Electric field intensity (E) at a point is defined as the force experienced per unit positive charge placed at that point. Formula: E = F/q, where F is the force in Newtons and q is the test charge in Coulombs. The SI unit of electric field intensity is Newton per Coulomb (N/C), which is equivalent to Volt per metre (V/m). Electric field is a vector quantity.
Electric field intensity E = F/q (force per unit positive charge).
SI unit: N/C (Newton/Coulomb) = V/m (Volt/metre).
Dimensional formula: [MLT⁻³A⁻¹].
Due to point charge: E = kq/r² = q/(4πε₀r²).
k = 9×10⁹ N·m²/C²; ε₀ = 8.854×10⁻¹² C²/N·m².
E is a vector — points away from +ve charge, toward -ve charge.
Uniform field between parallel plates: E = V/d.
Electric field lines start at +ve charges and end at -ve charges.
Definition: The electric field intensity at a point is the electrostatic force experienced per unit positive test charge placed at that point.
Formula: E = F/q
Where: • E = Electric field intensity (vector) • F = Electrostatic force on test charge (N) • q = Test charge (C) [must be small enough not to disturb the field]
SI Unit: • E = F/q → [N/C] or equivalently [V/m] • 1 N/C = 1 V/m
Dimensional formula of E: E = F/q = [MLT⁻²] / [AT] = [MLT⁻³A⁻¹]
Electric field due to point charge (Coulomb's Law): E = kq/r² = q/(4πε₀r²)
Where: • k = 9 × 10⁹ N·m²/C² (Coulomb's constant) • ε₀ = 8.854 × 10⁻¹² C²/N·m² (permittivity of free space) • q = source charge (C) • r = distance from charge (m)
Key properties of electric field:
Vector quantity (has direction) • For positive charge: E points away from the charge • For negative charge: E points toward the charge
Superposition principle: • Net E = vector sum of individual fields due to each charge
Electric field lines: • Start from positive charges, end at negative charges • The density of field lines indicates field strength • Field lines never intersect
Uniform electric field: • Between parallel plates: E = V/d (V = voltage, d = distance) • Direction: from +ve plate to −ve plate
Solved Example 1: A charge of 2 μC experiences a force of 0.1 N. Find E. E = F/q = 0.1 / (2 × 10⁻⁶) = 5 × 10⁴ N/C
Solved Example 2: Find E due to a 4 μC charge at 0.3 m distance. E = kq/r² = (9×10⁹ × 4×10⁻⁶) / (0.3)² = 36000 / 0.09 = 4 × 10⁵ N/C
Relation: E = V/d (uniform field) • If a 100 V potential difference across 2 m gap: E = 100/2 = 50 V/m = 50 N/C
Electric field intensity (E) at a point is the electrostatic force per unit positive test charge at that point: E = F/q. SI unit: N/C or V/m. Dimensional formula: [MLT⁻³A⁻¹]. E is a vector quantity — its direction is along the force on a positive test charge.
The SI unit of electric field intensity is Newton per Coulomb (N/C), which is equivalent to Volt per metre (V/m). These two units are identical: 1 N/C = 1 V/m.
E = kq/r² = q/(4πε₀r²), where k = 9×10⁹ N·m²/C², q = source charge, r = distance. The field points away from positive charges and toward negative charges.
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