PhysEduc › Class 12 Physics › Chapter 1: Electric Charges and Fields

The electric dipole

NCERT Class 12 Physics, Chapter 1: Electric Charges and Fields, §1.10–1.11. A short visual lesson: 2 narrated 3D videos, 2 interactive simulations and 2 quiz questions.

The field of a dipole

Narrated video. Transcript:

An electric dipole is a pair of equal and opposite charges, plus q and minus q, a small distance 2a apart. Its total charge is zero, but its field is not zero, because the two charges are not at the same place.

We describe it by its dipole moment: p equals q times 2a. It is a vector pointing from minus q to plus q, along the dipole’s axis. Its unit is the coulomb metre.

First, a point P on the axis, a distance r from the centre, on the side of plus q. Plus q is r minus a away, minus q is r plus a away. The nearer charge wins.

So E equals k q times 1 over r minus a squared, minus 1 over r plus a squared. This simplifies to k q times 4ar over r² minus a², all squared. It points along p.

Far away, where r is much larger than a, we can drop a² next to r². Then E equals 2 k p over r³, along p.

Now a point on the equatorial plane, a distance r from the centre. Both charges are the same distance away, so their fields have the same size. Their components across the axis cancel; their components along the axis add.

The result, for r much larger than a: E equals k p over r³, pointing opposite to p. At the same distance, the field on the axis is twice the field on the equator.

Both fall as 1 over r³: faster than the 1 over r² of a single charge, because far away the two charges nearly cancel each other.

Example: charges of plus and minus 10 microcoulombs, 5 millimetres apart. p is 10⁻⁵ times 5 × 10⁻³, which is 5 × 10⁻⁸ coulomb metre. At 15 centimetres on the axis, E is 2 times 9 × 10⁹ times 5 × 10⁻⁸, over 0.15 cubed: about 2.7 × 10⁵ newtons per coulomb.

At 15 centimetres on the equator, it is half that: about 1.3 × 10⁵ newtons per coulomb, opposite to p. Molecules such as water have a permanent dipole moment. In others, like carbon dioxide, a field can create one.

A dipole in a uniform field

Narrated video. Transcript:

Place a dipole in a uniform electric field E, with its moment p at an angle θ to the field.

Plus q feels a force q E along the field. Minus q feels a force q E against the field. The two forces are equal and opposite, so the net force on the dipole is zero.

But they act at different points. Together they form a couple, which turns the dipole. The torque is one force times the perpendicular distance between them: q E times 2a sin θ.

Since q times 2a is p, the torque is τ equals p E sin θ. As a vector, τ equals p cross E. It tries to turn p into line with E.

At θ equals 90°, sin θ is 1, and the torque is largest: p E. At θ equals 0°, the torque is zero, and the dipole sits pointing along the field.

At 180° the torque is zero too. But nudge it slightly, and the torque swings it all the way round. So θ equals 0° is stable, and 180° is unstable.

Example: a dipole of moment 4 × 10⁻⁹ coulomb metre is at 30° to a uniform field of 5 × 10⁴ newtons per coulomb. τ equals 4 × 10⁻⁹ times 5 × 10⁴ times sin 30°: 2 × 10⁻⁴ times 0.5, which is 10⁻⁴ newton metre.

If the field is not uniform, the two forces are no longer equal, so there is also a net force. When p points along the field, the dipole is pulled towards the stronger field.

That is why a charged comb picks up bits of paper. The comb’s field makes tiny dipoles in the paper, lined up with the field, and its non-uniform field pulls them in.

Try it: dipole in a uniform field

Interactive simulation. Things to try:

  • Drag either charge to turn the dipole. The two forces stay equal and opposite: the net force is always zero.
  • Watch the torque readout as θ goes from 0° to 90° to 180°. It follows sin θ.
  • Press Release from 170°. The dipole swings round and settles along E. Try 10°: it settles too. Which equilibrium is stable?
  • Double q, 2a or E, and the torque doubles.

Try it: the field of a dipole

Interactive simulation. Things to try:

  • P is on the axis, 8 cm from the centre of a dipole with 2a = 3 cm. Compare the net field with the formula 2kp/r³ in the readouts.
  • Move P further along the axis. The farther away, the closer the exact field gets to the formula, because r ≫ a.
  • Move P to (0, 6): the equator. The field is about half as big, and points opposite to p.
  • Double r: the field drops to about one eighth, not one quarter.

Torque on a dipole

Charges of ±2 μC are 3 cm apart. The dipole is at 30° to a uniform field of 5 × 10⁴ N/C. What is the torque on it?

  1. 1.5 × 10⁻³ N m
  2. 3.0 × 10⁻³ N m
  3. 2.6 × 10⁻³ N m
  4. 7.5 × 10⁻⁴ N m
Answer

A. p = q × 2a = 2 × 10⁻⁶ × 0.03 = 6 × 10⁻⁸ C m. τ = pE sin θ = 6 × 10⁻⁸ × 5 × 10⁴ × 0.5 = 1.5 × 10⁻³ N m. (3.0 × 10⁻³ leaves out sin θ; 2.6 × 10⁻³ uses cos 30°; 7.5 × 10⁻⁴ uses a instead of 2a.)

Axis and equator

The field of a short dipole at a distance r on its axis is E. What is the field at a distance 2r on its equatorial plane?

  1. E/2
  2. E/4
  3. E/8
  4. E/16
Answer

D. On the axis, E = 2kp/r³. On the equator at 2r, E′ = kp/(2r)³ = kp/8r³. So E′/E = (1/8)/2 = 1/16. Half from moving to the equator, and one eighth from doubling the distance.

Summary

  • Dipole moment (from −q to +q): p = q × 2a
  • On the axis, along p (r ≫ a): E = 2kp / r³
  • On the equator, opposite to p: E = kp / r³
  • Torque in a uniform field: τ = p × E = pE sin θ
  • A dipole is a pair of equal and opposite charges; its field falls as 1/r³.
  • At the same distance, the field on the axis is twice the field on the equator.
  • In a uniform field the net force on a dipole is zero, but the torque pE sin θ turns p towards E. θ = 0° is stable, θ = 180° unstable.
  • In a non-uniform field there is also a net force: this is how a charged comb attracts bits of paper.