#electric potential

Coulomb's Law and Force Between Point Charges

Two charges \( +5 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are 30 cm apart. What is the potential energy of the sys

**Electrostatic force** described by F = (1/4π ε₀)·q₁q₂/r² obeys Newton's third law. Magnitude depends on q₁q₂ and 1/r², enabling quantitative estimation at given separation, with sign indicating attraction or repulsion. Potential energy: U = k (q₁ q₂/r) . U = 9 × 10⁹ × ((5 × 10⁻⁶) × (-5 × 10⁻⁶)/0.3) = 9 × 10⁹ × (-25 × 10⁻¹²/0.3) = -0.75 J . Substituting values gives -0.75 J, which matches expected magnitude for this electrostatic configuration, confirming Coulomb's and Gauss's principles and charge quantization consistency.

Ref: NCERT > Physics Book > Electric Charges and Fields > Coulomb's Law and Force Between Point Charges

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PHYSICS

A dipole with p = 5 × 10⁻¹⁰ C m is along the x-axis. What is the potential at (2, 2, 0) m ? (Take 1/4 π varepsilo

Given: A dipole with p = 5 × 10⁻¹⁰ C m is along the x-axis. What is the potential at (2, 2, 0) m ? (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: r = sqrt2² + 2² = 2sqrt2 m, cos θ = frac22sqrt2 = frac1sqrt2. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: V = 9 × 10⁹ × frac5 × 10⁻¹⁰ × frac1sqrt2(2sqrt2)² = 9 × 10⁹ × frac5 × 10⁻¹⁰⁸ sqrt2 × frac1sqrt2 = 9 × 10⁹ × frac5 × 10⁻¹⁰¹⁶= 2.8125 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Moving Charges and Magnetism and Magnetism and Matter, Topic: Magnetic field due to current loop, solenoid and magnetic dipole moment. Page number should be added only after verification from the.

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Three charges +2 μC, -3 μC, and +1 μC are at (0, 0, 0), (2, 0, 0), and (0, 2, 0) m . What is the potential at (2, 2,

Given: Three charges +2 μC, -3 μC, and +1 μC are at (0, 0, 0), (2, 0, 0), and (0, 2, 0) m . What is the potential at (2, 2, 0) m ? (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: Distances: r_1 = sqrt2² + 2² = 2sqrt2, r_2 = 2 m, r_3 = 2 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: V = 9 × 10⁹( frac2 × 10⁻⁶² sqrt2 + frac-3 × 10⁻⁶²+ frac1 × 10⁻⁶²) . V = 9 × 10⁹( frac2 × 10⁻⁶².828 - frac3 × 10⁻⁶²+ frac1 × 10⁻⁶²) . V = 9 × 10⁹( 0.707 × 10⁻⁶- 1.5 × 10⁻⁶+ 0.5 × 10⁻⁶) = 9 × 10⁹ × (-0.293 × 10⁻⁶) = -2637 V . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Electric Charges and Fields and Electrostatic Potential, Topic: Electric field, potential and capacitance concepts.

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Two charges 7 μC and -2 μC are at (8, 0, 0) and (-8, 0, 0) cm . What is the potential at midpoint? (Take 1/4 π ε_0 = 9 ×

Given: Two charges 7 μC and -2 μC are at (8, 0, 0) and (-8, 0, 0) cm . What is the potential at midpoint? (Take 1/4 π ε_0 = 9 × 10⁹Nm² C^{-2 ). Formula: Distance to midpoint = 0.08 m. Substitution & Calculation: V = 9 × 10⁹( frac7 × 10⁻⁶⁰.08 + frac-2 × 10⁻⁶⁰.08 ) = 9 × 10⁹ × frac5 × 10⁻⁶⁰.08 = 5.625 × 10⁵V . Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Physics Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII, continuing as per NCERT advisory for 2026-27),Topic: Fundamental laws, definitions and applications as per latest NCERT. The section explains governing laws, formulas like μ₀ = 4π.

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