10⁹ × (9 × 10⁻⁶/(0.35)²) = 6.61 × 10⁵ N/C (towards -3 μC ). E₂ = 9 × 10⁹ × (3 × 10⁻⁶/(0.35)²) = 2.2 × 10⁵ N/C (towards -3 μC ). Net E = 6.61 × 10⁵ + 2.2 × 10⁵ = 8.81 × 10⁵

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

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Potential Energy of Mass Systems

Two masses 2kg and 8kg are separated by 4m. What is the gravitational potential at the midpoint? (G\=6.67×10−11N m2/kg2)

Distance to midpoint: r = 2m. U = −Gm1r−Gm2r. U = −6.67×10−11(22+82). U = −6.67×10−11×(1+4) = −3.335×10−10J/kg. As per NCERT, applying relevant law/formula with correct units and sign convention leads to -3.3 × 10⁻¹⁰ J/kg. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: Halliday & Resnick, Fundamentals of Physics, 11th ed., Chapter 13: Gravitation.

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