#Electrostatics

PHYSICS

Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the fo

Given: Two small charged spheres with charges 5 × 10⁻⁷ C and 7 × 10⁻⁷ C are placed 50 cm apart in air. What is the force between them? These values define the system as per NCERT data. Formula: Using Coulomb’s law: F = k |q_1 q_2|/r². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 9 × 10⁹ Nm²/C², q_1 = 5 × 10⁻⁷ C, q_2 = 7 × 10⁻⁷ C, r = 0.5 m . |q_1 q_2| = 5 × 7 × 10⁻¹⁴= 35 × 10⁻¹⁴ C² . r² = (0.5)² = 0.25 m² . F = 9 × 10⁹ × frac35 × 10⁻¹⁴⁰.25 = 9 × 10⁹ × 1.4 × 10⁻¹²= 0.0126 N . 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: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.

View Details

PHYSICS

A spherical shell has a net flux of 1.13 × 10⁵ Nm²/C through it. What is the charge enclosed?

Given: A spherical shell has a net flux of 1.13 × 10⁵ Nm²/C through it. What is the charge enclosed? These values define the system as per NCERT data. Formula: phi = q/varepsilon_0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: q = phi varepsilon_0 = 1.13 × 10⁵ × 8.854 × 10⁻¹²= 1.0 × 10⁻⁶ C = 1 μC . 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: Units and Measurements and Physical World, Topic: Dimensional analysis and fundamental principles.

View Details

PHYSICS

A charge of 4 μC is moved from infinity to a point where the potential is 250 V . What is the work done?

Given: A charge of 4 μC is moved from infinity to a point where the potential is 250 V . What is the work done? These values define the system as per NCERT data. Formula: Work done = Potential energy = q V. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: W = 4 × 10⁻⁶ × 250 = 10⁻³ J = 1 mJ . 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.

View Details

PHYSICS

A conductor has a surface charge density of 1.5 × 10⁻⁶C/m² . What is the electric field just outside it? (Take ε_0 = 8.8

Given: A conductor has a surface charge density of 1.5 × 10⁻⁶C/m² . What is the electric field just outside it? (Take ε_0 = 8.85 × 10⁻¹²C² N^{-1 m^{-2 ). Formula: E = sigma/ε_0 = frac1.5 × 10⁻⁶⁸.85 × 10⁻¹²approx 1.695 × 10⁵N/C .. Substitution: Substituting given values into formula as per NCERT 2026-27 method. Calculation: Simplifying step by step with proper SI units like m/s², J kg⁻¹ K⁻¹, 10⁻⁵, A m⁻¹ etc. 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.

View Details

PHYSICS

A dipole p = 6 × 10⁻⁹ C m is rotated from θ = 0° to 90° in a field E = 3 × 10⁵ N/C . What is the work done?

Given: A dipole p = 6 × 10⁻⁹ C m is rotated from θ = 0° to 90° in a field E = 3 × 10⁵ N/C . What is the work done? These values define the system as per NCERT data. Formula: Work done: W = p E (cos θ_0 - cos θ_1) = 6 × 10⁻⁹ × 3 × 10⁵ × (cos 0° - cos 90°). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: W = 6 × 10⁻⁹ × 3 × 10⁵ × (1 - 0) = 1.8 × 10⁻³ J . 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.

View Details

PHYSICS

A thin spherical shell of radius 25 cm has a charge of 15 μC . What is the electric field at a point 30 cm from the nte

Given: A thin spherical shell of radius 25 cm has a charge of 15 μC . What is the electric field at a point 30 cm from the nter? These values define the system as per NCERT data. Formula: Outside shell ( r > R ): E = k q/r². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: k = 9 × 10⁹ Nm²/C², q = 15 × 10⁻⁶ C, r = 0.3 m . E = 9 × 10⁹ × frac15 × 10⁻⁶(0.3)² = 9 × 10⁹ × frac15 × 10⁻⁶⁰.09 = 1.5 × 10⁶ N/C . 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.

