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Electric Field, Flux & Electric Dipole

Master Electric Field Flux and Dipole for JEE Main Advanced NEET and CBSE Class 12 Physics. Covers electric field definition due to point charge superposition principle, electric field lines all six properties, electric flux EA cosθ with angle cases, electric dipole moment p=q×2a, axial field 2kp/r³ and equatorial field kp/r³ directions, torque pE sinθ on dipole in uniform field stable and unstable equilibrium, potential energy −pE cosθ, general field formula for JEE Advanced, with fully verified practice questions.

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Coulomb's Law tells us the force between two charges — but how does one charge "know" about another charge sitting far away? The answer is the Electric Field: every charge creates a field in the space around it, and other charges respond to that field. This concept of field — an invisible vector quantity filling all of space — is one of the most powerful ideas in physics. Combined with the notion of Electric Flux (how much field passes through a surface) and the Electric Dipole (two equal-and-opposite charges separated by a distance), this topic forms the mathematical foundation on which all of electrostatics is built. For JEE and NEET, electric field calculations, dipole formulas, and flux problems are tested in nearly every paper.

1. Electric Field — Meaning and Definition

The Electric Field at a point is defined as the force experienced per unit positive test charge placed at that point, in the limit that the test charge → 0 (so it doesn't disturb the source).

$\vec{E} = lim_{q_{0} \to 0} \frac{\vec{F}}{q_{0}} \Rightarrow \vec{F} = q_{0} \vec{E}$

SI unit: N/C (= V/m) | Dimension: [MLT⁻³A⁻¹] | Vector quantity

Electric Field due to a Point Charge

$E = \frac{k Q}{r^{2}} = \frac{1}{4 π ε_{0}} \frac{Q}{r^{2}}$

Direction: radially outward from positive charge; radially inward toward negative charge.

Superposition of Electric Fields

The total electric field at a point due to multiple charges is the vector sum of the fields due to each individual charge:

${\vec{E}}_{t o t a l} = {\vec{E}}_{1} + {\vec{E}}_{2} + {\vec{E}}_{3} + \dots$

Electric Field Lines — Properties

Property	Explanation
Start and end	Start on positive charges; end on negative charges (or go to infinity)
Tangent direction	Tangent to a field line at any point gives the direction of E at that point
Density	Closer field lines = stronger field; farther apart = weaker field
No crossing	Two field lines never intersect (E at a point has a unique direction)
No closed loops	Electrostatic field lines never form closed loops (unlike magnetic field lines)
In conductors	Field lines are perpendicular to the surface of a conductor at every point

2. Electric Flux

Electric Flux (Φ) through a surface is the measure of how many electric field lines pass through that surface. It is a scalar quantity.

$Φ = \vec{E} \cdot \vec{A} = E A \cos θ$

Where θ is the angle between the electric field vector E and the area vector A (area vector is perpendicular to the surface, outward normal).

SI unit: N·m²/C (= V·m)

θ	Φ	Meaning
0°	EA (maximum)	Field perpendicular to surface — maximum flux
90°	0	Field parallel to surface — zero flux
180°	−EA	Field anti-parallel to outward normal — negative flux

Worked Example

A uniform electric field E = 500 N/C passes through a square surface of side 20 cm. Find flux when θ = 0° and θ = 60°.

A = 0.20 × 0.20 = 0.04 m²

Φ(0°) = 500 × 0.04 × cos 0° = 20 N·m²/C

Φ(60°) = 500 × 0.04 × cos 60° = 500 × 0.04 × 0.5 = 10 N·m²/C

3. Electric Dipole

An Electric Dipole consists of two equal and opposite point charges (+q and −q) separated by a small distance 2a. The dipole moment p is:

$\vec{p} = q \times 2 a (direction: from −q to +q)$

SI unit: C·m | Practical unit: Debye (D), 1 D = 3.336 × 10⁻³⁰ C·m

Electric Field due to a Dipole

Position	Formula (r ≫ a)	Direction
Axial point (on axis of dipole)	$E_{a x i a l} = \frac{2 k p}{r^{3}} = \frac{1}{4 π ε_{0}} \frac{2 p}{r^{3}}$	Along p (from −q to +q)
Equatorial point (on perpendicular bisector)	$E_{e q} = \frac{k p}{r^{3}} = \frac{1}{4 π ε_{0}} \frac{p}{r^{3}}$	Anti-parallel to p (from +q to −q)

Key ratio: E_axial = 2 × E_equatorial (at the same distance r ≫ a)

Both fall as 1/r³ — faster than a point charge (1/r²) because the dipole fields partially cancel.

