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#point charge

17 public questions tagged with this topic.

A point charge \( q = 5 \, \mu\text{C} \) is placed at the origin. What is the electric field magnitude at a point 2 m a

**Inverse-square law** for charges states F ∝ 1/r² while increasing with charge product. Using k = 9×10⁹ N·m²/C², force at distance r follows F = k q₁q₂/r², forming basis for pairwise force calculation. Electric field: E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 5 × 10⁻⁶ C , r = 2 m . E = 9 × 10⁹ × (5 × 10⁻⁶/(2)²) = 9 × 10⁹ × (5 × 10⁻⁶/4) = 1.125 × 10⁴ N/C . Substituting values gives 1.125 × 10⁴ N/C, 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

Why does the electric field due to a dipole exhibit a stronger angular dependence compared to a single point charge?

**Dipole moment** governs torque and energy in external field. Axial field stronger than equatorial, torque maximum at θ = 90°, zero when aligned. Work done rotating dipole relates to ΔU = pE(1 - cosθ), explaining stable equilibrium at θ = 0°. The dipole’s field arises from two opposite charges, creating a directional dependence. The angle between the dipole axis and the point affects the field’s magnitude and direction more significantly than a point charge’s isotropic 1/r² field. Substituting values gives Directional dependence, 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 > Electric Dipole - Moment, Field and Torque

Why does the electric field outside a charged conducting sphere resemble that of a point charge located at its center?

**Charge density formulation** allows integration over extended bodies, but highly symmetric cases yield simple expressions. Infinite line gives E ∝ λ/r, unlike point charge 1/r², reflecting different geometry of source. The spherical symmetry of the charge distribution on the conductor’s surface ensures that the field outside behaves as if all charge were at the center, as per Gauss’s law. This symmetry simplifies the field to a radial, point-charge-like pattern. Substituting values gives Spherical symmetry, 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 > Continuous Charge Distribution

A point charge \( -10 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 10 m along the

**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. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 10 × 10⁻⁶ C , r = 10 m . E = 9 × 10⁹ × (10 × 10⁻⁶/(10)²) = 9 × 10⁹ × (10 × 10⁻⁶/100) = 9 × 10² N/C . Substituting values gives 900 N/C, 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

A point charge \( -9 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 6 m along the x

**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. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 9 × 10⁻⁶ C , r = 6 m . E = 9 × 10⁹ × (9 × 10⁻⁶/(6)²) = 9 × 10⁹ × (9 × 10⁻⁶/36) = 2.25 × 10³ N/C . Substituting values gives 2.25 × 10³ N/C, 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 > Electric Charge, Quantization and Conservation

A point charge \( 10 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 5 m along the z

**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. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 10 × 10⁻⁶ C , r = 5 m . E = 9 × 10⁹ × (10 × 10⁻⁶/(5)²) = 9 × 10⁹ × (10 × 10⁻⁶/25) = 3.6 × 10³ N/C . Substituting values gives 3.6 × 10³ N/C, 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 > Electric Charge, Quantization and Conservation

A point charge \( -6 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 3 m along the y

**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. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 6 × 10⁻⁶ C , r = 3 m . E = 9 × 10⁹ × (6 × 10⁻⁶/(3)²) = 9 × 10⁹ × (6 × 10⁻⁶/9) = 6 × 10³ N/C . Substituting values gives 6 × 10³ N/C, which matches

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

A point charge \( 12 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 8 m along the x

**Fundamental property of charge** includes additivity and quantization, meaning net charge equals algebraic sum of constituents and each is multiple of e. When rod loses charge, electron removal is inferred, and n = q/e gives transferred count. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 12 × 10⁻⁶ C , r = 8 m . E = 9 × 10⁹ × (12 × 10⁻⁶/(8)²) = 9 × 10⁹ × (12 × 10⁻⁶/64) = 1.6875 × 10³ N/C . Substituting values gives 1.69 × 10³ N/C, 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 > Electric Charge, Quantization and Conservation

A point charge \( -8 \, \mu\text{C} \) is at the origin. What is the electric field magnitude at a point 4 m along the z

**Fundamental property of charge** includes additivity and quantization, meaning net charge equals algebraic sum of constituents and each is multiple of e. When rod loses charge, electron removal is inferred, and n = q/e gives transferred count. E = (k |q|/r²) . k = 9 × 10⁹ N·m²/C² , q = 8 × 10⁻⁶ C , r = 4 m . E = 9 × 10⁹ × (8 × 10⁻⁶/(4)²) = 9 × 10⁹ × (8 × 10⁻⁶/16) = 4.5 × 10³ N/C . Substituting values gives 4.5 × 10³ N/C, 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 > Electric Charge, Quantization and Conservation

A charge of \( 13 \, \mu\text{C} \) is at the center of a cube of edge 50 cm. What is the flux through one face?

**Flux definition** Φ = ∮ E·dA links field to area orientation. For uniform E perpendicular to surface, Φ = E A, with A = πr² for circle. Inclination reduces flux by cosθ factor, sign indicating outward or inward crossing. Total flux: Φ = (q/ε₀) = (13 × 10⁻⁶/8.854 × 10⁻¹²) = 1.468 × 10⁶ N·m²/C . Flux per face (6 faces): Φfₐcₑ = (1.468 × 10⁶/6) = 2.447 × 10⁵ N·m²/C . Substituting values gives 2.45 × 10⁵ N·m²/C, 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 > Electric Flux

A charge of \( 8 \, \mu\text{C} \) is at the center of a cube of edge 25 cm. What is the total flux through the cube?

**Gauss's theorem** states total flux through closed surface equals enclosed charge divided by free-space permittivity, Φ_total = q_enc/ε₀, ε₀ = 8.854×10⁻¹² C²/(N·m²). Result independent of shape or size, depends only on net enclosed charge, enabling charge determination from flux. Total flux: Φ = (q/ε₀) . Φ = (8 × 10⁻⁶/8.854 × 10⁻¹²) = 9.03 × 10⁵ N·m²/C . Substituting values gives 9.03 × 10⁵ N·m²/C, 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 > Gauss's Theorem and Total Flux

A charge of \( 4 \, \mu\text{C} \) is placed at the center of a cube of edge 10 cm. What is the electric flux through on

**Electric flux** through surface measures field lines crossing it, Φ = E·A = E A cosθ for uniform field, unit N·m²/C. For circular area in xy-plane with field along z, θ = 0°, cosθ = 1, so Φ = E·πR² directly, maximum when field normal to surface. Total flux: Φ = (q/ε₀) = (4 × 10⁻⁶/8.854 × 10⁻¹²) = 4.52 × 10⁵ N·m²/C . Flux through one face (cube has 6 faces): Φfₐcₑ = (Φ/6) = (4.52 × 10⁵/6) = 7.53 × 10⁴ N·m²/C . Substituting values gives 7.5 × 10⁴ N·m²/C, 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 > Electric Flux