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पाठ्यक्रम

Electric Potential Energy of an Electric Dipole in an Electrostatic Field

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Estimated time: 7 minutes
CISCE: Class 12

Introduction

Think of a compass needle in Earth's magnetic field. It naturally aligns itself along the field direction (like north-south) because that is its lowest-energy, most stable state. Rotating it away from this alignment requires external work — exactly how an electric dipole behaves in an electric field.

CISCE: Class 12

Definition: Electrostatic Potential Energy of a Dipole

The potential energy of an electric dipole in an electrostatic field is the amount of work done in bringing the dipole from infinity (or a reference position) into the field, or the work required to rotate it against the torque exerted by the field.

CISCE: Class 12

Derivation

Boxed Result:

Step Description Expression
1 Torque on dipole at angle θ τ = pE sin ⁡θ
2 Work done in rotating dipole by dθ dW = τ dθ = pE sin⁡ θ dθ
3 Integrate from initial angle θ1​ to final angle θ2 W = 
4 Solve integral W = pE(cos⁡θ1 − cos⁡θ2)
5 Define reference position at θ1 = 90° (U = 0) U = −pE cos⁡ θ
6 Express in vector form U = \[\vec p\cdot\vec E\]
U = −pE cos⁡ θ = −\[\vec p\] ⋅ \[\vec E\]
CISCE: Class 12

Special Cases

Angle (θ) Potential Energy (U) Physical Meaning Equilibrium Type
0° U = −pE (minimum) Dipole aligned with field Stable
90° U = 0 Dipole perpendicular to field Neutral
180° U = +pE (maximum) Dipole opposite to field Unstable
CISCE: Class 12

Example

Problem: Two charges +2e and −2e are separated by 2.4 Å, placed in a field E = 4.0 × 105 V m−1.

Step 1 - Calculate dipole moment:

  • p = q × d = (2e)(2.4 × 10−10) = 7.68 × 10−29 C m

Step 2: Potential energy at stable equilibrium (θ = 0°):

  • U0 = −pE = −(7.68 × 10−29)(4.0 × 105) = −3.07 × 10−23 J

Step 3: Work to rotate through 180°:

  • W = pE(1 − cos⁡180°) = 2pE = 6.14 × 10−23 J

Final Answers: p = 7.68 × 10−29 C m; U0 = −3.07 × 10−23 J; W = 6.14 × 10−23 J

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