Electric Field Intensity

Electric Field Intensity is the concept that allows electrostatics to be described in terms of properties of space rather than requiring a second charge to be present. It bridges Coulomb's Law and the broader theory of electric flux, potential, and energy. In GATE and university examinations, field calculations from point charges, dipoles, and charge distributions form a significant portion of the electromagnetics section.

Core Concept Explanation

The electric field intensity E at a point in space is defined as the force experienced by a small positive test charge placed at that point divided by the magnitude of that test charge. This definition removes the dependence on the test charge itself, making E a property of the source charge and the surrounding space. The mathematical definition is E = F / q, where F is the Coulomb force and q is the test charge.

For a single positive point charge Q located at the origin, the electric field at a distance r is given by E = Q / (4πε₀ r²) in the radially outward direction. If Q is negative, the field points radially inward toward Q. The unit of E is volts per metre (V/m) or equivalently newtons per coulomb (N/C).

The superposition principle applies directly to electric fields. If multiple point charges are present, the net field at any point is the vector sum of the individual fields produced by each charge. This is used to compute fields of dipoles, charge rings, and other distributions.

Mathematical Expression

For a point charge Q at position r', the electric field at observation point r is E(r) = Q(r − r') / (4πε₀ |r − r'|³). The numerator gives direction while the denominator handles both the magnitude scaling and unit vector normalization. For N discrete charges, E_total = sum of Qi(r − ri') / (4πε₀ |r − ri'|³) for i = 1 to N. This vector sum must account for x, y, and z components separately.

For continuous charge distributions, the sum converts to an integral. For a volume charge density ρv, the field is E = integral over V of [ρv(r − r') / (4πε₀ |r − r'|³)] dV'. Similar integrals apply to surface charge density ρs and line charge density ρL. These integrals are evaluated using coordinate system symmetry to simplify the problem.

Practical Understanding

A uniform electric field means E has the same magnitude and direction at every point. A parallel plate capacitor with small separation relative to plate size produces a nearly uniform field E = V/d, where V is the voltage across the plates and d is the separation. In contrast, a point charge produces a non-uniform field that weakens rapidly with distance.

The direction of the electric field at any point tells a positive test charge which way it would accelerate if released. Field lines originate from positive charges and terminate on negative charges. The density of field lines per unit area gives a visual indication of field strength.

Given:
Q1 = +2 µC at origin (0,0,0)
Q2 = −2 µC at position (4,0,0) m
Find: E at point P = (2,0,0) m

Why this formula applies:
Two point charges, field needed at midpoint — use superposition.

Formula:
E = sum [ Qi(r − ri) / (4πε₀ |r − ri|³) ]

Substitution:
E1 due to +Q1 at origin: r−r1 = (2,0,0), |r−r1| = 2 m
E1 = (2×10⁻⁶ × (2,0,0)) / (4π × 8.854×10⁻¹² × 8)
   = (2×10⁻⁶ × 2) / (4π × 8.854×10⁻¹² × 8) x̂

E2 due to −Q2 at (4,0,0): r−r2 = (−2,0,0), |r−r2| = 2 m
E2 = (−2×10⁻⁶ × (−2,0,0)) / (4π × 8.854×10⁻¹² × 8)
   = same magnitude as E1 in +x direction

Calculation:
Denominator = 4π × 8.854×10⁻¹² × 8 = 8.9×10⁻¹⁰
E1_x = (4×10⁻⁶) / (8.9×10⁻¹⁰) = 4494 V/m
E2_x = 4494 V/m (same direction)
E_total = E1 + E2

Final Answer:
E_total ≈ 8988 V/m in the +x̂ direction

Exam Tip: When finding E at the midpoint of two equal and opposite charges (a dipole), both charges contribute fields in the same direction at the midpoint — they do not cancel. A common trap is assuming they cancel because charges are equal and opposite. Always resolve direction carefully.

• E is defined as force per unit positive test charge: E = F/q. It characterises space, not the test charge.
• For a positive source charge, E points radially outward. For a negative source charge, E points radially inward.
• Magnitude falls as 1/r², same as Coulomb force, because the field originates from the same inverse-square law.
• Superposition allows computation of E from charge distributions by integrating contributions from all charge elements.
• Units are V/m or N/C — both are equivalent and interchangeable in all formulas.

Quick Revision

• E = F/q (definition); E = Q/(4πε₀r²) r̂ (point charge formula).
• Superposition: E_total = vector sum of individual E fields from each charge.
• Direction: away from positive charges, toward negative charges.
• E is independent of the test charge magnitude — it is a source-field quantity.
• For continuous distributions: E = integral of (dQ / 4πε₀ r²) r̂ over the distribution.
• Trap: at midpoint of a dipole, fields from both charges add, not cancel.
• Trap: always separate x, y, z components before adding multiple E vectors.