View Details

PHYSICS

Two charges 4 × 10⁻⁸ C and -2 × 10⁻⁸ C are 20 cm apart. At what distance from the positive charge on the line

Given: Two charges 4 × 10⁻⁸ C and -2 × 10⁻⁸ C are 20 cm apart. At what distance from the positive charge on the line joining them is the potential zero? (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: Total potential: V = 1/4 π varepsilon_0 ( frac4 × 10⁻⁸ x + frac-2 × 10⁻⁸⁰.2 - x ) = 0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Let distance from 4 × 10⁻⁸ C be x m, then distance from -2 × 10⁻⁸ C is 0.2 - x . . Simplify: 9 × 10⁹( frac4 × 10⁻⁸ x - frac2 × 10⁻⁸⁰.2 - x ) = 0 . 4/x = 2/0.2 - x Rightarrow 4 (0.2 - x) = 2x Rightarrow 0.8 - 4x = 2x Rightarrow 0.8 = 6x Rightarrow x = 0.8/6 = 0.133 m = 13.3 cm . 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.

View Details

PHYSICS

A point charge Q = 8 × 10⁻⁹C is placed at the origin. Calculate the potential at a point 4 m away from the charge. (Take

Given: A point charge Q = 8 × 10⁻⁹C is placed at the origin. Calculate the potential at a point 4 m away from the charge. (Take 1/4 π ε_0 = 9 × 10⁹Nm² C^{-2 ). Formula: Potential due to a point charge: V = 1/4 π ε_0 Q/r. Substitution & Calculation: Substitute: V = 9 × 10⁹ × frac8 × 10⁻⁹⁴= 9 × 10⁹ × 2 × 10⁻⁹= 18 V . 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.

View Details

PHYSICS

A body loses 3.2 × 10⁻⁸C of charge when rubbed. How many electrons were transferred from it?

Given: A body loses 3.2 × 10⁻⁸C of charge when rubbed. How many electrons were transferred from it? Formula: q = n e, e = 1.6 × 10⁻¹⁹C. Substitution & Calculation: Losing charge means electrons are removed, so charge is positive. . n = q/|e| = frac3.2 × 10⁻⁸¹.6 × 10⁻¹⁹= 2 × 10¹¹. 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.

View Details

PHYSICS

A thin spherical shell of radius 16 cm has a charge of 11 μC . What is the electric field at a point 20 cm from the nter

Given: A thin spherical shell of radius 16 cm has a charge of 11 μC . What is the electric field at a point 20 cm from the nter? These values define the system as per NCERT data. Formula: Outside shell ( r > R ): E = k q/r². This is standard NCERT relation. Substitution & Calculation: k = 9 × 10⁹Nm²/C², q = 11 × 10⁻⁶C, r = 0.2 m . E = 9 × 10⁹ × frac11 × 10⁻⁶(0.2)² = 9 × 10⁹ × frac11 × 10⁻⁶⁰.04 = 2.475 × 10⁶N/C . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Relevant Physics topic covering fundamental principles,

View Details

PHYSICS

A point charge Q = 18 × 10⁻⁹ C is placed at the origin. Calculate the potential at a point 6 m away from the charge

Given: A point charge Q = 18 × 10⁻⁹ C is placed at the origin. Calculate the potential at a point 6 m away from the charge. (Take 1/4 π varepsilon_0 = 9 × 10⁹ Nm² C^{-2 ). These values define the system as per NCERT data. Formula: Potential due to a point charge: V = 1/4 π varepsilon_0 Q/r. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Substitute: V = 9 × 10⁹ × frac18 × 10⁻⁹⁶= 9 × 10⁹ × 3 × 10⁻⁹= 27 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.

View Details

PHYSICS

A plane sheet has sigma = 8.854 × 10⁻¹¹ C/m² . What is the electric field near it?

Given: A plane sheet has sigma = 8.854 × 10⁻¹¹ C/m² . What is the electric field near it? These values define the system as per NCERT data. Formula: E = sigma/2 varepsilon_0. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: E = frac8.854 × 10⁻¹¹² × 8.854 × 10⁻¹²= 5 N/C . 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.

View Details