Electric Dipole
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Direct answer
Two equal and opposite charges ±q separated by 2a constitute an electric dipole of moment p = q × 2a. Its field falls as 1/r^3: on the axis E = 2 k p/r^3 (along p), on the perpendicular bisector E = k p/r^3 (opposite to p), so the axial field is twice the equatorial at the same distance. In a uniform external field, equal and opposite forces cancel and the dipole feels a pure torque tau = p E sin(theta) trying to align it, with potential energy U = −p E cos(theta). Small angular displacements execute SHM with period T = 2 pi sqrt(I/pE); in a non-uniform field a net force additionally pulls the dipole toward stronger field.
What you must remember
- Dipole moment: p = 2 a q from negative to positive; units C m; a water molecule's permanent dipole moment is about 6.2 × 10^-30 C m — the number quoted in NCERT.
- Field distances: axial E = 2 k p/r^3, equatorial E = k p/r^3; both die as 1/r^3, one power faster than a point charge's 1/r^2 — net charge zero leaves only the 1/r^3 signature.
- Torque and energy: tau = p E sin(theta) with U = −p E cos(theta); work to rotate from equilibrium (theta = 0) through angle theta is W = p E (1 − cos(theta)); a half-turn costs 2 p E.
- Uniform field verdict: net force zero, net torque generally nonzero — pure rotation without translation.
- Non-uniform field verdict: both force and torque; the dipole drifts toward the region of stronger field (the principle behind attracting paper bits with a comb).
- Dipole SHM: released from a small angle in a uniform field, tau = −pE theta gives T = 2 pi sqrt(I/(pE)) — a viva favourite.
- Flux note: a sphere enclosing a complete dipole has zero net flux (q(net) = 0) though E is nonzero on it.
- Pattern note: Main tests field ratios and torque–energy substitution; Advanced tests the SHM period, oscillating dipole radiation, and induced versus permanent dipoles.
Aligning, oscillating, and being pulled
A dipole at 60 degrees to a uniform field of 10^5 N/C has p = 10^-6 C m. Torque is tau = 10^-6 × 10^5 × sin(60) = 0.0866 N m, and the work to flip it from 60 degrees to 240 degrees follows from energies: U = −pE cos(theta), so the change is −pE(cos 240 − cos 60) = −pE(−0.5 − 0.5) = pE = 0.1 J. The same dipole nudged slightly from alignment and released oscillates: with moment of inertia I about its centre, T = 2 pi sqrt(I/(pE)) = 2 pi sqrt(I/0.1).
The non-uniform field completes the picture in one everyday line: a charged comb's field weakens with distance, a water-molecule dipole in a paper bit polarises, the near end feels a stronger attraction than the far end feels repulsion, and the net force pulls the paper to the comb.
Where students slip
Direction of p trips the first hurdle: it runs from negative to positive charge, opposite to the field of the pair itself along the axis; candidates reversing it also reverse the torque's sense. Second, the axial-versus-equatorial factor of two is a permanent exam fixture — the equatorial field at the same r is half and points antiparallel to p. Third, U = −p·E's sign discipline: the minimum energy is at theta = 0 (minus pE) and maximum at 180 degrees (plus pE), so work required to invert a dipole from alignment is 2 pE, not zero; mixing initial and final angles in W = U(f) − U(i) is where the sign errors breed.
Frequently asked questions
How do axial and equatorial fields of a dipole compare?
At the same distance, E(axial) = 2kp/r^3 along p and E(equatorial) = kp/r^3 opposite to p — the axial field is double and both fall as the cube of distance.
What torque acts on a dipole in a uniform electric field?
tau = pE sin(theta), tending to align p with E; the net force is zero, so the effect is purely a turning action.
How much work is needed to rotate a dipole from alignment to anti-alignment?
W = U(180°) − U(0°) = 2 pE, since potential energy U = −pE cos(theta) changes from −pE to +pE.
Why does a dipole oscillate when slightly displaced in a uniform field?
Restoring torque tau = −pE theta for small angles mimics SHM, giving period T = 2 pi sqrt(I/(pE)) for rotational inertia I.
Why is a dipole attracted toward a region of stronger field?
The two ends sit at different field strengths, so the attractive force on the nearer end exceeds the repulsive force on the farther end, leaving a net pull up the gradient.