Torque on a Dipole in Uniform Electric Field

$\vec{τ} = \vec{p} \times \vec{E} | τ | = p E \sin θ$

θ = 0° or 180°: τ = 0 (dipole aligned with or against field — equilibrium)
θ = 90°: τ = pE (maximum torque)
θ = 0° → stable equilibrium; θ = 180° → unstable equilibrium

Potential Energy of a Dipole in Uniform Field

$U = - \vec{p} \cdot \vec{E} = - p E \cos θ$

Minimum (most stable) when θ = 0°: U = −pE; Maximum (most unstable) when θ = 180°: U = +pE.

4. Worked Numericals

Example 1: A point charge Q = 4 μC. Find E at r = 60 cm.

E = kQ/r² = (9×10⁹ × 4×10⁻⁶) / (0.6)² = 36000 / 0.36 = 1.0 × 10⁵ N/C

Example 2: A dipole has p = 2×10⁻⁹ C·m. Find E at an axial point r = 10 cm.

E_axial = 2kp/r³ = (2 × 9×10⁹ × 2×10⁻⁹) / (0.1)³ = 36 / 10⁻³ = 3.6 × 10⁴ N/C

E_equatorial at same r = E_axial / 2 = 1.8 × 10⁴ N/C

Quick Formula Reference — Electric Field & Dipole

Quantity	Formula	Key Note
E (point charge)	kQ/r²	Falls as 1/r²
E_axial (dipole)	2kp/r³	Along p; falls as 1/r³
E_equatorial (dipole)	kp/r³	Anti-parallel to p; = E_axial/2
Flux Φ	EA cosθ	Scalar; max at θ=0°, zero at θ=90°
Torque on dipole	pE sinθ	Max at θ=90°; zero at θ=0°, 180°
PE of dipole	−pE cosθ	Min (stable) at θ=0°

JEE/NEET Power Tips — Avoid These Common Mistakes

E_axial is directed along p (from −q to +q); E_equatorial is anti-parallel to p (from +q to −q). This direction reversal at the equatorial point trips up most students. Draw the dipole and trace the field direction carefully.
Both dipole fields fall as 1/r³ — NOT 1/r². Point charge falls as 1/r²; dipole fields fall as 1/r³ because opposite charges partially cancel at large distances.
Flux through a closed surface containing zero net charge = 0 (Gauss's Law preview). Even if E ≠ 0 everywhere, if net enclosed charge = 0, net flux = 0.
Area vector direction: always outward normal for closed surfaces; for open surfaces, chosen by right-hand rule or given.
JEE ADV For a general angle θ, E_dipole = (kp/r³)√(3cos²θ + 1). This general formula is JEE Advanced level — not needed for CBSE/NEET.

Practice Questions

Q1 (JEE Main level): Two charges +3 μC and −3 μC are placed 4 cm apart. Find the electric field at a point on the axial line at a distance of 10 cm from the centre (r ≫ a approximation).

p = q × 2a = 3×10⁻⁶ × 0.04 = 1.2×10⁻⁷ C·m

E_axial = 2kp/r³ = (2 × 9×10⁹ × 1.2×10⁻⁷) / (0.1)³

= 2.16 / 10⁻³ = 2.16 × 10³ N/C, directed along p (from −q to +q)

Q2 (NEET level): An electric dipole of moment p is placed in a uniform electric field E at angle θ = 60°. Find the torque. What is the equilibrium orientation?

τ = pE sin 60° = pE × (√3/2) = (√3/2) pE

Equilibrium: θ = 0° (stable, τ = 0, U = −pE minimum) and θ = 180° (unstable, τ = 0, U = +pE maximum).

Q3 (1 mark MCQ): The electric field at a point on the equatorial line of a dipole is:

A) Along the dipole moment B) Opposite to the dipole moment C) Perpendicular to dipole moment D) Zero

Answer: B) Opposite to the dipole moment. On the equatorial line, both charges contribute field components that add up anti-parallel to p.

Q4 (Board level): A square surface of side 10 cm is placed in a uniform E = 1000 N/C with the normal to the surface making 45° with E. Find the electric flux through the surface.

A = 0.1 × 0.1 = 0.01 m²

Φ = EA cosθ = 1000 × 0.01 × cos 45° = 10 × (1/√2) = 5√2 ≈ 7.07 N·m²/C

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DifficultyIntermediate

Est. Read Time24 minutes

CategoryElectric Charges And Fields

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Electric Field, Flux & Electric Dipole

1. Electric Field — Meaning and Definition

Electric Field due to a Point Charge

Superposition of Electric Fields

Electric Field Lines — Properties

2. Electric Flux

Worked Example

3. Electric Dipole

Electric Field due to a Dipole

Torque on a Dipole in Uniform Electric Field

Potential Energy of a Dipole in Uniform Field

4. Worked Numericals

Quick Formula Reference — Electric Field & Dipole

Practice Questions

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